When 500μA curre nt is passed through a galvanometer of `20Omega` , it gives a full scale deflection. The external resistance need to be connected to measure 5V is

Options

A. `98Omega` parallel
B. `9980Omega` parallel
C. `98Omega` series
D. `9980Omega` series

Answer

To solve the problem, we need to find the external resistance that should be connected in series with a galvanometer to measure a voltage of 5V when a current of 500μA is flowing through it. The resistance of the galvanometer is given as 20Ω.

### Step-by-Step Solution:

1. **Identify the given values:**
- Current through the galvanometer, \( I_g = 500 \, \mu A = 500 \times 10^{-6} \, A \)
- Resistance of the galvanometer, \( G = 20 \, \Omega \)
- Voltage to be measured, \( V = 5 \, V \)

2. **Use Ohm's Law to find the total resistance:**
The total voltage across the circuit can be expressed using Ohm's Law:
\[ V = I_g \times (R + G) \]
where \( R \) is the external resistance we need to find.

3. **Rearranging the equation:**
We can rearrange the equation to solve for \( R \):
\[ R + G = \frac{V}{I_g} \]
Thus,
\[ R = \frac{V}{I_g} - G \]

4. **Substituting the values:**
Now, substituting the known values into the equation:
\[ R = \frac{5 \, V}{500 \times 10^{-6} \, A} - 20 \, \Omega \]

5. **Calculating \( \frac{5}{500 \times 10^{-6}} \):**
\[ \frac{5}{500 \times 10^{-6}} = \frac{5}{0.0005} = 10000 \, \Omega \]

6. **Finding \( R \):**
Now, substituting this back into the equation for \( R \):
\[ R = 10000 \, \Omega - 20 \, \Omega = 9980 \, \Omega \]

7. **Conclusion:**
The external resistance that needs to be connected in series to measure 5V is:
\[ R = 9980 \, \Omega \]

### Final Answer:
The external resistance needed is **9980 Ω in series**.

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