JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2019
- Hard
- Numerical
The galvanometer deflection, when key \(K_1\) is closed but \(K_2\) is open, equals \(\theta_0\) (see figure). On closing \(K_2\) also and adjusting \(R_2\) to \(5\,\Omega \) , the deflection in galvanometer becomes \(\frac{{\theta _0}}{5}\). The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]: .................. \(\Omega\)

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(22\)
Step-by-step Solution
Detailed explanation
When \(\mathrm{K}_{1}\) closed and \(\mathrm{K}_{2}\) is open
\(i=\frac{V}{220+R}\) .....\((i)\)
When \(\mathrm{k}_{1}\) and \(\mathrm{k}_{2}\) are closed.
\(\mathrm{KVL} \rightarrow\)
\(\left(\frac{\mathrm{iR}}{25}+\frac{\mathrm{i}}{5}\right) 220+\frac{\mathrm{i}}{5} \mathrm{R}=\mathrm{V}\)
\(\left(\frac{\mathrm{i} \mathrm{R}}{25}+\frac{\mathrm{i}}{5}\right) 220+\frac{\mathrm{i}}{5} \mathrm{R}=\mathrm{i}(220+\mathrm{R}) \quad\) From eq \((i)\)
\(\Rightarrow \quad R=22 \,\Omega\)
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