KCET · Physics · Current Electricity
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
- A \( 1 \Omega \)
- B - \( 2 \Omega \)
- C \( 4 \Omega \)
- D \( 0 \Omega \)
Answer & Solution
Correct Answer
(D) \( 0 \Omega \)
Step-by-step Solution
Detailed explanation
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
\(\frac{2}{2}=\frac{2}{\left(\frac{3 X}{3+X}\right)} \Rightarrow \frac{3 X}{3+X}=2 \Rightarrow 3 X=2(3+X) \)
\(\Rightarrow 3 X=6+2 X \Rightarrow X=6 \Omega\)
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
\(\frac{2}{2}=\frac{2}{\left(\frac{3 X}{3+X}\right)} \Rightarrow \frac{3 X}{3+X}=2 \Rightarrow 3 X=2(3+X) \)
\(\Rightarrow 3 X=6+2 X \Rightarrow X=6 \Omega\)
Therefore, \( \mathrm{S} \) is to be shunted with \( 6 \Omega \) resistance in order tobalance the bridge
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