Question

The current in the following circuit is

• A. 2/3 A
• B. 1 A
• C. 1/8 A
• D. 2/9 A

Answer 1

Answer: (B)

Applied voltage $(V)=2V$ and resistances $=3\Omega ,3\Omega ,3\Omega$.
From the given circuit, we find that two resistances are in series and third resistance is in parallel. Therefore equivalent resistance for series resistances $=3+3=6\Omega$.
Now it is connected parallel with $3\Omega$ resistance.
Therefore $\frac{1}{R}=\frac{1}{3}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$ or $R=2\Omega \text{. }$
And current flowing in the circuit (l)
$=\frac{V}{R}=\frac{2}{2}=1A$