Question

A moving coil galvanometer has a resistance of $50\,\Omega$ and gives a full-scale deflection for a current of $2 \text{ mA}$. To convert it into a voltmeter reading up to 10 V, the required series resistance to be connected is

• A. $4950\,\Omega$
• B. $5000\,\Omega$
• C. $4450\,\Omega$
• D. $5050\,\Omega$

Hint:

Think of it this way: Total resistance needed is $\frac{10\text{V}}{2\text{mA}} = 5000\,\Omega$. Since the galvanometer already provides $50\,\Omega$, you just need to add the remaining $5000 - 50 = 4950\,\Omega$ in series!

Answer 1

The Correct Option is A

Step 1: How to make a voltmeter.
Connect a big resistance $R$ in series with the galvanometer. This lets it read larger voltages safely.

Step 2: The series resistance formula.
\[ R = \frac{V}{I_{g}} - G \]where $V$ is the full-scale voltage, $I_{g}$ is the full-scale current, and $G$ is the coil resistance.

Step 3: List the values.
$G = 50\,\Omega$, $I_{g} = 2$ mA $= 2\times 10^{-3}$ A, $V = 10$ V.

Step 4: Work out $V/I_{g}$.
\[ \frac{V}{I_{g}} = \frac{10}{2\times 10^{-3}} = 5000 \,\Omega \]
Step 5: Subtract the coil resistance.
\[ R = 5000 - 50 = 4950 \,\Omega \]
Step 6: State the answer.
So the series resistance needed is $4950\,\Omega$, which is option 1.
\[ \boxed{R = 4950 \,\Omega} \]