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 V` 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`

Answer

To find the current drawn from the cell in the given Wheatstone bridge circuit, we will follow these steps:

### Step 1: Identify the resistances in the Wheatstone bridge
The resistances of the four arms are given as:
- P = 10 ohm
- Q = 30 ohm
- R = 30 ohm
- S = 90 ohm

### Step 2: Check if the Wheatstone bridge is balanced
For a Wheatstone bridge to be balanced, the following condition must hold:
\[ \frac{P}{Q} = \frac{R}{S} \]
Substituting the values:
\[ \frac{10}{30} = \frac{30}{90} \]
Simplifying both sides:
\[ \frac{1}{3} = \frac{1}{3} \]
Since both sides are equal, the bridge is balanced.

### Step 3: Determine the equivalent resistance of the circuit
When the Wheatstone bridge is balanced, the current through the galvanometer is zero. Therefore, we can treat the two pairs of resistors as being in series.

- The equivalent resistance of P and Q (in series) is:
\[ R_{PQ} = P + Q = 10 + 30 = 40 \, \text{ohm} \]

- The equivalent resistance of R and S (in series) is:
\[ R_{RS} = R + S = 30 + 90 = 120 \, \text{ohm} \]

### Step 4: Calculate the total equivalent resistance of the circuit
The equivalent resistances \( R_{PQ} \) and \( R_{RS} \) are in parallel:
\[ \frac{1}{R_{eq}} = \frac{1}{R_{PQ}} + \frac{1}{R_{RS}} \]
Substituting the values:
\[ \frac{1}{R_{eq}} = \frac{1}{40} + \frac{1}{120} \]
Finding a common denominator (120):
\[ \frac{1}{R_{eq}} = \frac{3}{120} + \frac{1}{120} = \frac{4}{120} = \frac{1}{30} \]
Thus, the equivalent resistance \( R_{eq} \) is:
\[ R_{eq} = 30 \, \text{ohm} \]

### Step 5: Add the internal resistance of the cell
The total resistance in the circuit is the equivalent resistance plus the internal resistance of the cell:
\[ R_{total} = R_{eq} + R_{internal} = 30 + 5 = 35 \, \text{ohm} \]

### Step 6: Calculate the current drawn from the cell
Using Ohm's law, the current \( I \) drawn from the cell can be calculated as:
\[ I = \frac{V}{R_{total}} = \frac{7 \, \text{V}}{35 \, \text{ohm}} = 0.2 \, \text{A} \]

### Final Answer
The current drawn from the cell is \( 0.2 \, \text{A} \).

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