Question

What is the current (I) in the circuit, as shown in figure

• A. 1.2 A
• B. 0.5 A
• C. 1 A
• D. 2 A

Answer 1

Answer: (D)

2 A

Answer 2

As $R_{2}, R_{3}$ and $R_{4}$ are in series, their equivalent resistance is $R_{2}+R_{3}+R_{4}=6 \Omega$. Now the $6 \Omega$ resistance is in parallel with $R_{1}=2 \Omega$ whose equivalent $_{j}$ risistance will be $\frac{2 \times 6}{2+6}=\frac{3}{2} \Omega$ $\therefore$ The current through the circuit, $i=\frac{3}{3 / 2}=2\, A$