Heating Effect of Current and Electrical Measuring Instruments | Physics JEE previous year questions with Solutions

Q 11 :

Two resistance of 100 $Ω$ and 200 $Ω$ are connected in series with a battery of 4V and negligible internal resistance. A voltmeter is used to measure voltage across $100 Ω$ resistance, which gives reading as 1V. The resistance of voltmeter must be ________ $Ω$ . [2024]

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(200)

Given potential across Voltmeter = 1V

$∴ Current in 100 Ω = \frac{1}{100} A = i_{1}$

$200 Ω = \frac{3}{100} A \Rightarrow$ as total potential $= 4 V_{1} & 1 V$ is drop across $100 Ω$

Let current in voltmeter = i

$i_{1} + i_{2} = i_{0}$

$i_{2} + \frac{1}{100} = \frac{3}{200}$

$i_{2} + \frac{3}{200} - \frac{1}{100} = \frac{1}{200} A_{1}$

$V = I R_{1} \Rightarrow 1 = \frac{1}{200} (r), \Rightarrow r = 200 Ω$

Q 12 :

Consider a moving coil galvanometer (MCG):

A. The torsional constant in moving coil galvanometer has dimensions [ ${ML}^{2} T^{- 2}$ ]

B. Increasing the current sensitivity may not necessarily increase the voltage sensitivity.

C. If we increase number of turns (N) to its double (2N), then the voltage sensitivity doubles.

D. MCG can be converted into an ammeter by introducing a shunt resistance of large value in parallel with galvanometer.

E. Current sensitivity of MCG depends inversely on number of turns of coil.

Choose the correct answer from the options given below: [2025]

A, B only
A, D only
B, D, E only
A, B, E only

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(1)

(A) $τ = C θ \Rightarrow [{ML}^{2} T^{- 2}] = [C] [1]$

(B) $C . S . = \frac{θ}{I} = \frac{BNA}{C};$

$V . S . = \frac{BNA}{RC}$ [R = also depends on 'N']

(C) $V . S . \propto \frac{NAB}{CR} R \to NR$

(D) Shunt of law of resistance, False [Theory]

(E) E [False] C.S $\propto$ N

$\Rightarrow C . S = \frac{NAB}{C}$

Q 13 :

A galvanometer having a coil of resistance 30 $Ω$ need 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be $\frac{30}{X} Ω$ , where X is [2025]

447
298
149
596

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(3)

$I_{g} R_{g} = (I - I_{g}) r_{s}$

$20 \times 10^{- 3} \times 30 = (3 - 0.02) \times r_{s}$

$r_{s} = (\frac{0.6}{2.98}) = \frac{30}{x}$

$x = (\frac{2.98 \times 30}{0.6}) = 149$

Q 14 :

In a moving coil galvanometer, two moving coils $M_{1}$ and $M_{2}$ have the following particulars:

$R_{1} = 5 Ω, N_{1} = 15, A_{1} = 3.6 \times 10^{- 3} m^{2}, B_{1} = 0.25 T$

$R_{2} = 7 Ω, N_{2} = 21, A_{2} = 1.8 \times 10^{- 3} m^{2}, B_{2} = 0.50 T$

Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of $M_{1}$ and $M_{2}$ ? [2025]

1 : 1
1 : 4
1 : 3
1 : 2

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(1)

$K θ = N I A B \Rightarrow \frac{θ}{I} = \frac{N A B}{K}$

Voltage sensitivity $= \frac{θ}{V} = \frac{θ}{R I} = \frac{N A B}{K R}$

Ratio of voltage sensitivity $= (\frac{N_{1} A_{1} B_{1}}{R_{1}}) (\frac{R_{2}}{N_{2} A_{2} B_{2}})$

Ratio $= (\frac{N_{1} A_{1} B_{1}}{N_{2} A_{2} B_{2}}) \frac{R_{2}}{R_{1}} = \frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.5} \times \frac{7}{5} = \frac{1}{1}$

Q 15 :

A cell of emf 90 V is connected across series combination of two resistors each of 100 $Ω$ resistance. A voltmeter of resistance 400 $Ω$ is used to measure the potential difference across each resistor. The reading of the voltmeter will be [2023]

45 V
40 V
80 V
90 V

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(2)

$R_{eq} = \frac{400 \times 100}{500} + 100 = 180 Ω$

[IMAGE 214]------------------

$i = \frac{90}{180} = \frac{1}{2} A$

$Reading = \frac{1}{2} \times \frac{400}{500} \times 100 = 40 V$

Q 16 :

The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by 50%. The percentage change in voltage sensitivity of the galvanometer will be [2023]

100%
50%
75%
0%

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(4)

$Current sensitivity = Voltage sensitivity \times R$

$Current sensitivity is made 1.5 times.$

$R also increases 1.5 times.$

$Hence voltage sensitivity = \frac{1.5 \times current sensitivity}{1.5 \times R} = no change$

Q 17 :

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. [2023]

Assertion A: For measuring the potential difference across a resistance of 600 $Ω$ , the voltmeter with resistance 1000 $Ω$ will be preferred over voltmeter with resistance 4000 $Ω$ .

Reason R: Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.

In the light of the above statements, choose the most appropriate answer from the options given below.

A is not correct but R is correct
Both A and R are correct and R is the correct explanation of A
Both A and R are correct but R is not the correct explanation of A
A is correct but R is not correct

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(1)

Error of voltmeter decreases with increase in its resistance.

Q 18 :

A student is provided with a variable voltage source $V$ , a test resistor $R_{r} = 10 Ω$ , two identical galvanometers $G_{1}$ and $G_{2}$ , and two additional resistors, $R_{1} = 10 M Ω$ and $R_{2} = 0.001 Ω$ . For conducting an experiment to verify Ohm's law, the most suitable circuit is [2023]

[IMAGE 215]
[IMAGE 216]
[IMAGE 217]
[IMAGE 218]

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(3)

To convert galvanometer into ammeter low resistance should be added in parallel and for voltmeter conversion, a very high resistance should be added in series.

Q 19 :

For designing a voltmeter of range 50 V and an ammeter of range 10 mA using a galvanometer which has a coil of resistance 54 $Ω$ showing a full-scale deflection for 1 mA as shown in figure. [2023]

[IMAGE 219]

(A) for voltmeter $R \approx 50 k Ω$
(B) for ammeter $r \approx 0.2 Ω$
(C) for ammeter $r \approx 6 Ω$
(D) for voltmeter $R \approx 5 k Ω$
(E) for voltmeter $R \approx 500 Ω$

Choose the correct answer from the options given below:

(C) and (E)
(C) and (D)
(A) and (C)
(A) and (B)

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(3)

$For voltmeter$

$R = \frac{V}{I_{g}} - G = \frac{50}{10^{- 3}} - 54 \approx 50 k Ω (A)$

$For ammeter$

$S = \frac{I_{g} G}{I - I_{g}} = \frac{10^{- 3} \times 54}{(10 - 1) \times 10^{- 3}} = 6 Ω (C)$

Q 20 :

As shown in the figure, the voltmeter reads 2 V across 5 $Ω$ resistor. The resistance of the voltmeter is ______ $Ω$ . [2023]

[IMAGE 220]

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(20)

$i_{1} = \frac{2 V}{5 Ω} = \frac{2}{5} A$

[IMAGE 221]--------------

$i = \frac{1 V}{2 Ω} = \frac{1}{2} A$

$∴$ $Current through voltmeter = i - i_{1}$

$= \frac{1}{2} - \frac{2}{5} = \frac{5 - 4}{10} = \frac{1}{10} A$

$∴$ $For voltmeter 2 = (\frac{1}{10}) R \Rightarrow R = 20 Ω$

Topic Question Set

Q 11 :

Two resistance of 100 $Ω$ and 200 $Ω$ are connected in series with a battery of 4V and negligible internal resistance. A voltmeter is used to measure voltage across $100 Ω$ resistance, which gives reading as 1V. The resistance of voltmeter must be ________ $Ω$ . [2024]

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Q 13 :

A galvanometer having a coil of resistance 30 $Ω$ need 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be $\frac{30}{X} Ω$ , where X is [2025]

Q 15 :

A cell of emf 90 V is connected across series combination of two resistors each of 100 $Ω$ resistance. A voltmeter of resistance 400 $Ω$ is used to measure the potential difference across each resistor. The reading of the voltmeter will be [2023]

Q 16 :

The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by 50%. The percentage change in voltage sensitivity of the galvanometer will be [2023]

Q 20 :

As shown in the figure, the voltmeter reads 2 V across 5 $Ω$ resistor. The resistance of the voltmeter is ______ $Ω$ . [2023]

[IMAGE 220]

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