Question

The terminals of battery of emf $12\, V$ and internal resistance $1 \,\Omega$ are connected to a circular coil of resistance $16 \,\Omega$ at two points distance a quarter of circumference of coil. The current flowing in smaller arc of circle is

• A. 2.4 A
• B. 2.25 A
• C. 0.75 A
• D. 1.66 A

Answer 1

Answer: (B)

Two parts have resistance of $12 \,\Omega$ and $4 \,\Omega$.
Equivalent resistance $=\frac{12 \times 4}{12+4}=3 \,\Omega$
Current $=\frac{\text { Applied emf }}{\text { Equivalent resistance }}=\frac{12}{4}=3\, A$

Current in $R_{2}$ resistor, $I=$ Total current $\times 3 / 4$
$=2.25 \,A$