Question

What is the current drawn from source in the circuit shown in figure.

• A. 10 A
• B. 1 A
• C. 5 A
• D. 2.5 A

Answer 1

The correct option is C 5 A

$R_{12}=\frac{6\times 3}{6+3}=2$

$R_{34}=\frac{4\times 4}{4+4}=2$

Now $R_{34}and{R}_5$ are in series so the equivalent resistance is=$4+2=6$,

$R_6$ $R_7$ are in series so $R_{67}=4+2=6$

Now the net equivalent resistance of the whole circuit is $\frac{6\times 6}{6+6}+2=5ohm$

$\therefore I=\frac{V}{R_{eq}}=\frac{25}{5}=5A$