The resistance of the four arms `P, Q, R` and `S` in a Wheatstone's bridge are `10 ohm 30 ohm` and `90 ohm` rerspectively. 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 for the cell will be

Options

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

Explanation

To solve the problem step by step, we will analyze the Wheatstone bridge and calculate the current drawn from the cell.

### Step 1: Identify the resistances in the Wheatstone bridge
The resistances are given as follows:
- \( P = 10 \, \Omega \)
- \( Q = 30 \, \Omega \)
- \( R = 30 \, \Omega \)
- \( S = 90 \, \Omega \)

### Step 2: Check if the Wheatstone bridge is balanced
The condition for a balanced Wheatstone bridge is:
\[ \frac{P}{Q} = \frac{R}{S} \]
Substituting the values:
\[ \frac{10}{30} = \frac{30}{90} \]
Both sides simplify to \( \frac{1}{3} \), confirming that the bridge is balanced.

### Step 3: Determine the equivalent resistance of the circuit
Since the bridge is balanced, no current flows through the galvanometer. We can simplify the circuit to just the resistances \( P, Q, R, \) and \( S \).

1. Calculate the equivalent resistance of \( P \) and \( Q \) in series:
\[ R_{PQ} = P + Q = 10 + 30 = 40 \, \Omega \]

2. Calculate the equivalent resistance of \( R \) and \( S \) in series:
\[ R_{RS} = R + S = 30 + 90 = 120 \, \Omega \]

3. Now, find the equivalent resistance of \( R_{PQ} \) and \( R_{RS} \) in parallel:
\[ \frac{1}{R_{eq}} = \frac{1}{R_{PQ}} + \frac{1}{R_{RS}} = \frac{1}{40} + \frac{1}{120} \]
Finding a common denominator (120):
\[ \frac{1}{R_{eq}} = \frac{3}{120} + \frac{1}{120} = \frac{4}{120} = \frac{1}{30} \]
Therefore,
\[ R_{eq} = 30 \, \Omega \]

### Step 4: Calculate the total resistance in the circuit
The total resistance in the circuit includes the internal resistance of the cell:
\[ R_{total} = R_{eq} + r = 30 + 5 = 35 \, \Omega \]
where \( r = 5 \, \Omega \) (internal resistance of the cell).

### Step 5: Calculate the current drawn from the cell using Ohm's Law
Using Ohm's Law:
\[ I = \frac{V}{R_{total}} \]
where \( V = 7 \, V \) (emf of the cell). Thus,
\[ I = \frac{7}{35} = 0.2 \, A \]

### Conclusion
The current drawn from the cell is \( 0.2 \, A \).

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