Question

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

• A. $\frac{9V}{7}$
• B. $\frac{5V}{18}$
• C. $\frac{9V}{7}$
• D. $\frac{18V}{5}$

Answer 1

Answer: (B)

$\frac{5V}{18}$

Answer 2

This circuit is a Wheatstone bridge. Since $\frac{P}{Q}$ = $\frac{R}{S}$ the four arms' resistances (P=4, Q=2, R=6, S=3) are in balance and no current flows through the center resistance (4). It is therefore taken off of the circuit.
Currently, the circuit's equivalent resistance is Req = ((4+2)(6+3)=6+9= 6+9 6+9 = $\frac{54}{15}$=$\frac{18}{5}$ Current, i=$\frac{V}{R}$ eq =$\frac{5V}{18}$.