Question 136: Current Electricity

Question

A galvanometer having a coil of resistance $30 \Omega$ need 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be $\frac{30}{X} \Omega$, where $X$ is

Accepted Answer

Given:
Galvanometer resistance:
$$ R_g = 30 \, \Omega $$
Full-scale deflection current:
$$ I_g = 20 \, mA = 0.02 \, A $$
Maximum current to be measured:
$$ I = 3 \, A $$
When using the galvanometer to measure 3 A, the extra current passing through the shunt resistor ($R_s$) is:
$$ I_s = I - I_g = 3 - 0.02 = 2.98 \, A. $$
Since the galvanometer and the shunt resistor are connected in parallel, their voltage drops must be equal. Therefore:
$$ I_g R_g = I_s R_s. $$
Substitute the given values:
$$ 0.02 	imes 30 = 2.98 	imes R_s \quad \Rightarrow \quad R_s = \frac{0.6}{2.98} \approx 0.2013 \, \Omega. $$
The shunt resistance is represented as:
$$ R_s = \frac{30}{X} \, \Omega. $$
Equate the two expressions:
$$ \frac{30}{X} = 0.2013. $$
Solving for $X$:
$$ X = \frac{30}{0.2013} \approx 149. $$
Thus, the value of $X$ is approximately $$149.$$

Answer

447

Answer

298

Answer

149

Answer

596

Current Electricity

A galvanometer having a coil of resistance 30 \Omega need 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be \frac{30}{X} \Omega, where X is

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