Question 142: Current Electricity

Question

A galvanometer has a coil of resistance $$200 \Omega$$ with a full scale deflection at $$20 \mu \mathrm{A}$$. The value of resistance to be added to use it as an ammeter of range $$(0-20) \mathrm{mA}$$ is :

Accepted Answer

To convert a galvanometer into an ammeter, we need to add a shunt resistance in parallel with the galvanometer's coil. The purpose of the shunt resistance is to bypass the majority of the current while allowing only a small fraction of it to pass through the galvanometer, thereby preventing it from being damaged by high currents.
Given parameters:
Resistance of the galvanometer's coil, $$R_g = 200 \Omega$$
Full-scale deflection current of the galvanometer, $$I_g = 20 \mu \mathrm{A} = 20 	imes 10^{-6} \mathrm{A}$$
Desired ammeter range, $$I = 20 \mathrm{mA} = 20 	imes 10^{-3} \mathrm{A}$$
The shunt resistance, $$R_s$$, can be calculated using the formula:
$$ \frac{R_s}{R_g + R_s} = \frac{I_g}{I} $$
Solving for $$R_s$$:
$$ R_s = R_g \left(\frac{I_g}{I - I_g}ight) $$
Substituting the given values:
$$ R_s = 200 \left(\frac{20 	imes 10^{-6}}{20 	imes 10^{-3} - 20 	imes 10^{-6}}ight) $$
Simplifying the expression:
$$ R_s = 200 \left(\frac{20 	imes 10^{-6}}{19.98 	imes 10^{-3}}ight) $$
$$ R_s \approx 200 \left(\frac{20 	imes 10^{-6}}{20 	imes 10^{-3}}ight) = 200 \left(\frac{1}{1000}ight) = 0.20 \Omega $$
Therefore, the value of the resistance to be added to use the galvanometer as an ammeter of range $$(0-20) \mathrm{mA}$$ is $$0.20 \Omega$$. Thus, the correct answer is:
Option C: $$0.20 \Omega$$

Answer

$$0.40\Omega$$

Answer

$$0.10\Omega$$

Answer

$$0.20\Omega$$

Answer

$$0.50\Omega$$

Current Electricity

A galvanometer has a coil of resistance $200 \Omega with a full scale deflection at 20 \mu \mathrm{A}. The value of resistance to be added to use it as an ammeter of range (0-20) \mathrm{mA}$ is :

Choose the correct answer: