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

Answer: (D)

$12 \, \Omega$ and $1 \, A$

Answer 2

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:

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