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:

• A. \( 12 \, \Omega \) and \( 11.4 \, A \)
• B. \( 10.5 \, \Omega \) and \( 1.14 \, A \)
• C. \( 10.5 \, \Omega \) and \( 1 \, A \)
• D. \( 12 \, \Omega \) and \( 1 \, A \)

Answer 1

The Correct Option is D

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:

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

Answer 2

Step 1: Identify series/parallel groups
The right network has three resistors between the same top and bottom nodes: R₂ = 8 Ω (vertical), R₃ = 4 Ω (right), and R₄ = 8 Ω (diagonal). Hence R₂, R₃, R₄ are in parallel.

Step 2: Equivalent of the parallel trio
\[ \frac{1}{R_p}=\frac{1}{8}+\frac{1}{4}+\frac{1}{8}=0.125+0.25+0.125=0.5 \;\Rightarrow\; R_p=2\ \Omega. \]

Step 3: Series with R₁
R₁ = 10 Ω is in series with R_p, so \[ R_{\text{eq}}=10+2=12\ \Omega. \]

Step 4: Battery current
With 12 V ideal source: \[ I=\frac{V}{R_{\text{eq}}}=\frac{12}{12}=1\ \text{A}. \]

Final answer

12 Ω and 1 A