Question

A galvanometer having variable shunt S used to measure current when connected with a series resistance 90 $\Omega$ and a battery of internal resistance 10 $\Omega$. It is observed that deflection in galvanometer are 9 divisions and 30 divisions for shunt resistances 10 $\Omega$ and 50 $\Omega$. The total divisions on the galvanometer is 50 and full scale deflection of the meter is 200 mA, then emf of battery (in volt) is

Answer 1

96

Answer 2

The problem involves a battery connected in series with an external resistance and a galvanometer with a shunt. We need to find the EMF of the battery using the given deflection data.

1. Calculate the current per division for the galvanometer: The full-scale deflection (FSD) current is $I_{fsd} = 200 \text{ mA} = 0.2 \text{ A}$. The total number of divisions on the galvanometer is 50. The current per division ($k$) is: $k = \frac{I_{fsd}}{\text{Total divisions}} = \frac{0.2 \text{ A}}{50 \text{ divisions}} = 0.004 \text{ A/division}$

2. Analyze the circuit: Let $E$ be the EMF of the battery and $r$ be its internal resistance ($r = 10 \Omega$). The external series resistance is $R = 90 \Omega$. The galvanometer has a resistance $G$. The shunt resistance is $S$.

The galvanometer and shunt are connected in parallel. Their equivalent resistance is $R_{GS} = \frac{GS}{G+S}$. The total external resistance in the circuit is $R_{ext} = R + R_{GS} = 90 + \frac{GS}{G+S}$. The total resistance of the circuit is $R_{total} = R_{ext} + r = 90 + \frac{GS}{G+S} + 10 = 100 + \frac{GS}{G+S}$. The total current flowing from the battery is $I_{total} = \frac{E}{R_{total}} = \frac{E}{100 + \frac{GS}{G+S}}$.

The total current $I_{total}$ splits between the galvanometer and the shunt. The current through the galvanometer ($I_g$) is given by: $I_g = I_{total} \times \frac{S}{G+S}$ Substitute $I_{total}$: $I_g = \frac{E}{100 + \frac{GS}{G+S}} \times \frac{S}{G+S} = \frac{ES}{(100 + \frac{GS}{G+S})(G+S)}$ $I_g = \frac{ES}{100(G+S) + GS}$

3. Set up equations for the two scenarios:

Scenario 1: Shunt resistance $S_1 = 10 \Omega$. Deflection $N_1 = 9$ divisions. Current through galvanometer $I_{g1} = N_1 \times k = 9 \times 0.004 \text{ A} = 0.036 \text{ A}$.

Using the $I_g$ formula: $0.036 = \frac{E \times 10}{100(G+10) + G \times 10}$ $0.036 = \frac{10E}{100G + 1000 + 10G}$ $0.036 = \frac{10E}{110G + 1000}$ $E = \frac{0.036(110G + 1000)}{10} = 0.0036(110G + 1000)$ $E = 0.396G + 3.6$ (Equation 1)

Scenario 2: Shunt resistance $S_2 = 50 \Omega$. Deflection $N_2 = 30$ divisions. Current through galvanometer $I_{g2} = N_2 \times k = 30 \times 0.004 \text{ A} = 0.120 \text{ A}$.

Using the $I_g$ formula: $0.120 = \frac{E \times 50}{100(G+50) + G \times 50}$ $0.120 = \frac{50E}{100G + 5000 + 50G}$ $0.120 = \frac{50E}{150G + 5000}$ $E = \frac{0.120(150G + 5000)}{50} = 0.0024(150G + 5000)$ $E = 0.36G + 12$ (Equation 2)

4. Solve the system of equations for G and E: Equate Equation 1 and Equation 2: $0.396G + 3.6 = 0.36G + 12$ $0.396G - 0.36G = 12 - 3.6$ $0.036G = 8.4$ $G = \frac{8.4}{0.036} = \frac{8400}{36} = \frac{700}{3} \Omega$

Substitute the value of $G$ into Equation 2 (or Equation 1) to find $E$: $E = 0.36 \left(\frac{700}{3}\right) + 12$ $E = (0.12 \times 700) + 12$ $E = 84 + 12$ $E = 96 \text{ V}$

The EMF of the battery is 96 V.