Question 462: Current Electricity

Question

The value of the shunt resistance that allows $10 \%$ of the main current through the galvanometer of $99 \Omega$ is

Accepted Answer

Step 1: Understand the situation
A galvanometer of resistance $R_g = 99 \,\Omega$.
It is shunted by some resistance $R_s$.
Only $10\%$ of the main current should go through the galvanometer, so the remaining $90\%$ goes through the shunt.
Step 2: Current ratio
Let the total current be $I$.
Galvanometer current: $I_g = 0.1I$.
Shunt current: $I_s = 0.9I$.
So, the ratio:
$ \frac{I_g}{I_s} = \frac{0.1I}{0.9I} = \frac{1}{9}. $
Step 3: Condition for parallel resistance
Since the galvanometer and the shunt are in parallel, their voltage drops must be equal:
$ I_g R_g = I_s R_s. $
So,
$ R_s = \frac{I_g}{I_s} \cdot R_g. $
Step 4: Substitute values
$ R_s = \left(\frac{1}{9}\right) \cdot 99 = 11 \,\Omega. $
Ã¢ÂÂ Final Answer:
The correct option is:
Option B: $11 \,\Omega$

Answer

$9 \Omega$

Answer

$11 \Omega$

Answer

$13 \Omega$

Answer

$15 \Omega$

Current Electricity

The value of the shunt resistance that allows 10 \% of the main current through the galvanometer of 99 \Omega is

Choose the correct answer: