KCET · Physics · Current Electricity
Each resistance in the given cubical network has resistance of \(1 \Omega\) and equivalent resistance between \(A\) and \(B\) is
- A \(\frac{5}{6} \Omega\)
- B \(\frac{6}{5} \Omega\)
- C \(\frac{5}{12} \Omega\)
- D \(\frac{12}{5} \Omega\)
Answer & Solution
Correct Answer
(A) \(\frac{5}{6} \Omega\)
Step-by-step Solution
Detailed explanation
The given network of resistances can be shown as follows,
Let \(6 I\) be the current through the cell, which is distributed as shown in the figure.
Applying Kirchhoff's law to the loop \(A B C C^{\prime} E A\), we have
\(\therefore \quad E=5 I\)
If \(R^{\prime}\) be the equivalent resistance of the network, then
\(R^{\prime}=\frac{\text { total emf }}{\text { total current }}=\frac{5 I}{6 I}=\frac{5}{6} \Omega\)
Let \(6 I\) be the current through the cell, which is distributed as shown in the figure.
Applying Kirchhoff's law to the loop \(A B C C^{\prime} E A\), we have
\(\therefore \quad E=5 I\)
If \(R^{\prime}\) be the equivalent resistance of the network, then
\(R^{\prime}=\frac{\text { total emf }}{\text { total current }}=\frac{5 I}{6 I}=\frac{5}{6} \Omega\)
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