Question

The scale of a galvanometer is divided into 160 equal divisions. The galvanometer shows full scale deflection of \(16\text{ mA}\) and maximum voltage is \(80\text{ mV}\) . Now the range is changed so that galvanometer reads \(160\text{ V}\) . The required resistance to be connected is

• A. \(9995\Omega\) in series.
• B. \(4995\Omega\) in series.
• C. \(9.5 \times 10^{-3}\Omega\) in parallel.
• D. \(4.95 \times 10^{-3}\Omega\) in parallel.

Hint:

Voltmeter → always connect high resistance in series.

Answer 1

The Correct Option is A

Concept:
To convert galvanometer to voltmeter: \[ R = \frac{V}{I} - G \] Step 1: Find galvanometer resistance. \[ G = \frac{80 \times 10^{-3}}{16 \times 10^{-3}} = 5 \Omega \]
Step 2: Required total resistance. \[ R_{total} = \frac{160}{16 \times 10^{-3}} = 10000 \Omega \]
Step 3: Series resistance. \[ R = 10000 - 5 = 9995 \Omega \]
Step 4: Conclusion. Required resistance = \(9995\Omega\) (series)