A voltmeter with a resistance of 50 × 103 Ω is used to measure voltage in a circuit. To increase its range to 3 times, the additional resistance to be put in series is

Question

A voltmeter with a resistance of 50 × 103 Ω is used to measure voltage in a circuit. To increase its range to 3 times, the additional resistance to be put in series is (A) 9 × 106 Ω (B) 105 Ω (C) 1.5 × 105 Ω (D) 9 × 105 Ω

Tardigrade Admin · Accepted Answer

Here, V=IgR and V'=IgR' or (R'/R)=(V'/V) or R'=(V'/V)R=(3V/V)×50×103 =1.5×105 Ω ∴ additional resistance, =1.5 ×105-0.5×105 =105 Ω

Q. A voltmeter with a resistance of $50 \times 1 0^{3} Ω$ is used to measure voltage in a circuit. To increase its range to $3$ times, the additional resistance to be put in series is

$9 \times 1 0^{6} Ω$

$1 0^{5} Ω$

$1.5 \times 1 0^{5} Ω$

$9 \times 1 0^{5} Ω$

Here, $V = I_{g} R$ and $V^{'} = I_{g} R^{'}$ or $\frac{R ^{'}}{R} = \frac{V ^{'}}{V}$
or $R^{'} = \frac{V ^{'}}{V} R = \frac{3 V}{V} \times 50 \times 1 0^{3}$
$= 1.5 \times 1 0^{5} Ω$
$∴$ additional resistance,
$= 1.5 \times 1 0^{5} - 0.5 \times 1 0^{5}$
$= 1 0^{5} Ω$

Q. A voltmeter with a resistance of 50×103Ω is used to measure voltage in a circuit. To increase its range to 3 times, the additional resistance to be put in series is

9×106Ω

105Ω

1.5×105Ω

9×105Ω