Question

The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be

• A. 1.0 A
• B. 0.2 A
• C. 0.1 A
• D. 2.0 A

Answer 1

The Correct Option is B

 The given problem involves solving for the current drawn from a cell in a Wheatstone's bridge circuit, which consists of four resistors and additional components like a cell and a galvanometer. The resistances for the arms of the bridge are given as follows:

• P = 10 Ω
• Q = 30 Ω
• R = 30 Ω
• S = 90 Ω

The electromotive force (EMF) and internal resistance of the cell are given as 7 V and 5 Ω, respectively. The resistance of the galvanometer is 50 Ω. We need to determine the current drawn from the cell.

In a Wheatstone bridge, when no current flows through the galvanometer, the condition for balance is:

\[\frac{P}{Q} = \frac{R}{S}\]

Substituting the given values:

\[\frac{10}{30} \neq \frac{30}{90}\]

This shows that the bridge is not balanced. Therefore, current will flow through the galvanometer, and we cannot apply the balance condition.

To find the total resistance in the circuit, we calculate the equivalent resistance of the Wheatstone bridge, because both arms P and Q are parallel to arms R and S:

Effective resistance of P and Q:

\[R_{PQ} = \frac{P \times Q}{P + Q} = \frac{10 \times 30}{10 + 30} = \frac{300}{40} = 7.5 \, \Omega\]

Effective resistance of R and S:

\[R_{RS} = \frac{R \times S}{R + S} = \frac{30 \times 90}{30 + 90} = \frac{2700}{120} = 22.5 \, \Omega\]

The total resistance in the circuit when these two parts are connected in series and including the internal resistance (R_int) and the galvanometer resistance (R_g) is:

\[R_{total} = R_{PQ} + R_{RS} + R_{int} = 7.5 + 22.5 + 5 = 35 \, \Omega\]

The total resistance with the galvanometer can be ignored in series as it will only affect the division in the arms, as we have full current calculated with series portion of P&Q and (R&S), and they would divide accordingly.

Finally, using Ohm's Law, we calculate the current drawn from the cell:

\[I = \frac{\text{EMF}}{R_{total}} = \frac{7}{35} = 0.2 \, \text{A}\]

Therefore, the current drawn from the cell is 0.2 A, which matches the correct answer.