KCET · Physics · Current Electricity
In the Wheatstone's network given, \(P=10 \Omega\), \(Q=20 \Omega, R=15 \Omega, S=30 \Omega\) the current have passing through the battery (of negligible internal resistance) is
- A \(0.36 \mathrm{~A}\)
- B zero
- C \(0.18 \mathrm{~A}\)
- D \(0.72 \mathrm{~A}\)
Answer & Solution
Correct Answer
(A) \(0.36 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
The balanced condition for Wheatstones bridge is
\(\frac{P}{Q}=\frac{R}{S}\)
as is obvious from the given values.
No, current flows through galvanometer is zero.
Now, \(P\) and \(R\) are in series, so
Resistance \(R_{1}=P+R=10+15=25 \Omega\)
Similarly, \(Q\) and \(S\) are in series, so
Resistance \(R_{2}=R+S=20+30=50 \Omega\)
Net resistance of the network as \(R_{1}\) and \(R_{2}\) are in parallel
\(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\)
\(\therefore R=\frac{25 \times 50}{25+50}=\frac{50}{3} \Omega \)
\(\text {Hence, } I=\frac{V}{R}=\frac{6}{50}=0.36 \mathrm{~A}\)
\(\frac{P}{Q}=\frac{R}{S}\)
as is obvious from the given values.
No, current flows through galvanometer is zero.
Now, \(P\) and \(R\) are in series, so
Resistance \(R_{1}=P+R=10+15=25 \Omega\)
Similarly, \(Q\) and \(S\) are in series, so
Resistance \(R_{2}=R+S=20+30=50 \Omega\)
Net resistance of the network as \(R_{1}\) and \(R_{2}\) are in parallel
\(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\)
\(\therefore R=\frac{25 \times 50}{25+50}=\frac{50}{3} \Omega \)
\(\text {Hence, } I=\frac{V}{R}=\frac{6}{50}=0.36 \mathrm{~A}\)
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