Question

The figure shown below, calculate the net current from the battery and net resistance of the circuit.

• A. $1.33 \, A, \, \frac{15}{2} \, \Omega $
• B. $1.33 \, A, \, 15 \, \Omega $
• C. $1.5 \, A, \, 10 \, \Omega $
• D. $1.5 \, A, \, 15 \, \Omega $

Answer 1

Answer: (A)

In the given circuit, $B$ and $E$ are at the same potential. Points $C$ and $F$ are also at the same potential.
Therefore, the resistance between $B$ and $E, \, C$ and $F$ will be ineffective and neglected. Now, the circuit becomes

In the above circuit, branch $ABCD$ and $AEFD$ are in the parallel combination.
Thus, the net resistance of the circuit.
$\frac{1}{R_{n e t}}=\frac{1}{5 + 5 + 5}+\frac{1}{5 + 5 + 5}=\frac{1}{15}+\frac{1}{15}=\frac{2}{15}$
$\Rightarrow R_{n e t}=\frac{15}{2} \, \Omega $
$\therefore $ Current, $I=\frac{V}{R_{n e t}}=\frac{10}{15 / 2}=\frac{20}{15}=\frac{4}{3}=1.33 \, A$