Question

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

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

Answer 1

The correct option is C $20\Omega$


Since, $R$ resistor is connected in parallel with the $20\Omega$ resistor, so effective resistance will be

$R_{eq}=\frac{20R}{20+R}=x\left(\text{let}\right)$

The potential difference between $\text{B}$ and $\text{B}$ will be zero only when bridge is balanced.

So, let’s balance Wheatstone bridge,

$\frac{20}{x}=\frac{10}{5}$

$\Rightarrow x=10\Omega$

$\Rightarrow \frac{20R}{20+R}=10\Rightarrow 10R=200$

$\therefore R=20\Omega$

Hence, option $\left(a\right)$ is correct.