Question

In the circuit shown in the figure, the current ' $I$ ' is

• A. 6 A
• B. 2 A
• C. 4 A
• D. 7 A

Answer 1

Answer: (A)

Applying junction law We have
$I =I_{1}+I_{2}$
$\frac{24-V}{3} =\frac{10-V}{2}+\frac{9-V}{1}$
$\Rightarrow \frac{24-V}{3}=\frac{28-3 V}{2}$
$\Rightarrow 2(24-V)=3(28-3 V)$
$\Rightarrow 48-2 V=84-9 V$
$\Rightarrow 7\, V =36$
$\Rightarrow V=5.14\, V$
From Ohm's law
$\Delta V=I R$
$\Delta V=24-5.14=18.86, R=3 \Omega$
$\therefore I=\frac{18.86}{3} \approx 6\, A$