Question 40: Current Electricity

Question

A galvanometer of resistance $50 \Omega$ is connected to a battery of 4 V along with a resistance of $3950 \Omega$ in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 10 divisions, the resistance in series should be equal to:

Accepted Answer

LetÃ¢ÂÂs work it out step by step:
Initial fullÃ¢ÂÂscale current
$$I_{m fs}=\frac{V}{R_g+R}=\frac{4m\,V}{50\Omega+3950\Omega}=\frac{4}{4000}m\,A=1\,mA.$$
Current for 10 divisions (oneÃ¢ÂÂthird of full scale)
$$I_{10}=\frac{10}{30}\,I_{m fs}=\frac13\cdot1\,mA=0.3333\,mA.$$
Total resistance needed to get 0.3333 mA from 4 V
$$R_{m tot}'=\frac{V}{I_{10}}=\frac{4m\,V}{0.3333	imes10^{-3}m\,A}=12000\Omega.$$
New series resistance
$$R'_{m series}=R_{m tot}'-R_g=12000\Omega-50\Omega=11950\Omega.$$
Answer: 11950 ÃÂ© (Option B).

Answer

8950 ohm

Answer

11950 ohm

Answer

7000 ohm

Answer

6000 ohm

Current Electricity

A galvanometer of resistance 50 \Omega is connected to a battery of 4 V along with a resistance of 3950 \Omega in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 10 divisions, the resistance in series should be equal to:

Choose the correct answer: