Question

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

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

Answer 1

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

Potential distribution of the given circuit will be as shown in the figure.


This arrangement can be modified as shown in figure.


From the above diagram, we can clearly see that this is a balanced Wheatstone bridge because,

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

So, circuit can be reduced as shown below.


So, effective resistance of the circuit will be
$R_{eq}=10\Omega$

So, from Ohm's law,

$I=\frac{V}{R_{eq}}=\frac{5}{10}=0.5\text{A}$

Hence, option $\left(d\right)$ is correct.