Question

What resistor should be connected in parallel with the $20\Omega$ resistor in branch $ADC$ in the circuit shown in Figure so that potential difference between $B$ and $D$ may be zero?

• A. $20\Omega$
• B. $10\Omega$
• C. $5\Omega$
• D. $15\Omega$

Answer 1

The correct option is A $20\Omega$

lets $R_1$ is kept in parallel and R is equivalent resistance between A and D in case of potential difference between B and D is zero, it is a balanced wheetstone bridge and -

$\frac{R_{AB}}{R_{AD}}=\frac{R_{BC}}{R_{DC}}\frac{20}{R}=\frac{10}{5}R=1020\parallel R_1=10\frac{20\times R_1}{20+R_1}=10R_1=20\Omega$