Q146 / 180 · KCET 2015 · Physics
In Wheatstones network \( P=2 \Omega, Q=2 \Omega, R=2 \Omega \) and \( S=3 \Omega \). The resistance with which \( S \) is
to shunted in order that the bridge may be balanced is
to shunted in order that the bridge may be balanced is
- A\( 1 \Omega \)
- B- \( 2 \Omega \)
- C\( 4 \Omega \)
- Correct answerD\( 0 \Omega \)
Solution
Given, \( P=2 \Omega ; Q=2 \Omega ; R=2 \Omega ; S=3 \Omega \)
Balance equation gives
\[ \frac{P}{Q}=\frac{R}{S} \]
Suppose the resistance \( S \) is shunted with resistance \( X \), then balance equation becomes
\[ \frac{P}{Q}=\frac{P}{\left(\frac{S X}{S+X}\right)} \]
Substitute the values, we get
\[ \begin{array}{l} \frac{2}{2}=\frac{2}{\left(\frac{3 X}{3+X}\right)} \Rightarrow \frac{3 X}{3+X}=2 \Rightarrow 3 X=2(3+X) \\ \Rightarrow 3 X=6+2 X \Rightarrow X=6 \Omega \end{array} \]
Therefore, \( \mathrm{S} \) is to be shunted with \( 6 \Omega \) resistance in order tobalance the bridge
Balance equation gives
\[ \frac{P}{Q}=\frac{R}{S} \]
Suppose the resistance \( S \) is shunted with resistance \( X \), then balance equation becomes
\[ \frac{P}{Q}=\frac{P}{\left(\frac{S X}{S+X}\right)} \]
Substitute the values, we get
\[ \begin{array}{l} \frac{2}{2}=\frac{2}{\left(\frac{3 X}{3+X}\right)} \Rightarrow \frac{3 X}{3+X}=2 \Rightarrow 3 X=2(3+X) \\ \Rightarrow 3 X=6+2 X \Rightarrow X=6 \Omega \end{array} \]
Therefore, \( \mathrm{S} \) is to be shunted with \( 6 \Omega \) resistance in order tobalance the bridge
Hence, the correct answer is option: 'D'.