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Step-by-Step Solution
Step 1: Verify if the Wheatstone Bridge is Balanced
The condition for a Wheatstone bridge to be balanced is that the ratio of the resistances in the opposite arms must be equal. Here, the four arms have resistances:
P = 10 Ω
Q = 30 Ω
R = 30 Ω
S = 90 Ω
Check the ratios:
Ratio 1: P/Q = 10/30 = 1/3
Ratio 2: R/S = 30/90 = 1/3
Since these two ratios are equal, the Wheatstone bridge is balanced. Hence, no current flows through the galvanometer (50 Ω). We can ignore the galvanometer's resistance when computing the total current drawn from the cell.
Step 2: Compute the Equivalent Resistance of the Balanced Bridge
For a balanced Wheatstone bridge, the equivalent resistance of the external circuit is given by the two series resistances in parallel. That is:
One branch: P + Q = 10 + 30 = 40 Ω
The other branch: R + S = 30 + 90 = 120 Ω
These two branches (40 Ω and 120 Ω) are in parallel, so their combined resistance $R_{\text{ext}}$ is:
$R_{\text{ext}} = \frac{40 \times 120}{40 + 120} = \frac{4800}{160} = 30 \, \Omega$
Step 3: Find the Total Circuit Resistance
The cell has an internal resistance $r = 5 \, \Omega$. Hence, the total resistance in the circuit is the sum of the equivalent external resistance and the internal resistance:
$R_{\text{total}} = 30 \, \Omega + 5 \, \Omega = 35 \, \Omega
Step 4: Calculate the Current Drawn from the Cell
The e.m.f. of the cell is $E = 7\, \text{V}$. Therefore, by Ohm’s law, the total current $I$ drawn from the cell is:
$I = \frac{E}{R_{\text{total}}} = \frac{7 \, \text{V}}{35 \, \Omega} = 0.2 \, \text{A}
Hence, the current drawn from the cell is 0.2 A.
