Question 515: Current Electricity

Question

When a resistance of $100 \Omega$ is connected in series with a galvanometer of resistance G , its range is V . To double its range a resistance of $1000 \Omega$ is connected in series. The value of G is

Accepted Answer

Given:
First scenario with a resistance of $ 100 \, \Omega $ in series leads to a range $ V $.
Second scenario, for doubling the range, a resistance of $ 1000 \, \Omega $ is connected.
The governing equation for resistance in series to reach the required voltage range is:
$ R_s = \frac{V}{I_s} - G $
For the first scenario:
$ 100 = \frac{V}{I_s} - G \quad 	ext{(Equation 1)} $
For the second scenario, where the range is doubled:
$ 1000 = \frac{2V}{I_s} - G \quad 	ext{(Equation 2)} $
Subtract Equation 1 from Equation 2:
$ 1000 - 100 = \left(\frac{2V}{I_s} - \frac{V}{I_s}ight) $
Solving the above, we find:
$ 900 = \frac{V}{I_s} \quad 	ext{(Equation 3)} $
Using Equation 3 in Equation 1:
$ 100 = 900 - G $
Solving for $ G $, we get:
$ G = 800 \, \Omega $
Therefore, the resistance of the galvanometer is $ 800 \, \Omega $.

Answer

$400 \Omega$

Answer

$300 \Omega$

Answer

$200 \Omega$

Answer

$100 \Omega$

Current Electricity

When a resistance of 100 \Omega is connected in series with a galvanometer of resistance G , its range is V . To double its range a resistance of 1000 \Omega is connected in series. The value of G is

Choose the correct answer: