In a Wheatstone bridge, the resistances \( R_1, R_2, R_3 \) and \( R_4 \) present in four arms are \( 10\,\Omega \), \( 30\,\Omega \), \( 30\,\Omega \) and \( 90\,\Omega \) respectively. If the galvanometer resistance is \( 50\,\Omega \) and the voltage of the battery is 7 V and its internal resistance is \( 5\,\Omega \), then the total current drawn from the battery is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.2 A
From the figure, \( R_2 \) (A to B) and \( R_4 \) (B to C) form the upper path, \( R_1 \) (A to D) and \( R_3 \) (D to C) the lower path, and the galvanometer joins B and D. A Wheatstone bridge is balanced when the ratios of the arms on the two sides are equal.
\( \displaystyle \frac{R_2}{R_4} = \frac{30}{90} = \frac{1}{3} \qquad \frac{R_1}{R_3} = \frac{10}{30} = \frac{1}{3} \)
The bridge is balanced, so B and D are at the same potential and no current flows through the galvanometer; its \( 50\,\Omega \) plays no part. The two paths are then in parallel.
\( R_{ABC} = 30 + 90 = 120\,\Omega \)
\( R_{ADC} = 10 + 30 = 40\,\Omega \)
\( \displaystyle R_{AC} = \frac{120 \times 40}{120 + 40} = 30\,\Omega \)
Adding the internal resistance of the battery,
\( \displaystyle I = \frac{E}{R_{AC} + r} = \frac{7}{30 + 5} = 0.2\,\mathrm{A} \)