JEE MAIN - Physics (2023 - 13th April Morning Shift - No. 20)
When a resistance of $$5 ~\Omega$$ is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of $$250 \mathrm{~mA}$$, however when $$1050 ~\Omega$$ resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is ____________ $$\Omega$$.
Answer
50
Explanation
Given:
$$\frac{250 \ \text{mA} \times 5}{5 + R_G} = i$$
$$i = \frac{25}{1050 + R_G}$$
Equating the two expressions for current, $$i$$:
$$\frac{250 \ \text{mA} \times 5}{5 + R_G} = \frac{25}{1050 + R_G}$$
This equation simplifies to:
$$100(5 + R_G) = 1050 \times 5 + R_G \times 5$$
Solving for the resistance of the galvanometer, $$R_G$$:
$$95 R_G = 4750$$
$$R_G = 50 \ \Omega$$
So, the resistance of the galvanometer is $$50 \ \Omega$$.
$$\frac{250 \ \text{mA} \times 5}{5 + R_G} = i$$
$$i = \frac{25}{1050 + R_G}$$
Equating the two expressions for current, $$i$$:
$$\frac{250 \ \text{mA} \times 5}{5 + R_G} = \frac{25}{1050 + R_G}$$
This equation simplifies to:
$$100(5 + R_G) = 1050 \times 5 + R_G \times 5$$
Solving for the resistance of the galvanometer, $$R_G$$:
$$95 R_G = 4750$$
$$R_G = 50 \ \Omega$$
So, the resistance of the galvanometer is $$50 \ \Omega$$.
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