Question 477: Current Electricity

Question

Only $4 \%$ of the total current in the circuit passes through a galvanometer. If the resistance of the galvanometer is G , then the shunt resistance connected to the galvanometer is

Accepted Answer

Let the total current in the circuit be $I$.
Current through the galvanometer, $I_G = 4\% 	ext{ of } I = \frac{4}{100}I = \frac{I}{25}$
Let the shunt resistance be $S$ (connected in parallel with the galvanometer).
The remaining current, $I_S = I - I_G = I - \frac{I}{25} = \frac{24I}{25}$, passes through the shunt.
Since $G$ and $S$ are in parallel, the voltage across both is the same.
So,
$ 	ext{Voltage across } G = 	ext{Voltage across } S $
$ I_G G = I_S S $
Substitute $I_G = \frac{I}{25}$ and $I_S = \frac{24I}{25}$:
$ \frac{I}{25} G = \frac{24I}{25} S $
Dividing both sides by $I$,
$ \frac{G}{25} = \frac{24S}{25} $
Multiply both sides by 25:
$ G = 24S $
$ S = \frac{G}{24} $
Correct Answer:
$\boxed{\frac{G}{24}}$
So, the answer is
Option B
.

Answer

$\frac{\mathrm{G}}{25}$

Answer

$\frac{\mathrm{G}}{24}$

Answer

24 G

Answer

25 G

Current Electricity

Only 4 \% of the total current in the circuit passes through a galvanometer. If the resistance of the galvanometer is G , then the shunt resistance connected to the galvanometer is

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