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GUJCET · Physics · Current Electricity
The equivalent resistance between A and B in the given circuit is ________.
- A \(3 \Omega\)
- B \(6 \Omega\)
- C \(12 \Omega\)
- D \(1.5 \Omega\)
Answer & Solution
Correct Answer
(D) \(1.5 \Omega\)
Step-by-step Solution
Detailed explanation
(D)
As per balanced condition of Wheatstone bridge.
\(\frac{P}{Q}=\frac{R}{S}\)
\(\therefore\) removing resistance from CD
\(\begin{aligned} R _{ ACB } & =3+3=6 \Omega \\ R _{ ADB } & =3+3=6 \Omega \\ R _{ AB } & =3 \Omega\end{aligned}\)
\(ACB , ADB\) and AB are in parallel.
\( \frac{1}{R_P} =\frac{1}{6}+\frac{1}{6}+\frac{1}{3} \\ \frac{1}{R_P} =\frac{1+1+2}{6} \\ \frac{1}{R_P} =\frac{4}{6} \\ R_P =\frac{6}{4}=\frac{3}{2}=1.5 \Omega\)
As per balanced condition of Wheatstone bridge.
\(\frac{P}{Q}=\frac{R}{S}\)
\(\therefore\) removing resistance from CD
\(\begin{aligned} R _{ ACB } & =3+3=6 \Omega \\ R _{ ADB } & =3+3=6 \Omega \\ R _{ AB } & =3 \Omega\end{aligned}\)
\(ACB , ADB\) and AB are in parallel.
\( \frac{1}{R_P} =\frac{1}{6}+\frac{1}{6}+\frac{1}{3} \\ \frac{1}{R_P} =\frac{1+1+2}{6} \\ \frac{1}{R_P} =\frac{4}{6} \\ R_P =\frac{6}{4}=\frac{3}{2}=1.5 \Omega\)
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