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
$I=I_1+I_2$
$\frac{24-V}{3}=\frac{10-V}{2}+\frac{9-V}{1}$
$\Rightarrow \frac{24-V}{3}=\frac{28-3V}{2}$
$\Rightarrow 2(24-V)=3(28-3V)$
$\Rightarrow 48-2V=84-9V$
$\Rightarrow 7V=36$
$\Rightarrow V=5.14V$
From Ohm's law
$\Delta V=IR$
$\Delta V=24-5.14=18.86,R=3\Omega$
$\therefore I=\frac{18.86}{3}\approx 6A$