Text Book Back Questions and Answers - Chapter 2 Current Electricity 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Current Electricity
Question 1.
The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor?

(a) $2 \mathrm{ohm}$
(b) $4 \mathrm{ohm}$
(c) $8 \mathrm{ohm}$
(d) $1 \mathrm{ohm}$
Answer:
(a) $2 \mathrm{ohm}$
Question 2.
A wire of resistance $2 \mathrm{ohms}$ per meter is bent to form a circle of radius $1 \mathrm{~m}$. The equivalent resistance between its two diametrically opposite points, $A$ and $B$ as shown in the figure is-

(a) $\pi \Omega$
(b) $\frac{\pi}{2} \Omega$
(c) $2 \pi \Omega$
(d) $\frac{\pi}{4} \Omega$
Answer:
(b) $\frac{\pi}{2} \Omega$
Question 3.
A toaster operating at $240 \mathrm{~V}$ has a resistance of $120 \Omega$. The power is
(a) $400 \mathrm{~W}$
(b) $2 \mathrm{~W}$
(c) $480 \mathrm{~W}$
(d) $240 \mathrm{~W}$
Answer:
(c) $480 \mathrm{~W}$
Question 4.
A carbon resistor of $(47 \pm 4.7) \mathrm{k} \Omega$ to be marked with rings of different colours for its identification. The colour code sequence will be
(a) Yellow - Green - Violet - Gold
(b) Yellow - Violet - Orange - Silver
(c) Violet - Yellow - Orange - Silver
(d) Green - Orange - Violet - Gold
Answer:
(b) Yellow - Violet - Orange - Silver
Question 5.
What is the value of resistance of the following resistor?

(a) $100 \mathrm{k} \Omega$
(b) $10 \mathrm{k} \Omega \mathrm{M}$
(c) $1 \mathrm{k} \Omega$
(d) $1000 \mathrm{k} \Omega$
Answer:
(a) $100 \mathrm{k} \Omega$
Question 6.
Two wires of $\mathrm{A}$ and $\mathrm{B}$ with circular cross section made up of the same material with equal lengths.

Suppose $R_A=3 R_B$, then what is the ratio of radius of wire $A$ to that of $B$ ?
(a) 3
(b) $\sqrt{3}$
(c) $\frac{1}{\sqrt{ } 3}$
(d) $\frac{1}{3}$
Answer:
(c) $\frac{1}{\sqrt{ } 3}$
Question 7.
A wire connected to a power supply of $230 \mathrm{~V}$ has power dissipation $P_1$ Suppose the wire is cut into two equal pieces and connected parallel to the same power supply. In this case power dissipation is $P_2$. The $\operatorname{ratio} \frac{p_2}{p_1}$ is.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4
Question 8.
In India electricity is supplied for domestic use at $220 \mathrm{~V}$. It is supplied at $110 \mathrm{~V}$ in USA. If the resistance of a $60 \mathrm{~W}$ bulb for use in India is $\mathrm{R}$, the resistance of a $60 \mathrm{~W}$ bulb for use in USA will be
(a) $\mathrm{R}$
(b) $2 R$
(c) $\frac{R}{4}$
(d) $\frac{R}{2}$
Answer:
(c) $\frac{R}{4}$
Question 9.
In a large building, there are 15 bulbs of $40 \mathrm{~W}, 5$ bulbs of $100 \mathrm{~W}, 5$ fans of $80 \mathrm{~W}$ and 1 heater of $1 \mathrm{~kW}$ are connected. The voltage of electric mains is $220 \mathrm{~V}$. The minimum capacity of the main fuse of the building will be (IIT-JEE 2014)
(a) $14 \mathrm{~A}$
(b) $8 \mathrm{~A}$
(c) $10 \mathrm{~A}$
(d) $12 \mathrm{~A}$
Answer:
(d) $12 \mathrm{~A}$

Question 10.
There is a current of $1.0 \mathrm{~A}$ in the circuit shown below. What is the resistance of $\mathrm{P}$ ?

(a) $1.5 \Omega$
(b) $2.5 \Omega$
(c) $3.5 \Omega$
(d) $4.5 \Omega$
Answer:
(c) $3.5 \Omega$
Question 11.

What is the current out of the battery?

(a) $1 \mathrm{~A}$
(b) $2 \mathrm{~A}$
(c) $3 \mathrm{~A}$
(d) $4 \mathrm{~A}$
Answer:
(a) $1 \mathrm{~A}$
Question 12 .
The temperature coefficient of resistance of a wire is 0.00125 per ${ }^{\circ} \mathrm{C}$. At $300 \mathrm{~K}$, its resistance is $1 \Omega$. The resistance of the wire will be $2 \Omega$. at
(a) $1154 \mathrm{~K}$
(ft) $1100 \mathrm{~K}$
(c) $1400 \mathrm{~K}$
(d) $1127 \mathrm{~K}$
Answer:
(d) $1127 \mathrm{~K}$
Question 13.
The internal resistance of a $2.1 \mathrm{~V}$ cell which gives a current of $0.2 \mathrm{~A}$ through a resistance of $10 \Omega$ is
(a) $0.2 \Omega$
(b) $0.5 \Omega$
(c) $0.8 \Omega$
(d) $1.0 \Omega$
Answer:
(b) $0.5 \Omega$
Question 14.
A piece of copper and another of germanium are cooled from room temperature to $80 \mathrm{~K}$. The resistance of
(a) each of them increases
(b) each of them decreases
(c) copper increases and germanium decreases
(d) copper decreases and germanium increases
Answer:
(d) copper decreases and germanium increases

Question 15.
In Joule's heating law, when I and $\mathrm{t}$ are constant, if the $\mathrm{H}$ is taken along the $\mathrm{y}$ axis and $\mathrm{I}^2$ along the $\mathrm{x}$ axis, the graph is
(a) straight line
(b) parabola
(c) circle
(d) ellipse
Answer:
(a) straight line
Short Answer Questions
Question 1.
Why current is a scalar?
Answer:
The current $\mathrm{I}$ is defined as the scalar product of current density and area vector in which the charges cross.
$\mathrm{I}=\vec{j} \cdot \vec{A}$
The dot product of two vector quantity is a scalar form. Hence, current is called as a scalar quantity.
Question 2.
Distinguish between drift velocity and mobility.
Answer:

Question 3.

State microscopic form of Ohm's law.
Answer:
Current density $\mathrm{J}$ at a point in a conductor is the amount of current flowing per unit area of the conductor around that point provided the area is held in a direction normal to the current.
$
\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}
$
Drift velocity, $\quad \mathrm{V}_d=\frac{e \mathrm{E}}{m} \tau$
$
\begin{aligned}
\mathrm{I} & =n e \mathrm{AV}_d=n e \mathrm{~A}\left(\frac{e \mathrm{E}}{m}\right) \tau \quad\left[\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}} ; \sigma=\frac{n e^2 \tau}{m}\right] \\
\frac{\mathrm{I}}{\mathrm{A}} & =\left(\frac{n e^2 \tau}{m}\right) \mathrm{E} \\
\mathrm{J} & =\sigma \mathrm{E}
\end{aligned}
$
Question 4.
State macroscopic form of Ohm's law.
Answer:
The macroscopic form of Ohm's Law relates voltage, current and resistance. Ohm's Law states that the current through an object is proportional to the voltage across it and inversely proportional to the object's resistance.
$
\mathrm{V}=\mathrm{IR}
$
Question 5.
What are ohmic and non-ohmic devices?
Answer:
Materials for which the current against voltage graph is a straight line through the origin, are said to obey Ohm's law and their behaviour is said to be ohmic. Materials or devices that do not follow Ohm's law are said to be non-ohmic.
Question 6.
Define electrical resistivity.
Answer:
Electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having unit area of cross section.

Question 7.
Define temperature coefficient of resistance.
Answer:
It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity at $T_0$
$
\therefore \quad \alpha=\frac{\rho_{\mathrm{T}}-\rho_{\mathrm{o}}}{\rho_{\mathrm{o}}\left(\mathrm{T}-\mathrm{T}_{\mathrm{o}}\right)}=\frac{\Delta \rho}{\rho_{\mathrm{o}} \Delta \mathrm{T}}
$
Question 8.
What is superconductivity?
Answer:
The ability of certain metals, their compounds and alloys to conduct electricity with zero resistance at very low temperatures is called superconductivity.
Question 9.
What is electric power and electric energy?
Answer:
1. Electric power:
It is the rate at which an electric appliance converts electric energy into other forms of energy. Or, it is the rate at which work is done by a source of emf in maintaining an electric current through a circuit.
$
\mathrm{P}=\frac{W}{t}=\mathrm{VI}=\mathrm{I}^2 \mathrm{R}=\frac{V_2}{R}
$
2. Electric energy:
It is the total workdone in maintaining an electric current in an electric circuit for a given time. $\mathrm{W}=\mathrm{Pt}=$ VIt joule $=\mathrm{I}^2 \mathrm{Rt}$ joule
Question 10.
Define current density.
Answer:
The current density (J) is defined as the current per unit area of cross section of the conductor $\mathrm{J}=\frac{1}{A}$
The S.I unit of current density.
$\frac{A}{m^2}$
$\mathrm{Am}^2$

Question 11.
Derive the expression for power $\mathrm{P}=\mathrm{VI}$ in electrical circuit.
Answer:
The electrical power $\mathrm{P}$ is the rate at which the electrical potential energy is delivered,
$
\mathrm{P}=\frac{d U}{d t}=\frac{d}{d t}(\mathrm{~V} \cdot \mathrm{dQ})=\mathrm{V} \frac{d Q}{d t}
$
Since the electric current $\mathrm{I}=\frac{d Q}{d t}$
So the equation can be rewritten as $\mathrm{P}=\mathrm{VI}$.
Question 12.
Write down the various forms of expression for power in electrical circuit.
Answer:
The electric power $P$ is the rate at which the electrical potential energy is delivered,
$
\mathrm{P}=\frac{d U}{d t}=\frac{1}{d t}(\mathrm{~V} \cdot \mathrm{dQ})=\mathrm{V} \cdot \frac{d Q}{d t}
$
[dU = V.dQ]
The electric power delivered by the battery to any electrical system.
$
\mathrm{P}=\mathrm{VI}
$
The electric power delivered to the resistance $\mathrm{R}$ is expressed in other forms.
$
\begin{aligned}
& \mathrm{P}=\mathrm{VI}=\mathrm{I}(\mathrm{IR})=\mathrm{I}^2 \mathrm{R} \\
& \mathrm{P}=\mathrm{IV}=\left(\frac{V}{R}\right) \mathrm{V}=\frac{V^2}{R} .
\end{aligned}
$
Question 13.
State Kirchhoff's current rule.
Answer:
It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a statement of conservation of electric charge.

Question 14.
State Kirchhoff's voltage rule.
Answer:
It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system.
Question 15 .
State the principle of potentiometer.
Answer:
The basic principle of a potentiometer is that when a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.
Question 16.
What do you mean by internal resistance of a cell?
Answer:
The resistance offered by the electrolyte of a cell to the flow of current between its electrodes is called internal resistance of the cell. An ideal battery has zero internal resistance and the potential difference across the battery equals to its emf. But a real battery is made of electrodes and electrolyte, there is resistance to the flow of charges within the battery. A freshly prepared cell has low internal resistance and it increases with ageing.
Question 17.
State Joule's law of heating.
Answer:
It states that the heat developed in an electrical circuit due to the flow of current varies directly as:
1. the square of the current
2. the resistance of the circuit and
3. the time of flow.
$
\mathrm{H}=\mathrm{I}^2 \mathrm{R} \text { ? }
$
Question 18 .
What is Seebeck effect?
Answer:
Seebeck discovered that in a closed circuit consisting of two dissimilar metals, when the junctions are maintained at different temperatures, an emf (potential difference) is developed.
Question 19.
What is Thomson effect?
Answer:
Thomson showed that if two points in a conductor are at different temperatures, the density of electrons at these points will differ and as a result the potential difference is created between these points. Thomson effect is also reversible.

Question 20.
What is Peltier effect?
Answer:
When an electric current is passed through a circuit of a thermocouple, heat is evolved at one junction and absorbed at the other junction. This is known as Peltier effect.
Question 21.
State the applications of Seebeck effect.
Answer:
Applications of Seebeck effect:
1. Seebeck effect is used in thermoelectric generators (Seebeck generators). These thermoelectric generators are used in power plants to convert waste heat into electricity.
2. This effect is utilised in automobiles as automotive thermoelectric generators for increasing fuel efficiency.
3. Seebeck effect is used in thermocouples and thermopiles to measure the temperature difference between the two objects.
Long Answer Questions
Question 1.
Describe the microscopic model of current and obtain genera! form of Ohm's Law.
Answer:
Microscopic model of current: Consider a conductor with area of cross-section A and an electric field E applied from right to left. Suppose there are $n$ electrons per unit volume in the conductor and assume that all the electrons move with the same drift velocity $\vec{v}_{\mathrm{d}}$.
The drift velocity of the electrons $=\mathrm{v}_{\mathrm{d}}$
The electrons move through a distance $\mathrm{dx}$ within a small interval of $\mathrm{dt}$

$
\mathrm{v}_{\mathrm{d}}=\frac{d x}{d t} ; \mathrm{dx}=\mathrm{v}_{\mathrm{d}} \mathrm{dt}
$
Since $\mathrm{A}$ is the area of cross section of the conductor, the electrons available in the volume or length $\mathrm{dx}$ is $=$ volume $x$ number per unit volume
$
=\mathrm{A} \mathrm{dx} \times \mathrm{n}
$
Substituting for $\mathrm{dx}$ from equation (1) in (2) $=\left(\mathrm{A} \mathrm{v}_{\mathrm{d}} \mathrm{dt}\right) \mathrm{n}$
Total charge in volume element $d Q=$ (charge) $x$ (number of electrons in the volume element) $\mathrm{dQ}=(\mathrm{e})\left(\mathrm{A} \mathrm{v_{d }} \mathrm{dt}\right) n$
Hence the current, $\mathrm{I}=\frac{d Q}{d t}=\frac{n e A v_d d t}{d t}$
$\mathrm{I}=$ ne $\mathrm{A} \mathrm{v}_{\mathrm{d}}$
Current denshy (J):
The current density (J) is defined as the current per unit area of cross section of the conductor $\mathrm{J}=\frac{I}{A}$
The S.I. unit of current density, $\frac{A}{\mathrm{~m}^2}$ (or) $\mathrm{Am}^{-2}$
$\mathrm{J}=\frac{n e A v_d}{A}$ (from equation 3)

$\mathrm{J}=\mathrm{nev}_{\mathrm{d}}$
The above expression is valid only when the direction of the current is perpendicular to the area A. In general, the current density is a vector quantity and it is given by $\vec{J}=\mathrm{ne} \vec{v}_{\mathrm{d}}$
Substituting $i$ from equation $\vec{v}_{\mathrm{d}}=\frac{-e \tau}{m} \vec{E}$
$\begin{aligned} \vec{J} & =-\frac{n \cdot e^2 \tau}{m} \vec{E} \\ \vec{J} & =-\sigma \vec{E}\end{aligned}$
But conventionally, we take the direction of (conventional) current density as the direction of electric field. So the above equation becomes
$
\vec{J}=\sigma \vec{E}
$
where $\sigma=\frac{n \cdot e^2 \tau}{m}$ is called conductivity.
The equation 6 is called microscopic form of ohm's law.
Question 2.
Obtain the macroscopic form of Ohm's law from its microscopic form and discuss its limitation.
Answer:
Ohm's law: The Ohm's law can be derived from the equation $\mathrm{J}=\sigma \mathrm{E}$. Consider a segment of wire of length 1 and cross sectional area $A$.

When a potential difference $\mathrm{V}$ is applied across the wire, a net electric field is created in the wire which constitutes the current. For simplicity, we assume that the electric field is uniform in the entire length of the wire, the potential difference (voltage $\mathrm{V}$ ) can be written as $\mathrm{V}=\mathrm{El}$
As we know, the magnitude of current density
$\mathrm{J}=\sigma \mathrm{E}=\sigma \frac{V}{l}$
But $\mathrm{J}=\frac{I}{A}$, so we write the equation as
$
\frac{I}{A} \sigma \frac{V}{l}
$
By rearranging the above equation, we get
$
\mathrm{V}=\mathrm{I}\left(\frac{l}{\sigma A}\right)
$
The quantity $\frac{l}{\sigma A}$ is called resistance of the conductor and it is denoted as $\mathrm{R}$. Note that the oA resistance is directly proportional to the length of the conductor and inversely proportional to area of cross section. Therefore, the macroscopic form of Ohm's law can be stated as $\mathrm{V}=\mathrm{IR}$
Question 3.
Explain the equivalent resistance of a series and parallel resistor network.
Answer:
1. Resistors in series:
When two or more resistors are connected end to end, they are said to be in series. The resistors could be
simple resistors or bulbs or heating elements or other devices. Fig. (a) shows three resistors $R_1, R_2$ and $R_3$ connected in series.

The amount of charge passing through resistor $R_1$ must also pass through resistors $R_2$ and $R_3$ since the charges cannot accumulate anywhere in the circuit. Due to this reason, the current I passing through all the three resistors is the same. According to Ohm's law, if same current pass through different resistors of different values, then the potential difference across each resistor must be different. Let $V_1, V_2$ and $V_3$ be the potential difference (voltage) across each of the resistors $R_1, R_2$ and $R_3$ respectively, then we can write $\mathrm{V}_1=\mathrm{IR}_1, \mathrm{~V}_2=\mathrm{IR}_2$ and $\mathrm{V}_3=\mathrm{IR}_3$. But the total voltage $\mathrm{V}$ is equal to the sum of voltages across each resistor.
$
\begin{aligned}
& \mathrm{V}=\mathrm{V}_1+\mathrm{V}_2+\mathrm{V}_3 \\
& =\mathrm{IR}_1+\mathrm{IR}_2+\mathrm{IR}_3 \ldots . \text { (1) } \\
& \mathrm{V}=\mathrm{I}\left(\mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3\right) \\
& \mathrm{V}=\mathrm{I} \cdot \mathrm{R}_{\mathrm{S}} \ldots \ldots \text { (2) } \\
& \text { where } \mathrm{R}_{\mathrm{S}}=\mathrm{R}_1+\mathrm{R}_2 \mathrm{R}_3 \ldots \ldots \text { (3) }
\end{aligned}
$
When several resistances are connected in series, the total or equivalent resistance is the sum of the individual resistances as shown in fig. (b).

Note:
The value of equivalent resistance in series connection will be greater than each individual resistance.
2. Resistors in parallel:
Resistors are in parallel when they are connected across the same potential difference as shown in figer.

In this case, the total current $I$ that leaves the battery in split into three separate paths. Let $I_1, I_2$ and $I_3$ be the current through the resistors $R_1, R_2$ and $R_3$ respectively. Due to the conservation of charge, total current in the circuit $\mathrm{I}$ is equal to sum of the currents through each of the three resistors.
$
\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3
$
Since the voltage across each resistor is the same, applying Ohm's law to each resistor, we have
$
\mathrm{I}_1=\frac{V}{R_1} \mathrm{I}_2=\frac{V}{R_2}, \mathrm{I}_1=\frac{V}{R_3}
$
Substituting these values in equation (1), we get
$
\begin{aligned}
& \mathrm{I}=\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}=\mathrm{V}\left[\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right] \\
& \mathrm{I}=\frac{V}{R_p} \\
& \frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}
\end{aligned}
$
Here $\mathrm{R}_{\mathrm{P}}$ is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination as shown in the fig. (b).

Note:
The value of equivalent resistance in parallel connection will be lesser than each individual resistance.
Question 4.
Explain the determination of the internal resistance of a cell using voltmeter.
Answer:
Determination of internal resistance:
The emf of cell $\xi$ is measured by connecting a high resistance voltmeter across it without connecting the external resistance $R$. Since the voltmeter draws very little current for deflection, the circuit may be considered as open. Hence, the voltmeter reading gives the emf of the cell. Then, external resistance $R$ is included in the circuit and current $\mathrm{I}$ is established in the circuit. The potential difference across $\mathrm{R}$ is equal to the potential difference across the cell (V).

The potential drop across the resistor $\mathrm{R}$ is $\mathrm{V}=\mathrm{IR} \ldots .(1)$
Due to internal resistance $\mathrm{r}$ of the cell, the voltmeter reads a value $\mathrm{V}$, which is less than the emf of cell $\xi$. It is because, certain amount of voltage (Ir) has dropped across the internal resistance $r$.

$\mathrm{V}=\xi-\mathrm{Ir}$ $\mathrm{Ir}=\xi-\mathrm{V}$
Dividing equation (2) by equation (1), we get
$\frac{I r}{I R}=\frac{\xi-V}{V}$
$\mathrm{r}=\left|\frac{\xi-V}{V}\right| \mathrm{R} \ldots \ldots(3)$
Since $\zeta, \mathrm{V}$ and $\mathrm{R}$ are known, internal resistance $\mathrm{r}$ can be determined.
Question 5.
State and explain Kirchhoff's rules.
Answer:
Kirchhoff's first rule (current rule or junction rule):
Statement: It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a
statement of conservation of electric charge.

Explanation:
All charges that enter a given junction in a circuit must leave that junction since charge cannot build up or disappear at a junction. Current entering the junction is taken as positive and current leaving the junction is taken as negative.
Applying this law to the junction A, $\mathrm{I}_1+\mathrm{I}_2-\mathrm{I}_3-\mathrm{I}_4-\mathrm{I}_5=0$
Or
$\mathrm{I}_1+\mathrm{I}_2=+\mathrm{I}_3 \mathrm{I}_4+\mathrm{I}_5$
Kirchhoff's second rule (voltage rule or loop rule):
Statement: It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system. (The energy supplied by the emf sources is equal to the sum of the energy delivered to all resistors).
Explanation:
The product of current and resistance is taken as positive when the direction of the current is followed. Suppose if the direction of current is opposite to the direction of the loop, then product of current and voltage across the resistor is negative. It is shown in Fig. (a) and (b). The emf is considered positive when proceeding from the negative to the positive terminal of the cell. It is shown in Fig. (c) and (d).

Kirchhoff voltage rule has to be applied only when all currents in the circuit reach a steady state condition (the current in various branches are constant).
Question 6.
Obtain the condition for bridge balance in Wheatstone's bridge.
Answer:
An important application of Kirchhoff's rules is the Wheatstone's bridge. It is used to compare resistances and also helps in determining the unknown resistance in electrical network. The - bridge consists of four resistances $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ and $\mathrm{S}$ connected. A galvanometer $\mathrm{G}$ is connected between the points $\mathrm{B}$ and $\mathrm{D}$. The battery is connected between the points $\mathrm{A}$ and $\mathrm{C}$. The current through the galvanometer is $\mathrm{I}_{\mathrm{G}}$ and its resistance is $\mathrm{G}$.

Applying KirchhofFs current rule to junction B,
$\mathrm{I}_1-\mathrm{I}_{\mathrm{G}}-\mathrm{I}_3=0$
Applying Kirchhoff's current rule to junction D,
$\mathrm{I}_2-\mathrm{I}_{\mathrm{G}}-\mathrm{I}_4=0$
Applying Kirchhoff's voltage rule to loop ABDA,
$\mathrm{I}_1 \mathrm{P}+\mathrm{I}_{\mathrm{G}} \mathrm{G}-\mathrm{I}_2 \mathrm{R}=0$
(3)
Applying Kirchhoff's voltage rule to loop ABCDA,
$\mathrm{I}_1 \mathrm{P}+\mathrm{I}_3 \mathrm{Q}-\mathrm{I}_4 \mathrm{~S}-\mathrm{I}_2 \mathrm{R}=0$
(4)
When the points $\mathrm{B}$ and $\mathrm{D}$ are at the same potential, the bridge is said to be balanced. As there is no potential difference between $\mathrm{B}$ and $\mathrm{D}$, no current flows through galvanometer $\left(\mathrm{I}_{\mathrm{G}}=0\right)$. Substituting $\mathrm{I}_{\mathrm{G}}=$ 0 in equation, (1), (2) and (3), we get
$\mathrm{I}_1=\mathrm{I}_3$
$(5)$
$\mathrm{I}_2=\mathrm{I}_4$
$\mathrm{I}_1 \mathrm{P}=\mathrm{I}_2 \mathrm{R}$
(7)
Substituting the equation (5) and (6) in equation (4)
$\mathrm{I}_1 \mathrm{P}+\mathrm{I}_1 \mathrm{Q}-\mathrm{I}_2 \mathrm{R}=0$ $\mathrm{I}_1(\mathrm{P}+\mathrm{Q})=\mathrm{I}_2(\mathrm{R}+\mathrm{S})$
Dividing equation (8) by equation (7), we get
$
\begin{aligned}
& \frac{P+Q}{P}=\frac{R+S}{R} \\
& 1+\frac{Q}{P}=1+\frac{S}{R} \\
& \Rightarrow \frac{Q}{P}=\frac{S}{R} \\
& \frac{P}{Q}=\frac{R}{S} \ldots \ldots . .(9)
\end{aligned}
$
This is the bridge balance condition. Only under this condition, galvanometer shows null deflection. Suppose we know the values of two adjacent resistances, the other two resistances can be compared. If three of the resistances are known, the value of unknown resistance (fourth one) can be determined.

Question 7.
Explain the determination of unknown resistance using meter bridge.
Answer:
The meter bridge is another form of Wheatstone's bridge. It consists of a uniform manganin wire $\mathrm{AB}$ of one meter length. This wire is stretched along a meter scale on a wooden board between two copper strips C and D. Between these two copper strips another copper strip $E$ is mounted to enclose two gaps $G_1$ and $\mathrm{G}_2$ An unknown resistance $P$ is connected in $\mathrm{G}_1$ and a standard resistance $\mathrm{Q}$ is connected in $\mathrm{G}_2$.
A jockey (conducting wire) is connected to the terminal $\mathrm{E}$ on the central copper strip through a galvanometer (G) and a high resistance (HR). The exact position of jockey on the wire can be read on the scale. A Lechlanche cell and a key $(\mathrm{K})$ are connected across the ends of the bridge wire.

The position of the jockey on the wire is adjusted so that the galvanometer shows zero deflection. Let the point be $J$. The lengths $A J$ and $J B$ of the bridge wire now replace the resistance $R$ and $S$ of the Wheatstone's bridge. Then
$
\frac{P}{Q}=\frac{R}{S}=\frac{R^{\prime} \cdot A J}{R^{\prime} . J B}
$
where $\mathrm{R}^{\prime}$ is the resistance per unit length of wire
$
\begin{aligned}
& \frac{P}{Q}=\frac{A J}{J B}=\frac{l_1}{l_2} \ldots \ldots . . \text { (2) } \\
& \mathrm{P}=\mathrm{Q} \frac{l_1}{l_2} \ldots \ldots . . \text { (3) }
\end{aligned}
$
The bridge wire is soldered at the ends of the copper strips. Due to imperfect contact, some resistance might be introduced at the contact. These are called end resistances. This error can be eliminated, if another set of readings are taken with $P$ and $Q$ interchanged and the average value of $\mathrm{P}$ is found. To find the specific resistance of the material of the wire in the coil $\mathrm{P}$, the radius $\mathrm{r}$ and length 1 of the wire is measured. The specific resistance or resistivity $\mathrm{r}$ can be calculated using the relation
Resistance $=\rho-\frac{l}{A}$
By rearranging the above equation, we get
$\rho=$ Resistance $\mathrm{x} \frac{A}{l}$
If $P$ is the unknown resistance, equation (4) becomes
$
\rho=\mathrm{P} \frac{\pi r^2}{l} \text {. }
$
Question 8.
How the emf of two cells are compared using potentiometer?
Answer:
Comparison of emf of two cells with a potentiometer:
To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer wire $\mathrm{CD}$ is connected to a battery Bt and a key $\mathrm{K}$ in series. This is the primary circuit. The end $\mathrm{C}$ of the wire is
connected to the terminal $\mathrm{M}$ of a DPDT (Double Pole Double Throw) switch and the other terminal $\mathrm{N}$ is connected to a jockey through a galvanometer $\mathrm{G}$ and a high resistance HR. The cells whose emf $\zeta_1$ and $\zeta_2$ to be compared are connected to the terminals $\mathrm{M}_1, \mathrm{~N}_1$ and $\mathrm{M}_2, \mathrm{~N}_2$ of the DPDT switch.

The positive terminals of Bt, $\xi_1$ and $\zeta_2$ should be connected to the same end C. The DPDT switch is pressed towards $M_1, N_1$ so that cell $\zeta_1$ is included in the secondary circuit and the balancing length $l_1$ is found by adjusting the jockey for zero deflection, Then the second cells $\zeta_2$ is included in the circuit and the balancing length $l_2$ is determined. Let $\mathrm{r}$ be the resistance per unit length of the potentiometer wire and I be the current flowing through the wire. we have.
$\zeta_1=\operatorname{Irl}_1$
$\breve{\zeta}_2=\operatorname{Irl}_2 \ldots \ldots(2)$
By dividing (1) by (2)
$
\frac{\xi^1}{\xi^2}=\frac{l^1}{l^2} \ldots \ldots(3)
$
By including a rheostat $(\mathrm{Rh})$ in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.

Question 13.
Two cells each of $5 \mathrm{~V}$ are connected in series across a $8 \Omega$ resistor and three parallel resistors of $4 \Omega, 6 \mathrm{SI}$ and $12 \Omega$. Draw a circuit diagram for the above arrangement. Calculate (i) the current drawn from the cell (ii) current through each resistor.
Solution:

$
\begin{aligned}
& \mathrm{V}_1=5 \mathrm{~V} ; \mathrm{V}_2=5 \mathrm{~V} \\
& \mathrm{R}_1=8 \Omega ; \mathrm{R}_2=4 \Omega ; \mathrm{R}_3=6 \Omega ; \mathrm{R}_4=12 \Omega
\end{aligned}
$
Three resistors $\mathrm{R}_2, \mathrm{R}_3$ and $\mathrm{R}_4$ are connected parallel combination

$
\begin{aligned}
\frac{1}{\mathrm{R}_p} & =\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3}+\frac{1}{\mathrm{R}_4} \\
& =\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=\frac{3}{12}+\frac{2}{12}+\frac{1}{12}=\frac{6}{12}
\end{aligned}
$
Resistors $R_1$ and $\mathrm{R}_{\mathrm{p}}$ are connected in series combination
$
\begin{aligned}
& \mathrm{R}_{\mathrm{S}}=\mathrm{R}_1+\mathrm{R}_{\mathrm{p}}=8+2=10 \\
& \mathrm{R}_{\mathrm{S}}=10 \Omega
\end{aligned}
$
Total voltage connected series to the circuit
$
\begin{aligned}
& \mathrm{V}=\mathrm{V}_1+\mathrm{V}_2 \\
& =5+5=10 \\
& \mathrm{~V}=10 \mathrm{~V} .
\end{aligned}
$
(i) Current through the circuit, $\mathrm{I}=\frac{V}{R_s}=\frac{10}{10}$
$
\mathrm{I}=1 \mathrm{~A}
$
Potential drop across the parallel combination,
$
\mathrm{V}^{\prime}=\mathrm{IR}_{\mathrm{p}}=1 \times 2
$
V' $2 \mathrm{~V}$
(ii) Current in $4 \Omega$ resistor, $\mathrm{I}=\frac{V^{\prime}}{R_2}=\frac{2}{4}=0.5 \mathrm{~A}$
Current in $6 \Omega$ resistor, $\mathrm{I}=\frac{V^{\prime}}{R_3}=\frac{2}{6}=0.33 \mathrm{~A}$
Current in $12 \Omega$ resistor, $\mathrm{I}=\frac{V^{\prime}}{R_4}=\frac{2}{12}=0.17 \mathrm{~A}$
Question 14.
Four light bulbs $\mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{S}$ are connected in a circuit of unknown arrangement. When each bulb is removed one at a time and replaced, the following behavior is observed.

Draw the circuit diagram for these bulbs.
Solution:

Question 15 .
In a potentiometer arrangement, a cell of emf $1.25 \mathrm{~V}$ gives a balance point at $35 \mathrm{~cm}$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $63 \mathrm{~cm}$, what is the emf of the second cell?
Solution:
Emf of the cell $_1, \zeta_1=1.25 \mathrm{~V}$
Balancing length of the cell, $1_1=35 \mathrm{~cm}=35 \times 10^{-2} \mathrm{~m}$
Balancing length after interchanged, $1_2=63 \mathrm{~cm}=63 \times 10^{-2} \mathrm{~m}$
Emf of the cell $_1, \xi_2=$ ?
The ration of emf's, $\frac{\xi_1}{\xi_2}=\frac{l_1}{l_2}$
The ration of emf's, $\xi_2=\xi_1=\left(\frac{l_2}{l_1}\right)$ $=1.25 \times\left(\frac{63 \times 10^{-2}}{35 \times 10^{-2}}\right)=12.5 \times 1.8$
$\zeta_2=2.25 \mathrm{~V}$.