Text Book Back Questions and Answers - Chapter 4 Electricity 10th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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Chapter 4 - Electricity - 10th Science Guide Samacheer Kalvi Solutions

Question $1 .$
A charge of 12 coulomb flows through a bulb in 5 seconds. What is the current through the bulb?
Solution:
Charge $\mathrm{Q}=12 \mathrm{C}$, Time $\mathrm{t}=5 \mathrm{~s}$. Therefore,
Current $\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}=\frac{12}{5}=2.4 \mathrm{~A}$.
 

Question 2.
The work done in moving a charge of $10 \mathrm{C}$ across two points in a circuit is $100 \mathrm{~J}$. What is the potential difference between the points?
Solution:
Charge $Q=10 \mathrm{C}$, Work Done $W=100 \mathrm{~J}$
Potential Difference $\mathrm{V}=\frac{\mathrm{W}}{\mathrm{Q}}=\frac{100}{10}$
Therefore, $V=10$ volt.
 

Question 3.
Calculate the resistance of a conductor through which a current of 2 A passes, when the potential difference between its ends is $30 \mathrm{~V}$.
Solution:
Current through the conductor $\mathrm{I}=2 \mathrm{~A}$,
Potential Difference $\mathrm{V}=30 \mathrm{~V}$
From Ohm's Law: $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$
Therefore, $R=\frac{30}{2}=15 \Omega$.

Question $4 .$
The resistance of a wire of length $10 \mathrm{~m}$ is $2 \mathrm{ohm}$. If the area of cross-section of the wire is $2 \times 10^{-7} \mathrm{~m}^{2}$, determine its
1. Resistivity
2. Conductance
3. Conductivity
Solution:
Given: Length, L $=10 \mathrm{~m}$, Resistance, $R=2 \mathrm{ohm}$
and Area, $\mathrm{A}=2 \times 10^{-7} \mathrm{~m}^{2}$
$\begin{aligned} \text { Resistivity } \rho=\frac{\mathrm{RA}}{\mathrm{L}} &=\frac{2 \times 2 \times 10^{-7}}{10} \\ &=4 \times 10^{-8} \Omega \mathrm{m} \end{aligned}$
Conductance, $\mathrm{G}=\frac{1}{\mathrm{R}}=\frac{1}{2}=0.5 \mathrm{mho}$
Conductivity, $\sigma=\frac{1}{\rho}=\frac{1}{4 \times 10^{-8}}$
$$
=0.25 \times 10^{-8} \mathrm{mho} \mathrm{m}^{-1}
$$

Question $5 .$
Three resistors of resistances $5 \mathrm{ohms}, 3 \mathrm{ohms}$ and $2 \mathrm{ohms}$ are connected in series with $10 \mathrm{~V}$ battery.
Calculate their effective resistance and the current flowing through the circuit.
Solution:
$$
\begin{aligned}
&\mathrm{R}_{1}=5 \Omega, \mathrm{R}_{2}=3 \Omega, \mathrm{R}_{3}=2 \Omega, \mathrm{V}=10 \mathrm{~V} \\
&\mathrm{R}_{\mathrm{S}}=\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}=5+3+2=10 \Omega
\end{aligned}
$$
The current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{s}}=\frac{10}{10}=1 \mathrm{~A}$.
 

Question 6.
An electric heater of resistance $5 \Omega$ is connected to an electric source. If a current of $6 \mathrm{~A}$ flows through the heater, then find the amount of heat produced in 5 minutes.
Solution:
Given resistance $R=5 \Omega$, Current $I=6 \mathrm{~A}$,
Time $\mathrm{t}=5$ minutes $=5 \times 60 \mathrm{~s}=300 \mathrm{~s}$

Amount of heat produced, $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}=6^{2} \times 5 \times 300=54000 \mathrm{~J}$
Hence, H = 54000 J.
 

Question $7 .$
Two bulbs are having the ratings as $60 \mathrm{~W}, 220 \mathrm{~V}$ and $40 \mathrm{~W}, 220 \mathrm{~V}$ respectively. Which one has a greater resistance?
Solution:
Electric power $P=\frac{V^{2}}{R}$
For the same value of $V, R$ is inversely proportional to $P$.
Therefore, the lesser the power, the greater the resistance
Hence, the bulb with $40 \mathrm{~W}, 220 \mathrm{~V}$ rating has greater resistance.
 

Question $8 .$
Calculate the current and the resistance of a $100 \mathrm{~W}, 200 \mathrm{~V}$ electric bulb in an electric circuit.
Solution:
Power $P=100 \mathrm{~W}$ and Voltage $\mathrm{V}=200 \mathrm{~V}$
Boyserument $\overline{V I}=\frac{P}{V}=\frac{100}{200}=0.5 \mathrm{~A}$
Resistance, $R=\frac{V}{I}=\frac{200}{0.5}=400 \Omega$.
 

Question $9 .$
In the circuit diagram is given below resistors $R_{1}, R_{2}$ and $R_{3}$ of $5 \Omega, 10 \Omega$ and $20 \Omega$ respectively are connected as shown in fig.
Calculate:
(A) Current through each resistor
(B) The total current in the circuit
(C) The total resistance in the circuit

Sol
(A) Since the resistors are connected in parallel, the potential difference across each resistor is same (i.e. V $=10 \mathrm{~V}$ )
Therefore, the current through $\mathrm{R}_{1}$ is, $\mathrm{I}_{1}=\frac{\mathrm{V}}{\mathrm{R}_{1}}=\frac{10}{5}=2 \mathrm{~A}$
Current through $\mathrm{R}_{2}, \mathrm{I}_{2}=\frac{\mathrm{V}}{\mathrm{R}_{2}}=\frac{10}{10}=1 \mathrm{~A}$
Current through $\mathrm{R}_{3}, I_{3}=\frac{V}{R} \quad \frac{10}{20}=0.5 \mathrm{~A}$

(B) Total current in the circuit, $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}=2+1+0.5=3.5 \mathrm{~A}$
(C) Total resistance in the circuit
$$
\begin{aligned}
\frac{1}{R_{\mathrm{P}}} &=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}} \\
&=\frac{1}{5}+\frac{1}{10}+\frac{1}{20} \\
&=\frac{4+2+1}{20} \\
&=\frac{1}{R_{P}}=\frac{7}{20}
\end{aligned}
$$
Hence, $\mathrm{R}_{\mathrm{p}}=\frac{20}{7}=2.857 \Omega$.
 

Question 10 .
Three resistors of $1 \Omega, 2 \Omega$ and $4 \Omega$ are connected in parallel in a circuit. If a $1 \Omega$ resistor draws a current of $1 \mathrm{~A}$, find the current through the other two resistors.
Solution:
$$
\mathrm{R}_{1}=1 \Omega, \mathrm{R}_{2}=2 \Omega, \mathrm{R}_{3}=4 \Omega, \text { Current } \mathrm{I}_{1}=1 \mathrm{~A}
$$
The potential difference across the $1 \Omega$ resistor $=I_{1} R_{1}=1 \times 1=1 \mathrm{~V}$
Since, the resistors are connected in parallel in the circuit, the same potential difference will exist across the other resistors also.
So, the current in the $2 \Omega$ resistor, $\frac{V}{R_{2}}=\frac{1}{2}=0.5 \mathrm{~A}$
Similarly, the current in the $4 \Omega$ resistor, $\frac{V}{R_{3}}=\frac{1}{4}=0.25 \mathrm{~A}$.

I. Choose the best answer
Question $1 .$
Which of the following is correct?
(a) Rate of change of charge is electrical power.
(b) Rate of change of charge is current
(c) Rate of change in energy is current.
(d) Rate of change of current is a charge.
Answer:
(b) Rate of change of charge is current
 

Question $2 .$
SI unit of resistance is________

(a) mho
(b) joule
(c) ohm
(d) ohmmeter.
Answer:
(c) ohm
 

Question $3 .$
In a simple circuit, why does the bulb glow when you close the switch?
(a) The switch produces electricity.
(b) Closing the switch completes the circuit.
(c) Closing the switch breaks the circuit.
(d) The bulb is getting charged.
Answer:
(d) The bulb is getting charged.
 

Question $4 .$
Kilowatt-hour is the unit of
(a) resistivity
(b) conductivity
(c) electrical energy
(d) electrical power.
Answer:
(c) electrical energy
 

II. Fill in the blanks
Question $1 .$
When a circuit is open, _______cannot pass through it.
Answer:
Current.

Question $2 .$
The ratio of the potential difference to the current is known as_______
Answer:
Ohm's law.
 

Question 3 .
The wiring in a house consists of _______ circuits.
Answer:
domestic electric
 

Question $4 .$
The power of an electric device is a product of _______ and _______
Answer:
electric current, the potential difference.

Question $5 .$
$\mathrm{LED}$ stands for
Answer:
Light Emitting Diode.
 

III. State whether the following statements are true or false: If false correct the statement.
Question $1 .$
Ohm's law states the relationship between power and voltage.
Answer:
False.
Correct Statement: Ohm's law states the relationship between potential difference and current.
 

Question $2 .$
MCB is used to protect household electrical appliances.
Answer:
True.
 

Question $3 .$
The SI unit for electric current is the coulomb.
Answer:
False.
Correct Statement: The SI unit of electric current is Ampere.
 

Question $4 .$
One unit of electrical energy consumed is equal to 1000 kilowatt-hour.
Answer:
True.

Question $5 .$
The effective resistance of three resistors connected in series is lesser than the lowest of the individual resistances.
Answer:
False.
Correct Statement: The effective resistance of three resistors connected in series is greater than the highest of the individual resistance.
 

IV. Match the items in column-I to the items in column-II:
Question $1 .$

Answer:
1. (e) Ampere
2. (a) Volt
3. (b) Ohmmeter
4. (c)Watt
5. (d) Joule
 

V. Assertion and Reason Type Questions
Mark the correct choice as
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
 

Question $1 .$
Assertion: Electric appliances with a metallic body have three wire connections.
Reason: Three - pin connections reduce heating of the connecting wires
Answer:
(c) If the assertion is true, but the reason is false.
Correct Reason: Three - pin connections to protect the electrical shocking. Because appliances carry high electric current.
 

Question $2 .$
Assertion: In a simple battery circuit the point of highest potential is the positive terminal of the battery.
Reason: The current flows towards the point of the highest potential.
Answer:
(c) If the assertion is true, but the reason is false.
Correct Reason: The current flows towards the points of the lower potential.

Question $3 .$
Assertion: LED bulbs are far better than incandescent bulbs.
Reason: LED bulbs consume less power than incandescent bulbs.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
 

VI. Very Short Answer Questions.
Question $1 .$
Define the unit of current.
Answer:
The SI unit of electric current is ampere (A). The current flowing through a conductor is said to be one ampere when a charge of one coulomb flows across any cross-section of a conductor in one second.

Hence,
1 ampere $=\frac{1 \text { coulomb }}{1 \text { second }} .$
 

Question $2 .$
What happens to the resistance, when the conductor is made thicker?
Answer:
Resistance of a conductor is inversely proportional to the area of cross section of the conductors. Hence, resistance is inversely proportional to square of the radius of the conductor.
$\mathrm{R} \propto \frac{1}{A}$ $\mathrm{R} \propto \frac{1}{r^{2}}$
When a conductor is made thicker, radius will be increased and so its resistance will be decreased.
 

Question $3 .$
Why is tungsten metal used in bulbs, but not infuse wires?
Answer:
Because tungsten has a high melting point it is used in blubs. It cannot be used in fuse wires because in fuse wires the material must have a low melting point.
 

Question 4 .
Name any two devices, which are working on the heating effect of the electric current.
Answer:
(i) Electric heater
(ii) Fuse wire
 

VII. Short Answer Questions.

Question $1 .$
Define electric potential and potential difference.
Answer:
Electric Potential: The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against the electric force.

Electric Potential Difference: The electric potential difference between two points is defined as the amount of work done in moving a unit positive charge from one point to another point against the electric force.
 

Question $2 .$
What is the role of the earth wire in domestic circuits?
Answer:
This wire provides a low resistance path to the electric current. The earth wire sends the current from the body of the appliance to the Earth, whenever a live wire accidentally touches the body of the metallic electric appliance. Thus, the earth wire serves as a protective conductor, which saves us from electric shocks.
 

Question $3 .$
State Ohm's law.
Answer:

According to Ohm's law, at a constant temperature, the steady current ' $\mathrm{I}$ ' flowing through a conductor is directly proportional to the potential difference ' $\mathrm{V}$ ' between the two ends of the conductor.
$I \propto V$
$\mathrm{V}=\mathrm{IR}$.
 

Question $4 .$
Distinguish between the resistivity and conductivity of a conductor.
Answer:

Question $5 .$
What connection is used in domestic appliances and why?
Answer:
Parallel connection is used in domestic appliances. When any disconnection of one circuit in our home, does not affect the other circuit.
 

VIII. Long Answer Questions.
Question $1 .$
With the help of a circuit diagram derive the formula for the resultant resistance of three resistances connected:
(a) in series and
(b) in parallel
Answer:
(a) Resistors in Series:
A series circuit connects the components one after the other to form a 'single-loop'. A series circuit has only one loop through which current can pass. If the circuit is interrupted at any point in the loop, no
current can pass through the circuit and hence no electric appliances connected in the circuit will work. Series circuits are commonly used in devices such as flashlights. Thus, if
(i) Resistors are connected end to end so that the same current passes through each of them, then they are said to be connected in series.

(ii) Let, three resistances $R_{1}, R_{2}$ and $R_{3}$ be connected in series.
(iii) Let the current flowing through them be I.
(iv) According to Ohm's Law, the potential differences $V_{1}, V_{2}$ and $V_{3}$ across $R_{1}, R 2$ and $R_{3}$ respectively, are given by:
$\mathrm{V}_{1}=\mathrm{IR}_{1} \ldots$ (1)
$\mathrm{V}_{2}=\mathrm{IR}_{2} \ldots$ (2)
$\mathrm{V}_{3}=\mathrm{IR}_{3} \ldots$ (3)
The sum of the potential differences across the ends of each resistor is given by:
$\mathrm{V}=\mathrm{V}_{1}+\mathrm{V}_{2}+\mathrm{V}_{3}$
using equations $(1),(2)$ and (3), we get
$\mathrm{V}=\mathrm{IR}_{1}+\mathrm{IR}_{2}+\mathrm{IR}_{3} \ldots$ (4)
(v) The effective resistor is a single resistor, which can replace the resistors effectively, so as to allow the same current through the electric circuit.
(vi) Let, the effective resistance of the series-combination of the resistors, be $R_{S}$. Then,
$\mathrm{V}=\mathrm{IR}_{\mathrm{S}} \ldots$.(5)
Combining equations (4) and (5)
$\mathrm{IR}_{\mathrm{S}}=\mathrm{IR}_{1}+\mathrm{IR}_{2}+\mathrm{IR}_{3}$
$\mathrm{R}_{\mathrm{S}}=\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$
Thus, we can understand that
(vii) When a number of resistors are connected in series, their equivalent resistance or effective resistance is equal to the sum of the individual resistances.
(viii) When $V$ resistors of equal resistance $R$ are connected in series, the equivalent resistance is ' $n R$ '. i.e., $\mathrm{R}_{\mathrm{S}}=\mathrm{nR}$

(ix) The equivalent resistance in a series combination is greater than the highest of the individual resistances.
(b) Resistors in Parallel:
A parallel circuit has two or more loops through which current can pass. If the circuit is disconnected in one of the loops, the current can still pass through the other loop(s). The wiring in a house consists of parallel circuits.

(i) Consider that three resistors $R_{1}, R_{2}$ and $R_{3}$ are connected across two common points $A$ and $B$.
(ii) The potential difference across each resistance is the same and equal to the potential difference between $A$ and $B$. This is measured using the voltmeter.
(iii) The current $I$ arriving at A divides into three branches $I_{1}, I_{2}$ and $I_{3}$ passing through $R_{1}, R_{2}$ and $R_{3}$ respectively.
According to the $\mathrm{Ohm}$ 's law, we have,
$$
\begin{aligned}
&I_{1}=\frac{V}{R_{1}} \ldots(1) \\
&I_{2}=\frac{V}{R_{2}} \ldots(2) \\
&I_{3}=\frac{V}{R_{3}} \ldots \text { (3) }
\end{aligned}
$$
The total current through the circuit is given by
$$
\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}
$$
Using equations (1), (2) and (3), we get
$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{1}}+\frac{\mathrm{V}}{\mathrm{R}_{2}}+\frac{\mathrm{V}}{\mathrm{R}_{3}} \ldots$ (4)
Let the effective resistance of the parallel combination of resistors be $R_{P}$ Then,
$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{p}}} \ldots$ (5)
Combining equations (4) and (5), we have
$\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{p}}}=\frac{\mathrm{V}}{\mathrm{R}_{1}}+\frac{\mathrm{V}}{\mathrm{R}_{2}}+\frac{\mathrm{V}}{\mathrm{R}_{3}}$ $\frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{\mathrm{s}}}$
(iv) When a number of resistors are connected in parallel, the sum of the reciprocals of the individual resistances is equal to the reciprocal of the effective or equivalent resistance.

(v) When ' $n$ ' resistors of equal resistances $R$ are connected in parallel, the equivalent resistance is $\frac{R}{n}$ $\frac{1}{R_{P}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R} \ldots+\frac{1}{R}=\frac{n}{R}$
Hence, $\mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}}{n}$
(vi) The equivalent resistance in a parallel combination is less than the lowest of the individual resistances.
 

Question $2 .$
(a) What is meant by electric current?
(b) Name and define its unit.
(c) Which instrument is used to measure the electric current? How should it be connected in a circuit?
Answer:
(a) Electric current is the rate of flow of charges in a conductor.
(b) The unit of current is ampere. The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross-section of a conductor, in one second. Hence,
1 ampere $=\frac{1 \text { coulomb }}{1 \text { second }}$
(c) Ammeter is used to measure electric current. Ammeter is always connected in series.
 

Question 3 .
(a) State Joule's law of heating.
(b) An alloy of nickel and chromium is used as the heating element. Why?
(c) How does a fuse wire protect electrical appliances?
Answer:
(a) The heat produced in the resistor is $\mathrm{H}=\mathrm{W}=\mathrm{VQ}$
We know that the relation between the charge and current is $Q=I$.
Using this, we get $\mathrm{H}=\mathrm{VIt}$
From Ohm's Law, V IR. Hence, we have $\mathrm{H}=\mathrm{I}^{2} \mathrm{R}$ t
This is known as Joule's law of heating.
Joule's law of heating states that the heat produced in any resistor is:

- Directly proportional to the square of the current passing through the resistor. $\mathrm{H}=\mathrm{VIt}$
- Directly proportional to the resistance of the resistor. $V=I R$
- Directly proportional to the time for which the current is passing through the resistor. $H=I^{2} R t$
(b) An alloy of nickel and chromium is used as the heating element. Because:
- it has high resistivity
- it has a high melting point
- it is not easily oxidized.
(c) (i) The fuse wire is connected in series, in an electric circuit.
(ii) When a large current passes through the circuit, the fuse wire melts due to Joule's heating effect and hence the circuit gets disconnected.
(iii) Therefore, the circuit and the electric appliances are saved from any damage.
(iv) The fuse wire is made up of a material whose melting point is relatively low.

Question $4 .$
Explain about domestic electric circuits, (circuit diagram not required)
Answer:
In our homes, electricity is distributed through the domestic electric circuits wired by the electricians. The first stage of the domestic circuit is to bring the power supply to the main-box from a distribution panel, such as a transformer. The important components of the main-box are: (i) a fuse box and (ii) a meter. The meter is used to record the consumption of electrical energy. The fuse box contains either a fuse wire or a Miniature Circuit Breaker $(\mathrm{MCB})$. The function of the fuse wire or an MCB is to protect the household electrical appliances from overloading due to excess current.

An MCB is a switching device, which can be activated automatically as well as manually. It has a spring attached to the switch, which is attracted by an electromagnet when an excess current passes through the circuit. Hence, the circuit is broken and the protection of the appliance is ensured.

The electricity is brought to houses by one wire has a red insulation and is called the "live wire'. The other wire has a black insulation and is called the "neutral wire'. The electricity supplied to your house is actually an alternating current having an electric potential of $220 \mathrm{~V}$. Both, the live wire and the neutral wire enter into a box where the main fuse is connected with the live wire. After the electricity meter, these wires enter into the main switch, which is used to discontinue the electricity supply whenever required. After the main switch, these wires are connected to live wires of two separate circuits.
Out of these two circuits, one circuit is of a 5 A rating, which is used to run the electric appliances with a lower power rating, such as tube lights, bulbs and fans. The other circuit is of a 15 A rating, which is used to two insulated wires. Out of these two wires, run electric appliances with a high power rating, such as air-conditioners, refrigerators, electric iron and heaters. It should be noted that all the circuits in a house are connected in parallel, so that the disconnection of one circuit does not affect the other circuit. One more advantage of the parallel connection of circuits is that each electric appliance gets an equal voltage.

Question $5 .$
(a) What are the advantages of LED TV on normal TV?
(b) List the merits of the LED bulb.
Answer:
(a) (i) It has a brighter picture quality.
(ii) It is thinner in size.
(iii) It uses less power and consumes very less energy.
(iv) Its life span is more
(v) It is more reliable
(b) (i) As there is no filament, there is no loss of energy in the form of heat. It is cooler than the incandescent bulb.
(ii) In comparison with the fluorescent light, the LED bulbs have a significantly low power requirement.
(iii) It is not harmful to the environment.
(iv) A wide range of colours is possible here.
(v) It is cost-efficient and energy-efficient.
(vi) Mercury and other toxic materials are not required.

IX. Numerical Problems:

Question $1 .$
An electric iron consumes energy at the rate of $420 \mathrm{~W}$ when heating is at the maximum rate and $180 \mathrm{~W}$ when heating is at the minimum rate. The applied voltage is $220 \mathrm{~V}$. What is the current in each case?
Solution:
Case: 1
Power $(\mathrm{P})=420 \mathrm{~W}$
Applied Voltage $(\mathrm{V})=220 \mathrm{~V}$
Current $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{420}{220}=1.9 \mathrm{~A}$
Case: 2
Power $(\mathrm{P})=180 \mathrm{~W}$
Applied Voltage $(\mathrm{V})=2.20 \mathrm{~V}$
Current $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{180}{220}=0.8 \mathrm{~A}$
 

Question 2.
A 100 -watt electric bulb is used for 5 hours daily and four 60 watt bulbs are used for 5 hours daily.
Calculate the energy consumed (in $\mathrm{kWh}$ ) in the month of January.
Answer:
Energy used by $100 \mathrm{~W}$ bulb is $\mathrm{E}=\mathrm{P} \times \mathrm{t}$
$=100 \times 5=500 \mathrm{Wh}$
Energy used by four $60 \mathrm{~W}$ bulbs $\mathrm{E}=4 \times 60 \times 5=1200 \mathrm{Wh}$
Total energy per day $=500+1200=1700 \mathrm{Wh}$
$=1.7 \mathrm{kWh}$
Number of days in January $=31$ days.
Energy consumed in January $=31 \times 1.7=52.7 \mathrm{kWh}$.

Question $3 .$
A torch bulb is rated at $3 \mathrm{~V}$ and $600 \mathrm{~mA}$. Calculate it's?
(a) power
(b) resistance
(c) energy consumed if it is used for 4 hours.
Solution:
$\mathrm{V}=3 \mathrm{~V}$ and $\mathrm{I}=600 \mathrm{~mA}=600 \times 10^{-3} \mathrm{~A}$
(a) Power (P) $=\mathrm{VI}$
$=3 \times 600 \times 10^{-3}$
$=1800 \times 10^{-3}$
$=1.8 \mathrm{~W}$ (or) watt.
(b) Resistance $(R)=\frac{V}{I}=\frac{3}{600 \times 10^{-3}}=5 \Omega$.
(c) Power $(P)=1.8 \mathrm{~W}$ and
Time $=4$ hours $=4 \times 60 \times 60=14400$ second
Energy consumed $\mathrm{E}=\mathrm{P} \times \mathrm{t}=1.8 \times 14400=25920$ joules $\mathrm{E}=25.9 \mathrm{KJ}$.

Question $4 .$
A piece of wire having a resistance $R$ is cut into five equal parts.
(a) How will the resistance of each part of the wire change compare with the original resistance?
(b) If the five parts of the wire are placed in parallel, how will the resistance of the combination change?
(c) What will be the ratio of the effective resistance in series connection to that of the parallel connection?
Solution:
(a) Consider a piece of wire having resistance R. It cut into 5 equal parts. So number of equal resistros are 5. $n=5$
When ' $n$ ' resistors of equal resistance $R$ are connected in series, the equivalent resistance is ' $n R$ '
$$
\begin{aligned}
&\mathrm{R}_{\mathrm{S}}=\mathrm{nR} \\
&\Rightarrow \mathrm{R}_{\mathrm{S}}=5 \mathrm{R} \\
&\Rightarrow \mathrm{R}=\frac{\mathrm{R}_{\mathrm{s}}}{5} \\
&\Rightarrow \mathrm{R}=0.2 \mathrm{R}_{\mathrm{S}}
\end{aligned}
$$
Each part of resistance ' $R$ ' is equal to $0.2$ times of original resistance.
(b) Effective Resistance of 5 Resistors
$$
\begin{aligned}
&\frac{1}{R_{P}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{5}{R} \\
&R_{P}=\frac{R}{5}=0.2 R=0.2 R .
\end{aligned}
$$
(c) Effective resistance of series combination $R_{S}=5 R$
Effective resistance of parallel combination $\mathrm{R}_{\mathrm{p}}=\frac{\mathrm{R}}{5}$
The ratio of series connection to the parallel connection
$$
\begin{aligned}
&\frac{\mathrm{R}_{s}}{\mathrm{R}_{\mathrm{p}}}=\frac{5 \mathrm{R}}{(\mathrm{R} / \mathrm{s})} \\
&\frac{\mathrm{R}_{\mathrm{s}}}{\mathrm{R}_{\mathrm{p}}}=5 \mathrm{R} \times \frac{5}{\mathrm{R}}=25 \\
&\mathrm{R}_{\mathrm{S}}: \mathrm{R}_{\mathrm{P}}=25: 1
\end{aligned}
$$

XI. HOTS Questions
Question $1 .$
Two resistors when connected in parallel give the resultant resistance of $2 \mathrm{ohms}$, but when connected in series the effective resistance becomes $9 \mathrm{ohms}$. Calculate the value of each resistance.
Solution:
Resultant resistance of parallel combination $R_{P}=2 \Omega$
Resultant resistance of series combination $R_{\mathrm{S}}=9 \Omega$

\begin{aligned}
&\frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\\
&\frac{1}{2}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\\
&\frac{1}{2}=\frac{R_{1}+R_{2}}{R_{1} R_{2}}\\
&2\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)=\mathrm{R}_{1} \mathrm{R}_{2} \ldots \text {. (1) }\\
&\mathrm{R}_{\mathrm{S}}=\mathrm{R}_{1}+\mathrm{R}_{2}\\
&\Rightarrow 9=R_{1}+R_{2}\\
&\Rightarrow \mathrm{R}_{1}=9-\mathrm{R}_{2} \ldots \text {... (2) }\\
&\text { Substitute equation (2) in equation (1) }\\
&2\left(9-R_{2}+R_{2}\right)=\left(9-R_{2}\right) R_{2}\\
&\Rightarrow 18=9 R_{2}-R_{2}^{2}\\
&\Rightarrow \mathrm{R}_{2}^{2}-9 \mathrm{R}_{2}+18=0\\
&\Rightarrow\left(\mathrm{R}_{2}-3\right)\left(\mathrm{R}_{2}-6\right)=0\\
&\Rightarrow \mathrm{R}_{2}=3,6\\
&\text { (i) If } R_{2}=3 ; R_{1}=9-R_{2}=9-3=6 \Omega\\
&\text { (ii) If } R_{2}=6 ; R_{1}=9-R_{2}=9-6=3 \Omega \text {. }
\end{aligned}

Question $2 .$
How many electrons are passing per second in a circuit in which there is a current of $5 \mathrm{~A}$ ?
Answer:
A current $10^{18}$ electrons
A current of $5 \mathrm{~A}$ consists
$=5 \times 6.25 \times 10^{18}$ electrons $=31.25 \times 10^{18}$
Number of electrons passing per second
$=31.25 \times 10^{18}$ electrons
 

Question $3 .$
A piece of wire of resistance $10 \mathrm{ohms}$ is drawn out so that its length is increased to three times its original length. Calculate the new resistance.
Solution:
Specific resistance $(\rho)=\frac{\mathrm{RA}}{l}$
$\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$

When the length is increased by three and then the area of cross-section is reduced by three,
Resistance of wire $\mathrm{R}=10 \Omega$
$$
\text { New resistance } \begin{aligned}
\mathrm{R}^{\prime} &=\frac{\rho(3 l)}{(\mathrm{A} / 3)} \\
&=\rho(3 l) \times \frac{3}{\mathrm{~A}} \\
&=\frac{9 \rho l}{\mathrm{~A}} \\
&=9 \times \mathrm{R} \\
&=9 \times 10 \\
\mathrm{R}^{\prime} &=90 \Omega
\end{aligned}
$$

I. Choose the Correct Answer
Question $1 .$
The motion of electric charges through a conductor will constitute
(a) electric current
(b) electric potential
(c) electric field
(d) none.
Answer:
(a) electric current
 

Question $2 .$
Electric charge is expressed in:
(a) Volt
(b) Joule
(c) Coulomb
(d) $\mathrm{Ohm}$
Answer:
(c) Coulomb
 

Question 3 .
A charge of 60 coulomb flows through a bulb in 5 minutes what is the current through the bulb?
(a) $2 \mathrm{~A}$
(b) $0.2 \mathrm{~A}$
(c) $12 \mathrm{~A}$
(d) $0.12 \mathrm{~A}$.

Answer:
(b) $0.2 \mathrm{~A}$
Hint: $\mathrm{I}=\frac{q}{t}=\frac{60}{5 \times 60}=\frac{1}{5}=0.2 \mathrm{~A}$
 

Question $4 .$
1 coulomb of charge is equivalent to the charge of:
(a) $6.25 \times 10^{18}$ electrons
(b) $6 \times 10^{18}$ protons
(c) $1.6 \times 10^{-19}$ electrons
(d) $1.6 \times 10^{-19}$ protons
Answer:
(a) $6.25 \times 10^{18}$ electrons
 

Question $5 .$
The S.I. unit of electric potential difference is
(a) Ampere
(b) Joule
(c) Watt
(d) Volt.
Answer:
(d) Volt
 

Question $6 .$
The potential difference $V$ is proportional to the current $I$, the graph between $V$ and $I$ is
(a) straight line
(b) parabola
(c) ellipse
(d) none.
Answer:
(a) straight line

Question $7 .$
In an electric circuit, voltmeter reads $24 \mathrm{~V}$ and ammeter reads $6 \mathrm{~A}$. The value of resistance is:
(a) $40 \Omega$
(b) $2 \Omega$
(c) $0.25 \Omega$
(d) $4 \Omega$
Answer:
(d) $4 \Omega$
 

Question $8 .$
What is the potential difference between the ends of a resistor of $15 \Omega$ when a current of 2 A passes through it?
(a) $30 \mathrm{~V}$
(b) $7.5 \mathrm{~V}$

(c) $3 \mathrm{~V}$
(d) $300 \mathrm{~V}$
Answer:
(a) $30 \mathrm{~V}$
 

Question $9 .$
How much heat is generated when current $I$ is passing through a resistor for time $t$ ?
(a) $\mathrm{I}^{2} \mathrm{Rt}$
(b) $\operatorname{IR}^{2} \mathrm{t}$
(c) $\mathrm{VI}$
(d) $I^{2} R$
Answer:
(a) $\mathrm{I}^{2} \mathrm{R} t$
 

Question 10
The unit of conductance is
(a) ohm ${ }^{-1}$
(b) volt ${ }^{-1}$ ampere
(c) both (a) and (b)
(d) ohm.
Answer:
(c) both (a) and (b)

Question $11 .$
Two electric bulbs have resistances in the ratio $1: 2$. If they are joined in series, the energy consumed in these are in the ratio:
(a) $1: 2$
(b) $2: 1$
(c) $4: 1$
(d) $1: 1$
Answer:
(a) $1: 2$
 

Question $12 .$
The resistance of a wire of length $10 \mathrm{~cm}$ is $2 \mathrm{ohm}$, then its conductance is
(a) $0.5 \mathrm{ohm}$
(b) $5 \mathrm{ohm}^{-1}$
(c) $0.5 \mathrm{ohm}^{-1}$
(d) $20 \mathrm{ohm}^{-1}$.
Answer:
(c) $0.5 \mathrm{ohm}^{-1}$
 

Question 13 .
The value of current flowing through a circuit consisting of two resistance $6 \Omega$ and $18 \Omega$ in series with a battery of $3 \mathrm{~V}$ is:

(a) $0.5 \mathrm{~A}$
(b) $0.125 \mathrm{~A}$
(c) $6 \mathrm{~A}$
(d) $0.25 \mathrm{~A}$
Answer:
(b) $0.125 \mathrm{~A}$
 

Question 14
When two $2 \Omega$ resistors are connected in parallel, the effective resistance is
(a) $4 \Omega$
(b) $1 \Omega$
(c) $0.5 \Omega$
(d) $5 \Omega$.
Answer:
(b) $1 \Omega$
 

Question 15
When two $2 \Omega$ resistors are connected in series, the effective resistance is
(a) $1 \Omega$
(b) $4 \Omega$
(c) $5 \Omega$
(d) $2 \Omega$.
Answer:
(b) $4 \Omega$

Question $16 .$
One kilowatt-hour is:
(a) $3.6 \times 10^{6} \mathrm{~J}$
(b) $1000 \mathrm{~W}$
(c) $3600 \mathrm{~W}^{-1}$
(d) $2.778 \times 10^{3} \mathrm{~J}$
Answer:
(a) $3.6 \times 10^{6} \mathrm{~J}$
 

Question $17 .$
When ' $n$ ' number of resistors are connected in parallel, the effecive resistance for parallel is
(a) $\mathrm{nR}$
(b) $\frac{n}{\mathrm{R}}$
(c) $\frac{\mathbb{R}}{n}$
(d) none of these.
Answer:
(c) $\frac{\mathbf{R}}{n}$
 

Question $18 .$
The effective resistance for the given circuit in $\mathrm{AB}$

(a) $1 \Omega$
(b) $2 \Omega$
(c) $3 \Omega$
(d) $1.5 \Omega$.
Answer:
(d) $1.5 \Omega$.
 

Question $19 .$
The effective resistance between $A B$ in the given circuit

(a) $\frac{1}{4} \Omega$
(b) $\frac{1}{2} \Omega$
(c) $\frac{3}{4} \Omega$
(d) $\frac{4}{3} \Omega$.
Answer:
(d) $\frac{4}{3} \Omega$.
 

Question $20 .$
The symbol of Ammeter is:
(a) $\mathrm{V}$
(b) $\mathrm{A}$
(c) $\mathrm{G}$
(d) I
Answer:
(b) $\mathrm{A}$
 

Question $21 .$
Tungsten material is used in
(a) Fuse wire
(b) bulbs
(c) batteries
(d) none.
Answer:
(b) bulbs

Question $22 .$
When a charge $24 \mathrm{C}$ flow through a bulb in 10 second then the current flowing through the bulb is:
(a) $1.2 \mathrm{~A}$
(b) $2.4 \mathrm{~A}$
(c) $4.8 \mathrm{~A}$
(d) $0.06 \mathrm{~A}$
Answer:
(b) $2.4 \mathrm{~A}$
 

Question 23 .
The unit of electric power is
(a) Volt-ampere
(b) Watt
(c) both (a) and (b)
(d) only (b).
Answer:
(c) both (a) and (b)
 

Question 24.
One horsepower is equal to
(a) 764 watt
(b) 746 watt
(c) $647 \mathrm{watt}$
(d) 674 watt.
Answer:
(b) 746 watt

Question $25 .$
Unit of specific resistance is:
(a) mho
(b) mho $\mathrm{m}$
(c) ohm m
(d) ampere
Answer:
(c) ohm m
 

Question 26.
To protect the horse hold electrical appliances from overloading due to excess current
(a) Fuse wire
(b) $\mathrm{MCB}$
(c) both (a) and (b)
(d) none.
Answer:
(c) both (a) and (b)
 

Question $27 .$
In India, domestic circuits are supplied with a frequency of

(a) $60 \mathrm{~Hz}$
(b) $50 \mathrm{~Hz}$
(c) $220 \mathrm{~Hz}$
(d) $230 \mathrm{~Hz}$
Answer:
(b) $50 \mathrm{~Hz}$
 

Question $28 .$
What is the unit of conductivity?
(a) ohm $\mathrm{m}^{-1}$
(b) mho $\mathrm{m}$
(c) ohm m
(d) $\mathrm{mho} \mathrm{m}^{-1}$
Answer:
(d) $m$ ho $\mathrm{m}^{-1}$
 

Question $29 .$
The colour of the LED will depend on the type of ______ used.
(a) circuit
(b) materials
(c) display
(d) segment.
Answer:
(b) materials

Question 30 .
One way of overcoming the energy crisis is to use more _______ bulbs.
(a) filament
(b) glass
(c) $\mathrm{LCD}$
(d) LED.
Answer:
(d) $\mathrm{LED}$
 

Question 31.
When resistances are connected in parallel current is more as effective resistance is:
(a) maximum
(b) less
(c) more
(d) minimum
Answer:
(b) less
 

Question $32 .$
The current in the electric bulb of $100 \mathrm{~W}$ and $200 \mathrm{~V}$ electric circuit is
(a) $5 \mathrm{~A}$

(b) $0.5 \mathrm{~A}$
(c) $50 \mathrm{~A}$
(d) $500 \mathrm{~A}$.
Answer:
(b) $0.5 \mathrm{~A}$
 

II. Fill in the Blanks
Question $1 .$
Rate of flow of charges in a conductor is_______
Answer:
electric current.
 

Question $2 .$
_______is used to fix the magnitude of the current through a circuit.
Answer:
Resistor.
 

Question $3 .$
_______is used to select the magnitude of the current through a circuit.
Answer:
Rheostat.
 

Question $4 .$
_______is used to measure the current.
Answer:
Ammeter.

Question $5 .$
_______is used to measure the potential difference.
Answer:
Voltmeter.
 

Question $6 .$
_______is used to indicate the direction of the current.
Answer:
Galvanometer.
 

Question $7 .$
Electric current passes in the circuit from_______ terminal to the_______ terminal.
Answer:
positive, negative.
 

Question $8 .$
Electrical resistivity is_______ for different materials.
Answer:
different.

Question $9 .$
Reciprocal of resistance is_______
Answer:
conductance.
 

Question $10 .$
Reciprocal of electrical resistivity is_______
Answer:
electrical conductivity.
 

Question $11 .$
The equivalent resistance in a series combination is_______ than the highest of the individual resistances.
Answer:
greater.
 

Question $12 .$
The equivalent resistance in a parallel combination is_______ than the lowest of the individual resistances.
Answer:
less.
 

Question 13 .
The heating effect of current is used in devices like _______ and_______.
Answer:
electric heater, electric iron.
 

Question $14 .$
Nichrome is an alloy of _______ and_______.
Answer:
Nickel, Chromium.

Question $15 .$
The filament is made up of a material whose_______ is very high.
Answer:
melting point.
 

Question $16 .$
Electric power is the product of _______and_______.
Answer:
electric current, a potential difference.
 

Question $17 .$
The important components of main box are_______ and_______.
Answer:
fuse box, meter.
 

Question $18 .$
LED is_______.
Answer:
Light Emitting Diode.

Question $19 .$
$\mathrm{LCD}$ is_______
Answer:
Liquid Crystal Display.
 

Question 20 .
An array of LEDs act as_______
Answer:
pixels.
 

Question 21.
$\mathrm{A}$ _______is the display device used to give an output in the form of_______
Answer:
Seven Segment display, numbers or text.
 

III. Match the following.
Question $1 .$

Answer:
1. (c) semiconductor device
2. (e) MCB
3. (d) Filament
4. (b) alloy
5. (a) heating device
 

Question $2 .$

Answer:
1. (c) $P=V I$
2. (e) $\mathrm{H}=\mathrm{I}^{2} \mathrm{RT}$
3. (a) $V=I R$
4. (b) $\rho=\frac{R A}{l}$
5. (e) $I=\frac{Q}{t}$
 

Question $3 .$

Answer:
1. (d) To measure current
2. (c) To measure voltage
3. (b) Direction of current
4. (a) To measure resistance
 

Question $4 .$

Answer:
1. (b) $\Omega^{-1} \mathrm{~m}$
2. (d) $\mathrm{m}^{-1} \Omega^{-1}$
3. (e) Kilowatt-hour
4. (a) watt
5. (c) coulomb
 

Question $5 .$

Answer:
1. (c) Resistor
2. (e) Rheostat
3. (b) Voltmeter
4. (a) Ammeter
5. (d) Diode
 

IV. State whether the following statements are true or false: If false correct the statement.
Question $1 .$
The equivalent resistance in a series combination in lesser than the highest of the individual resistances.
Answer:
False.
Correct Statement: The equivalent resistance in a series combination is greater than the highest of the individual resistances.
 

Question $2 .$
When ' $n$ ' number of resistors of equal resistance $R$ connected in series, the equivalent resistance is
$\mathrm{R}_{\mathrm{P}}=\frac{\mathbf{R}}{n}$.
Answer:
False.
Correct Statement: $\mathrm{R}_{\mathrm{S}}=\mathrm{nR}$
 

Question 3 .
When resistors are connected in series, a current is less as effective resistance is more.
Answer:
False.
Correct Statement: Current is more as effective resistance is less.

Question $4 .$
Nichrome is an alloy of copper and chromium.
Answer:
False.
Correct Statement: Nichrome is an alloy of nickel and chromium.
 

Question 5.
The fuse wire is made up of a material whose melting point is relatively high.
Answer:
False.
Correct Statement: The fuse wire is made up of a material whose melting point is relatively low.

V. Very Short Answer Questions
Question $1 .$
When a steady current flows in a conductor of non-uniform cross-section will the charge passing per unit time depend on area of cross section of the conductor?
Answer:
No, the charge passing per unit time through the conductor is independent of area of cross-section of the wire.
 

Question $2 .$
Name any two - component and give its use?

Answer:

Question $3 .$
Define the unit of electric potential?
Answer:
1. The SI unit of electric potential or potential difference is volt $(V)$.
2. The potential difference between two points is one volt if one joule of work is done in moving one coulomb of charge from one point to another against the electric force.
1 volt $=\frac{1 \text { joule }}{1 \text { coulomb }}$
 

Question $4 .$
Draw a graph between $V$ and I for a conductor by ohm's law?
Answer:

Question $5 .$
What is the relationship between conductance and resistance? State the Unit of conductance.
Answer:
conductance $=\frac{1}{\text { resistance }}$
$\mathrm{G}=\frac{1}{R}$
Unit of conductance is mho.
 

Question $6 .$
Define the unit of electric power.
Answer:
The SI unit of electric power is watt. When a current of 1 ampere passes across the ends of a conductor, which is at a potential difference of 1 volt, then the electric power is
$\mathrm{P}=1$ volt $\times 1$ ampere $=1$ watt.
 

Question $7 .$
Define one kilowatt-hour? Give its value?
Answer:
One kilowatt-hour is otherwise known as one unit of electrical energy. One kilowatt-hour means that electric power of 1000 watt has been utilized for an hour.
$1 \mathrm{kWh}=1000$ watt-hour $=1000 \times(60 \times 60)$ watt-second $=3.6 \times 10^{6} \mathrm{~J}$.
 

Question $8 .$
Define electric current.
Answer:
Electric current is defined as the rate of flow of charge in a conductor.
 

VI. Short Answer Questions.

Question $1 .$
Define electrical resistivity? Give its unit.
Answer:
The electrical resistivity of a material is defined as the resistance of a conductor of unit length and unit area of cross - section.
Its unit is ohm metre.
 

Question $2 .$
Define specific resistance electrical conductivity? Give its unit.
Answer:
The reciprocal of electrical resistivity of a material is called its electrical conductivity.
$\sigma=\frac{1}{\rho}$
Its unit is $\mathrm{ohm}^{-1} \mathrm{metre}^{-1}$. It is also represented as mho metre ${ }^{-1}$.
 

Question 3 .
Differences between series and parallel circuit?
Answer:

Question $4 .$
Distinguish electric power and electric energy?
Answer:

Question $5 .$
Define electric potential difference.
Answer:
The electric potential difference between two points is defined as the amount of work done in moving a unit positive charge from one point to another point against the electric force.
 

Question $6 .$
Define Volt.
Answer:
The potential difference between two points is one volt, if one joule of work is done in moving one
coulomb of charge from one point to another against the electric force.
1 volt $=\frac{1 \text { joule }}{1 \text { coulomb }}$

Question $7 .$
What is the unit of resistance? Define its unit of resistance is $\mathrm{ohm}$.
Answer:
Resistance of a conductor is said to be one ohm if a current of one ampere flows through it when a potential difference of one volt is maintained across its ends.
$1 \mathrm{ohm}=\frac{1 \text { volt }}{1 \text { amphre }}$
 

VII. Answer in detail.
Question $1 .$
Explain the series connection of parallel resistors.
Answer:
(i) If we consider the connection of a set of parallel resistors that are connected in series, we get a series parallel circuit.
(ii) Let $R_{1}$ and $R_{2}$ be connected in parallel to give an effective resistance of $R_{P 1}$.
(iii) Similarly, let $R_{3}$ and $R_{4}$ be connected in parallel to give an effective resistance of $R_{P 2}$.
(iv) Then, both of these parallel segments are connected in series.

Finally, using the equation the $R_{S}=R_{1}+R_{2}+R_{3}$ net effecitve resistance is given by $R_{\text {total }}=R_{P 1}+R_{P 2}$
 

Question $2 .$
Explain the parallel connection of series resistors.
Answer:
(i) Let $R_{1}$ and $R_{2}$ be connected in parallel to give an effective resistance of $R_{P 1}$.
(ii) If you consider a connection of a set of series resistors connected in a parallel circuit, we get a parallelseries circuit.
(iii) $L e t R_{1}$ and $R_{2}$ be connected in series to give an effective resistance of $R_{S 1}$.
(iv) Similarly, let $R_{3}$ and $R_{4}$ be connected in series to give an effective resistance of $R_{S 2}$.
(v) Then, both of these serial segments are connected in parallel.

Parallel - Series combination of resistors
Using equation $R_{S}=R_{1}+R_{2}+R_{3}$ we get
$$
\begin{aligned}
&R_{S}=R_{1}+R_{2}+R_{3} \\
&R_{S 1}=R_{1}+R_{2} \\
&R_{S 2}=R_{3}+R_{4}
\end{aligned}
$$
Finally, using equation, $\frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}$ the net effective resistance is given by $\frac{1}{\mathrm{R}_{\text {total }}}=\frac{1}{\mathrm{R}_{s 1}}+\frac{1}{\mathrm{R}_{s 2}}$.
 

VIII. Numerical Problems
Question 1.
Show that one ampere is equivalent to a flow of $6.25 \times 10^{18}$ elementary charges per second.
Solution:
$$
\begin{aligned}
&\mathrm{I}=1 \mathrm{~A}, \mathrm{t}=1 \mathrm{~s}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C} \\
&\mathrm{q}=\mathrm{ne} \\
&\mathrm{I}=\frac{q}{t}=\frac{n e}{t}
\end{aligned}
$$
Number of electrons, $n=\frac{\mathrm{I} t}{e}=\frac{1 \times 1}{1.6 \times 10^{-19}}=6.25 \times 10^{18}$.
 

Question $3 .$
Calculate the resistivity of the material of wire $10 \mathrm{~m}$ long, $0.4 \mathrm{~mm}$ in diameter and having a resistance of $20 \Omega$.
Solution:
$$
\begin{aligned}
&1=10 \mathrm{~m} ; \mathrm{d}=0.4 \mathrm{~mm}=0.4 \times 10^{-3} \mathrm{~m} \\
&\mathrm{r}=0.2 \times 10^{-3} \mathrm{~m} \\
&\mathrm{R}=20 \Omega
\end{aligned}
$$
Resistivity (or) specific resistance of a material

Question $4 .$
Find the effective resistance between $\mathrm{A}$ and $\mathrm{B}$ in the given circuit.
Solution:
Effective resistance in parallel combination $3 \Omega$ and $2 \Omega$
$$
\begin{aligned}
&\frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}=\frac{1}{3}+\frac{1}{2}=\frac{2+3}{6}={ }_{\frac{5}{6}} \Omega \\
&R_{P}=\frac{6}{5}=1.2 \Omega \\
&\mathrm{R}_{\mathrm{S}}=\mathrm{R}_{1}+\mathrm{R}_{2}=1.2+2=3.2 \Omega
\end{aligned}
$$