WELCOME TO SaraNextGen.Com

Page No 221: - 12) ELECTRICITY


Question 1:

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is −

(a) https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1199/Chapter%2012_html_m144e062f.gif

(b) https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1199/Chapter%2012_html_3d0b22e5.gif

(c) 5

(d) 25

Answer:

(d) Resistance of a piece of wire is proportional to its length. A piece of wire has a resistance R. The wire is cut into five equal parts.

Therefore, resistance of each part = https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1199/Chapter%2012_html_mc70695b.gif

All the five parts are connected in parallel. Hence, equivalent resistance (R’) is given as

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1199/Chapter%2012_html_mff9f4bb.gif

Therefore, the ratiohttps://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1199/Chapter%2012_html_75e0f633.gif  is 25.

 

 

Question 2:

Which of the following terms does not represent electrical power in a circuit?

(a) I2R

(b) IR2

(c) VI

(d) https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1200/Chapter%2012_html_45bb165c.gif

Answer:

(b) Electrical power is given by the expression, P = VI … (i)

According to Ohm’s law, V = IR … (ii)

Where,

V = Potential difference

I = Current

R = Resistance

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1200/Chapter%2012_html_2925806e.gif

From equation (i), it can be written

P = (IR) × I

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1200/Chapter%2012_html_2a3795e3.gif

From equation (ii), it can be written

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1200/Chapter%2012_html_7f821d6b.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1200/Chapter%2012_html_m77f32199.gif

Power P cannot be expressed as IR2.

 

 

Question 3:

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be −

(a) 100 W

(b) 75 W

(c) 50 W

(d) 25 W

Answer:

(d)Energy consumed by an appliance is given by the expression,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1201/Chapter%2012_html_1170d3bd.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1201/Chapter%2012_html_m2938f725.gif

Where,

Power rating, = 100 W

Voltage, = 220 V

Resistance, https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1201/Chapter%2012_html_35b6eed5.gif

The resistance of the bulb remains constant if the supply voltage is reduced to 110 V. If the bulb is operated on 110 V, then the energy consumed by it is given by the expression for power as

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1201/Chapter%2012_html_m1631ea65.gif

Therefore, the power consumed will be 25 W.

 

Question 4:

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be −

(a) 1:2

(b) 2:1

(c) 1:4

(d) 4:1

Answer:

(c) The Joule heating is given by, H = i2Rt

Let, R be the resistance of the two wires.

The equivalent resistance of the series connection is RS = R + R = 2R

If V is the applied potential difference, then it is the voltage across the equivalent resistance.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/2/156/1204/NCERT_Gr10_Ch12_Pg221_Q4_html_211e4ba0.gif

The heat dissipated in time t is,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/2/156/1204/NCERT_Gr10_Ch12_Pg221_Q4_html_m15f7f766.gif

The equivalent resistance of the parallel connection is RP = https://img-nm.mnimgs.com/img/study_content/curr/1/10/2/156/1204/NCERT_Gr10_Ch12_Pg221_Q4_html_4788d82.gif

V is the applied potential difference across this RP.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/2/156/1204/NCERT_Gr10_Ch12_Pg221_Q4_html_d9a3acf.gif

The heat dissipated in time t is,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/2/156/1204/NCERT_Gr10_Ch12_Pg221_Q4_html_29b913da.gif

So, the ratio of heat produced is,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/2/156/1204/NCERT_Gr10_Ch12_Pg221_Q4_html_4c1d1b9e.gif

Note: https://img-nm.mnimgs.com/img/study_content/curr/1/10/2/156/1204/NCERT_Gr10_Ch12_Pg221_Q4_html_4929f8da.gif  R also https://img-nm.mnimgs.com/img/study_content/curr/1/10/2/156/1204/NCERT_Gr10_Ch12_Pg221_Q4_html_4929f8da.gif  i2 and https://img-nm.mnimgs.com/img/study_content/curr/1/10/2/156/1204/NCERT_Gr10_Ch12_Pg221_Q4_html_4929f8da.gif  t. In this Questiont is same for both the circuit. But the current through the equivalent resistance of both the circuit is different. We could have solved the Question directly using H https://img-nm.mnimgs.com/img/study_content/curr/1/10/2/156/1204/NCERT_Gr10_Ch12_Pg221_Q4_html_4929f8da.gif  R if in case the current was also same. As we know the voltage and resistance of the circuits, we have calculated i in terms of voltage and resistance and used in the equation H = i2Rt to find the ratio.

 

Question 5:

How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer:

To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.

 

Question 6:

A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10−8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Answer:

Resistance (R) of a copper wire of length l and cross-section A is given by the expression,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1208/Chapter%2012_html_mab70f9d.gif

Where,

Resistivity of copper, https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1208/Chapter%2012_html_21420f04.gif

Area of cross-section of the wire, A = https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1208/Chapter%2012_html_21735016.gif

Diameter= 0.5 mm = 0.0005 m

Resistance, R = 10 Ω

Hence, length of the wire, https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1208/Chapter%2012_html_m445dc9de.gif

If the diameter of the wire is doubled, new diameter https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1208/Chapter%2012_html_76b7ac6d.gif https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1208/Chapter%2012_html_m4c171919.gif

Therefore, resistance https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1208/Chapter%2012_html_349dbe4.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1208/Chapter%2012_html_m626d1b2f.gif

Therefore, the length of the wire is 122.7 m and the new resistance is https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1208/Chapter%2012_html_m38b4958.gif

Question 7:

The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below −

(amperes )

0.5

1.0

2.0

3.0

4.0

V (volts)

1.6

3.4

6.7

10.2

13.2

Plot a graph between V and I and calculate the resistance of that resistor.

Answer:

The plot between voltage and current is called IV characteristic. The voltage is plotted on x-axis and current is plotted on y-axis. The values of the current for different values of the voltage are shown in the given table.

V (volts)

1.6

3.4

6.7

10.2

13.2

(amperes )

0.5

1.0

2.0

3.0

4.0

The IV characteristic of the given resistor is plotted in the following figure.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1212/Chapter%2012_html_4011400d.jpg

The slope of the line gives the value of resistance (R) as,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1212/Chapter%2012_html_114fa046.gif

Therefore, the resistance of the resistor ishttps://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1212/Chapter%2012_html_64989b47.gif .

 

 

Question 8:

When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Answer:

Resistance (R) of a resistor is given by Ohm’s law as,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1214/Chapter%2012_html_381fd0a5.gif

Where,

Potential difference, V = 12 V

Current in the circuit, I = 2.5 mA = https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1214/Chapter%2012_html_3e147763.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1214/Chapter%2012_html_3ccf8ba1.gif

Therefore, the resistance of the resistor is https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1214/Chapter%2012_html_m413b2023.gif .

 

 

Question 9:

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Answer:

There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1216/Chapter%2012_html_m582fe138.gif

Where,

R is the equivalent resistance of resistanceshttps://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1216/Chapter%2012_html_m368cbf97.gif . These are connected in series. Hence, the sum of the resistances will give the value of R.

R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω

Potential difference, V = 9 V

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1216/Chapter%2012_html_3e3c0750.gif

Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.

 

 

Question 10:

How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer:

For x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is given by Ohm’s law as

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1219/Chapter%2012_html_381fd0a5.gif

Where,

Supply voltage, V = 220 V

Current, I = 5 A

Equivalent resistance of the combination = R,given as

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1219/Chapter%2012_html_47dea442.gif

From Ohm’s law,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1219/Chapter%2012_html_m642fb624.gif

Therefore, four resistors of 176 Ω are required to draw the given amount of current.

 

 

Question 11:

Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Answer:

If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1220/Chapter%2012_html_m3cbe57b4.gif  Hence, we should either connect the two resistors in series or parallel.

(i) Two resistors in parallel

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1220/Chapter%2012_html_43fc98fc.jpg

Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1220/Chapter%2012_html_442bd4cb.gif

The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω + 3 Ω = 9 Ω.

(ii) Two resistors in series

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1220/Chapter%2012_html_m7d55780c.jpg

Two 6 Ω resistors are in series. Their equivalent resistance will be the sum https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1220/Chapter%2012_html_3de49b0f.gif

The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1220/Chapter%2012_html_me3e7a67.gif

Therefore, the total resistance ishttps://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1220/Chapter%2012_html_77914a3b.gif .

 

 

Question 12:

Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Answer:

Resistance R1 of the bulb is given by the expression,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1222/Chapter%2012_html_m4d630b3c.gif

Where,

Supply voltage, V = 220 V

Maximum allowable current, I = 5 A

Rating of an electric bulb https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1222/Chapter%2012_html_2423a4bd.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1222/Chapter%2012_html_m59b28dbe.gif

According to Ohm’s law,

V = I R

Where,

R is the total resistance of the circuit for x number of electric bulbs

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1222/Chapter%2012_html_m46859c93.gif

Resistance of each electric bulb, https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1222/Chapter%2012_html_1c422f61.gif Ω

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1222/Chapter%2012_html_m3c97189a.gif

Therefore, 110 electric bulbs are connected in parallel.

 

Question 13:

A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistances, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer:

Supply voltage, V = 220 V

Resistance of one coil, R =https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_4b4cdc08.gif

(i) Coils are used separately

According to Ohm’s law,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_7c7134e.gif

Where,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_m3507c365.gif is the current flowing through the coil

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_11030834.gif

Therefore, 9.16 A current will flow through the coil when used separately.

(ii) Coils are connected in series

Total resistance, https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_23c554a3.gif https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_4bb624b7.gif

According to Ohm’s law,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_33932c8f.gif

Where,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_13d89b8c.gif is the current flowing through the series circuit

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_m5ee3518a.gif

Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.

(iii) Coils are connected in parallel

Total resistance, https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_m5553d0fe.gif  is given ashttps://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_7009623c.gif

According to Ohm’s law,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_m5bb0b0e4.gif

Where,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_48d481b3.gif  is the current flowing through the circuit

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1224/Chapter%2012_html_mecf1951.gif

Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.

Question 14:

Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer:

(i) Potential difference, V = 6 V

1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω

According to Ohm’s law,

V = IR

Where,

I is the current through the circuit

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1226/Chapter%2012_html_m4adf195b.gif

This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the 2 Ω resistor ishttps://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1226/Chapter%2012_html_5d6570d1.gif . Power is given by the expression,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1226/Chapter%2012_html_39318718.gif

(ii) Potential difference, V = 4 V

12 Ω and 2 Ω resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across 2 Ω resistor will be 4 V.

Power consumed by 2 Ω resistor is given by

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1226/Chapter%2012_html_m7511a21e.gif

Therefore, the power used by 2 Ω resistor is 8 W.