Question:

In the given figure \( R_1 = 10 \, \Omega \), \( R_2 = 8 \, \Omega \), \( R_3 = 4 \, \Omega \), and \( R_4 = 8 \, \Omega \). The battery is ideal with an EMF of 12V.
The equivalent resistance of the circuit and the current supplied by the battery are respectively:
Circuit

Updated On: Nov 21, 2024
  • \( 12 \, \Omega \) and \( 11.4 \, A \)
  • \( 10.5 \, \Omega \) and \( 1.14 \, A \)
  • \( 10.5 \, \Omega \) and \( 1 \, A \)
  • \( 12 \, \Omega \) and \( 1 \, A \)
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The Correct Option is D

Solution and Explanation

Here, \(R_2\), \(R_3\), and \(R_4\) are in parallel. The equivalent resistance of these resistors is given by:

\[ \frac{1}{R_{234}} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{8} + \frac{1}{4} + \frac{1}{8} \]

\[ R_{234} = 2 \, \Omega \]

This resistance is in series with \(R_1\), so the total equivalent resistance is:

\[ R_{\text{total}} = R_1 + R_{234} = 10 \, \Omega + 2 \, \Omega = 12 \, \Omega \]

The current supplied by the battery is:

\[ I = \frac{V}{R_{\text{total}}} = \frac{12 \, V}{12 \, \Omega} = 1 \, \text{A} \]

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