Q162 / 180 · COMEDK 2020 · Physics
If $P=Q=R=10 \Omega$ and $S=20 \Omega$, then what resistance should be joined with $S$ to balance the Wheatstone's network?
- AJoin a resistance of $10 \Omega$ in series with $S$
- BJoin a resistance of $10 \Omega$ in parallel with $S$
- Correct answerCJoin a resistance of $20 \Omega$ in parallel with $S$
- DJoin a resistance of $20 \Omega$ in series with $S$
Solution
The given Wheatstone bridge eircuit is shown belin.
Wheatstune is bulanced only when $$ \frac{P}{Q}=\frac{k}{\mathrm{~S}} $$ Thus, the value of resistance $S$ should be 10 $\Omega$. Hut it is given to $20 \Omega$. Hence, to decresese its vulue upto $10 \Omega$, suppose $x \Omega$ tesistunce is comnected pantlel to it. $\therefore \quad \frac{1}{10}=\frac{1}{20}+\frac{1}{x}$ $\Rightarrow \quad \frac{1}{x}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$ $\Rightarrow \quad x=20 \Omega$
Wheatstune is bulanced only when $$ \frac{P}{Q}=\frac{k}{\mathrm{~S}} $$ Thus, the value of resistance $S$ should be 10 $\Omega$. Hut it is given to $20 \Omega$. Hence, to decresese its vulue upto $10 \Omega$, suppose $x \Omega$ tesistunce is comnected pantlel to it. $\therefore \quad \frac{1}{10}=\frac{1}{20}+\frac{1}{x}$ $\Rightarrow \quad \frac{1}{x}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$ $\Rightarrow \quad x=20 \Omega$
Hence, the correct answer is option: 'C'.