Current Electricity

JEE Main level, single correct, medium

311 of 380

Question 311 of 380

A voltmeter with internal resistance of x Ωx\ \Omega can be used to measure upto 2020 V. In order to increase its measuring range to 3030 V, the required modification is to _____.

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Answer: (A) connect resistor of x2 Ω\dfrac{x}{2}\ \Omega, in series with voltmeter.

Let the maximum current for full-scale deflection of the voltmeter be IgI_g.

Ig=V1Rv=20xI_g = \dfrac{V_1}{R_v} = \dfrac{20}{x}

To increase the measuring range to V2=30V_2 = 30 V, a resistance RR must be connected in series with the voltmeter.

V2=Ig(Rv+R)V_2 = I_g(R_v + R)

30=20x(x+R)30 = \dfrac{20}{x}(x + R)

30x=20x+20R30x = 20x + 20R

10x=20R10x = 20R

R=x2 ΩR = \dfrac{x}{2}\ \Omega

Thus, a resistor of x2 Ω\dfrac{x}{2}\ \Omega should be connected in series with the voltmeter.

Answer: connect resistor of x2 Ω\dfrac{x}{2}\ \Omega, in series with voltmeter.

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