Question

A galvanometer of resistance $50 \Omega$ is converted to a battery of 3 V along with a resistance of 2950 $\Omega$ in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be.

• A. 6050 $\Omega$
• B. 4450$\Omega$
• C. 5050 $\Omega$
• D. 5550$\Omega$

Answer 1

Answer: (B)

4450$\Omega$

Answer 2

Total initial resistance $= \mathrm { G } + \mathrm { R } = 50 \Omega + 2950 \Omega$

$= 3000 \Omega$

Current

If the deflection has to be reduced to 20 divisions then current

$\mathrm { I } ^ { \prime } = \frac { 1 \mathrm {~mA} } { 30 } \times 20 = \frac { 2 } { 3 } \mathrm {~mA}$

Let x be the effective resistance of the circuit

$3 \mathrm {~V} = 3000 \Omega \times 1 \mathrm {~mA} = \mathrm { x } \Omega \times \frac { 2 } { 3 } \mathrm {~mA}$

Or $x = 3000 \times 1 \times \frac { 3 } { 2 } = 4500 \Omega$

$\therefore$ Resistance to be added = $= ( 4500 \Omega - 50 \Omega )$

$= 4450 \Omega$