"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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"

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:

$$ 0.02 \times 30 = 2.98 \times R_s \quad \Rightarrow \quad R_s = \frac{0.6}{2.98} \approx 0.2013 \, \Omega. $$

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 $149.$

"

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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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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