Question 149: Current Electricity

Question

A galvanometer of resistance $$100 \Omega$$ when connected in series with $$400 \Omega$$ measures a voltage of upto $$10 \mathrm{~V}$$. The value of resistance required to convert the galvanometer into ammeter to read upto $$10 \mathrm{~A}$$ is $$x 	imes 10^{-2} \Omega$$. The value of $$x$$ is :

Accepted Answer

To convert a galvanometer into an ammeter to measure larger currents, a low resistance known as the
shunt resistance
($$R_{	ext{sh}}$$) is connected in parallel with the galvanometer. The value of this shunt resistance can be calculated using the principles of parallel circuits and the desired maximum current the ammeter should read.
The original configuration of the galvanometer allows it to measure up to $$10\,	ext{V}$$, and it has a resistance of $$100\,\Omega$$. When connected in series with a $$400\,\Omega$$ resistor, the total resistance in the circuit is $$100\,\Omega + 400\,\Omega = 500\,\Omega$$. Given this configuration measures up to $$10\,	ext{V}$$, we can calculate the maximum current it is designed to measure using Ohm's law:
$I = \frac{V}{R} = \frac{10\,	ext{V}}{500\,\Omega} = 0.02\,	ext{A}$
Now, to recalibrate the device to measure up to $$10\,	ext{A}$$, we require the calculation of the shunt resistor $$R_{	ext{sh}}$$ that needs to be connected in parallel with the galvanometer. The total current $$I$$ will now be $$10\,	ext{A}$$, and the part of this current flowing through the galvanometer ($$I_g$$) remains $$0.02\,	ext{A}$$ (as before, to ensure we do not exceed the device's original maximum measuring capability), leaving the rest to flow through the shunt. Thus, $$I - I_g$$ flows through the shunt.
Since the voltage drop across both the shunt and the galvanometer must be the same for parallel components, we use OhmÃ¢ÂÂs Law $$V = IR$$ for both and set up an equation to calculate $$R_{	ext{sh}}$$:
$I_g R_g = (I - I_g) R_{	ext{sh}}$
Substituting known values ($$R_g = 100\,\Omega$$, $$I = 10\,	ext{A}$$, and $$I_g = 0.02\,	ext{A}$$):
$0.02\,	ext{A} 	imes 100\,\Omega = (10\,	ext{A} - 0.02\,	ext{A}) R_{	ext{sh}}$
This simplifies to:
$2\,	ext{V} = 9.98\,	ext{A} 	imes R_{	ext{sh}}$
Solving for $$R_{	ext{sh}}$$ gives:
$R_{	ext{sh}} = \frac{2\,	ext{V}}{9.98\,	ext{A}} \approx 0.2004\,\Omega$
Expressing this in terms of $$	imes 10^{-2} \Omega$$ gives $$R_{	ext{sh}} \approx 20.04 	imes 10^{-2} \Omega$$. Therefore, the value of $$x$$ is approximately $$20.04$$, and rounding it according to the provided options leads to the closest value:
Option B: 20

Answer

2

Answer

20

Answer

800

Answer

200

Current Electricity

A galvanometer of resistance $100 \Omega when connected in series with 400 \Omega measures a voltage of upto 10 \mathrm{~V}. The value of resistance required to convert the galvanometer into ammeter to read upto 10 \mathrm{~A} is x \times 10^{-2} \Omega. The value of x$ is :

Choose the correct answer: