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A non-zero current passes through the galvanometer $G$ shown in the circuit when the key $K$ is closed and its value does not change when the key is opened. Then, which of the following statement(s) is/are true?
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Verified Answer
The correct answers are:
The current through the galvanometer is $40 \mathrm{mA}$., After the key is closed, the current through the $200 \Omega$ resistor is same as the current through the $300 \Omega$ resistor., The galvanometer resistance is 150$\Omega$
$\therefore I_{C}=\frac{10}{100+G}$
$300 I_{3}=\frac{10}{100+G} G$
$I_{3}=\frac{G}{30}\left[\frac{1}{100+G}\right]$
$\therefore \quad I_{3}+I_{G}=\frac{1}{(100+G)}\left[10+\frac{G}{30}\right]=\frac{(300+G)}{(100+G) \times 30}$
According to the problem, $10=\left(\frac{200}{3}+\frac{300 G}{300+G}\right)\left[\frac{(300+G)}{(100+6) 30}\right]$
$Y=R C$
$10=\frac{60000+1100 G}{3} \times \frac{1}{100+G} \times \frac{1}{30}$
$\Rightarrow 900(100+G)=60000+1100 G$
$\Rightarrow \quad 30000=200 G$
$\therefore$
$G=150 \Omega$
$I_{G}=\frac{10}{100+150}=\frac{10}{250}=40 \mathrm{mA}$
As $\quad \frac{200}{100}=\frac{300}{150}$
So, $\quad I_{200}=I_{300}$
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