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)

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, $i=\frac{3}{3/2}=2A$