Question

The current $I$ drawn from the $5\text{V}$ source will be

• A. $0.33\text{A}$
• B. $0.5\text{A}$
• C. $0.67\text{A}$
• D. $0.17\text{A}$

Answer 1

The correct option is B $0.5\text{A}$

The given circuit can be redrawn using Wheatstone bridge principle,

$\frac{P}{Q}=\frac{R}{S}$

$\frac{10}{5}=\frac{20}{10}$

Which is a balanced Wheatstone's bridge, and hence, no current flows in the middle resistor, so an equivalent circuit would be as shown below.


$R_{\text{eq}}=\frac{30\times 15}{15+30}=10\Omega$

By Ohm's law $V=IR$

So, $I=\frac{V}{R}=\frac{5}{10}=0.5\text{A}$