A galvanometer of 100 Ω resistance can give full scale deflection for Ig = 1 mA. Find the value of shunt resistance Rs to get the 5 mA range ammeter.

Correct Option: 25 Ω

Understanding the Circuit from the Diagram:
As shown in the provided circuit diagram, to convert a galvanometer into an ammeter capable of measuring larger currents, a small resistance called a shunt resistance ($R_s$) is connected in parallel with the galvanometer ($G$).
The main current entering the system is labeled $I$. A small fraction of this current, $I_g$, flows through the galvanometer, while the remaining current ($I-I_g$) passes through the parallel shunt resistor $R_s$.

Given Data:
Resistance of the galvanometer ($G$) = $100\Omega$
Full-scale deflection current of the galvanometer ($I_g$) = $1\mathrm{mA}$
Total range of the converted ammeter ($I$) = $5\mathrm{mA}$

Formula and Derivation:
Since the galvanometer and the shunt resistor are connected in parallel, the potential difference across both components must be equal:

$V_{\text{galvanometer}}=V_{\text{shunt}}$

$I_g\times G=(I-I_g)\times R_s$

Rearranging the equation to solve for the shunt resistance ($R_s$):

$R_s=\frac{I_g\times G}{I-I_g}$

Calculation:
Substitute the given values into the formula:

$R_s=\frac{1\mathrm{mA}\times 100\Omega }{5\mathrm{mA}-1\mathrm{mA}}$

$R_s=\frac{1\times 100}{4}$

$R_s=25\Omega$

Therefore, the value of the shunt resistance required is 25 Ω.