Question

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

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

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.
$\frac{1}{R_{net}}=\frac{1}{5+5+5}+\frac{1}{5+5+5}=\frac{1}{15}+\frac{1}{15}=\frac{2}{15}$
$\Rightarrow R_{net}=\frac{15}{2}\Omega$
$\therefore$ Current, $I=\frac{V}{R_{net}}=\frac{10}{15/2}=\frac{20}{15}=\frac{4}{3}=1.33A$