Question 45: Current Electricity

Question

The potential difference between the points $\mathrm{P}$ and $\mathrm{Q}$ of the arrangement shown in figure is

Accepted Answer

To convert the galvanometer into a voltmeter, we need to find the additional series resistance required. The full-scale deflection voltage for the galvanometer itself is given by:
$ V_g = I_g 	imes R_g = 2 	imes 10^{-6} \, 	ext{A} 	imes 10 \, \Omega = 2 	imes 10^{-5} \, 	ext{V} $
To achieve a full-scale deflection of 10 mV, the total resistance $R_v$ of the voltmeter should satisfy:
$ 10 	imes 10^{-3} \, 	ext{V} = I_g 	imes (R_g + R_s) $
Where $R_s$ is the series resistance. Solving for $R_s$, we have:
$ R_s = \frac{10 	imes 10^{-3}}{2 	imes 10^{-6}} - 10 = 5000 - 10 = 4990 \, \Omega $
Thus, the resistance of the voltmeter is:
$ R_v = R_g + R_s = 10 + 4990 = 5000 \, \Omega $
To convert the galvanometer into an ammeter of range 1 mA, we need to find the shunt resistance $R_a$. For the galvanometer to show full-scale deflection at 1 mA, the remaining current (beyond the 2 ÃÂµA for full deflection) should pass through the shunt:
$ I_{shunt} = 1 \, 	ext{mA} - 2 \, \mu 	ext{A} = 0.998 \, 	ext{mA} $
The voltage across the galvanometer (and hence the shunt) is:
$ V = I_g 	imes R_g = 2 	imes 10^{-6} \, 	ext{A} 	imes 10 \, \Omega = 2 	imes 10^{-5} \, 	ext{V} $
Then $R_a$, the shunt resistance, can be calculated using:
$ R_a = \frac{V}{I_{shunt}} = \frac{2 	imes 10^{-5}}{0.998 	imes 10^{-3}} \approx 0.02004 \, \Omega $
Finally, to find the ratio of the resistance of the converted voltmeter to that of the ammeter:
$ 	ext{Ratio} = \frac{R_v}{R_a} = \frac{5000}{0.02004} \approx 249800 \approx 250000 $
Therefore, the ratio is closest to:
Option C: 250000 : 1

Answer

$-9.3$ V

Answer

42 V

Answer

$-42$ V

Answer

$-3.9$ V

Current Electricity

The potential difference between the points \mathrm{P} and \mathrm{Q} of the arrangement shown in figure is

Choose the correct answer: