Four resistors, \( 5\,\Omega \) (between A and B), \( 10\,\Omega \) (between B and C), \( 10\,\Omega \) (between A and D) and \( 20\,\Omega \) (between D and C), form a Wheatstone bridge, and a \( 10\,\Omega \) resistor is connected between B and D. A 5 V battery of negligible internal resistance is connected between A and C. The current drawn from the battery is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.5 A
A Wheatstone bridge is balanced when the ratio of the resistances in the two arms on one side equals the ratio on the other side. Then B and D are at the same potential, and no current flows through the resistor joining them.
The two ratios are:
\( \begin{aligned} \frac{R_{AB}}{R_{BC}} &= \frac{5\,\Omega}{10\,\Omega} \\ &= \frac{1}{2} \end{aligned} \)
\( \begin{aligned} \frac{R_{AD}}{R_{DC}} &= \frac{10\,\Omega}{20\,\Omega} \\ &= \frac{1}{2} \end{aligned} \)
From the above, the two ratios are equal. Hence, the bridge is balanced. The \( 10\,\Omega \) resistor between B and D carries no current, so it can be removed from the circuit.
The circuit is then two branches in parallel between A and C, each a series pair:
\( \begin{aligned} R_{ABC} &= R_{AB}+R_{BC} \\ &= 5\,\Omega+10\,\Omega \\ &= 15\,\Omega \end{aligned} \)
\( \begin{aligned} R_{ADC} &= R_{AD}+R_{DC} \\ &= 10\,\Omega+20\,\Omega \\ &= 30\,\Omega \end{aligned} \)
The equivalent resistance of the two branches in parallel is:
\( \begin{aligned} R_{\mathrm{eq}} &= \frac{R_{ABC}\,R_{ADC}}{R_{ABC}+R_{ADC}} \\ &= \frac{15\,\Omega\times30\,\Omega}{15\,\Omega+30\,\Omega} \\ &= 10\,\Omega \end{aligned} \)
The current drawn from the battery is:
\( \begin{aligned} I &= \frac{V}{R_{\mathrm{eq}}} \\ &= \frac{5\,\mathrm{V}}{10\,\Omega} \\ &= 0.5\,\mathrm{A} \end{aligned} \)