KCET · Physics · Current Electricity
In the circuit shown, the end \(A\) is at potential \(V_0\) and end \(B\) is grounded. The electric current \(I\) indicated in the circuit is
- A \(V_0 / R\)
- B \(2 V_0 / R\)
- C \(3 V_0 / R\)
- D \(V_0 / 3 R\)
Answer & Solution
Correct Answer
(D) \(V_0 / 3 R\)
Step-by-step Solution
Detailed explanation
1st and 2nd networks form balanced Wheatstone bridge.
Hence, circuit diagram can be redrawn as
\(\therefore R_{A R}=2 R\|2 R+4 R\| 4 R\)
\(=\frac{2 R \times 2 R}{2 R+2 R}+\frac{4 R \times 4 R}{4 R+4 R}=R+2 R=3 R\)
\(\therefore \quad I=\frac{V_0}{R_{A B}}=\frac{V_0}{3 R}\)
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