Question

For the network shown in the figure the value of the current $i$ is : -

• A. $\frac{18V}{5}$
• B. $\frac{5V}{9}$
• C. $\frac{9V}{35}$
• D. $\frac{5V}{18}$

Answer 1

The correct option is D $\frac{5V}{18}$

The given diagram can be redrawn as shown below.

From the above diagram, we observe that

$\frac{R_{AC}}{R_{AD}}=\frac{R_{CB}}{R_{DB}}=\frac{2}{3}$

So it is a balanced Wheatstone bridge.

Hence, the middle $4\Omega$ can be neglected. So circuit become

Equivalent resistance of the circuit will be

$R_{eq}=(4+2)||(6+3)=\frac{6\times 9}{6+9}$

$R_{eq}=\frac{18}{5}\Omega$

So current drawn in circuit

$i=\frac{V}{R_{eq}}=\frac{5V}{18}\text{A}$

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