Question 514: Current Electricity

Question

When a resistance of $200 \Omega$ is connected in series with a galvanometer of resistance ' $G$ ', its range is ' $V$ '. To triple its range, a resistance of $2000 \Omega$ is connected in series. The value of ' $G$ ' is

Accepted Answer

To solve for the galvanometer resistance ' $G$ ', we use the formula:
$ R_s = \frac{V}{I_g} - G $
For the two cases provided:
First Case:
A resistance of $200 \Omega$ is connected in series. Using the formula, we get:
$    200 = \frac{V}{I_g} - G 	ag{1}    $
Second Case:
A resistance of $2000 \Omega$ is added to triple the range, leading to:
$    2000 = \frac{3V}{I_g} - G 	ag{2}    $
By subtracting equation (1) from equation (2), we eliminate $G$ and solve for $\frac{V}{I_g}$:
$ 2000 - 200 = \frac{3V}{I_g} - \frac{V}{I_g} $
$ 1800 = \frac{2V}{I_g} $
$ 	herefore \frac{V}{I_g} = 900 $
Substituting $\frac{V}{I_g} = 900$ back into equation (1):
$ 200 = 900 - G $
Solving for $G$ gives:
$ G = 700 \Omega $

Answer

$200 \Omega$

Answer

$400 \Omega$

Answer

$600 \Omega$

Answer

$700 \Omega$

Current Electricity

When a resistance of 200 \Omega is connected in series with a galvanometer of resistance ' G ', its range is ' V '. To triple its range, a resistance of 2000 \Omega is connected in series. The value of ' G ' is

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