Question

In the electric circuit shown in the figure, current flowing through 4 $\Omega$ resistor is :

• A. $3A.$
• B. $2A$
• C. $1A$
• D. $4A$

Answer 1

The correct option is C $1A$

The circuit can be reduced, step by step, to a single equivalent resistance. The 8 ohm and the 8 ohm are connected in parallel , and so they can be replaced by an equivalent resistor Rp of 4 ohms, using the below equation.

$Rp=\frac{(8\times 8)}{(8+8)}=4ohms$

This resistor is connected in series with the 4 ohm resistor R1. The total resistance Rt of the circuit is then

$R_t=R_1+R_p=4+4=8ohms.$

Since the 4 ohm and the 4 ohm are connected in series, they have the same current I1, which must be equal to the current of the battery. Using Ohms law we get,

$I=\frac{V}{R}=\frac{8}{8}=1A$.

Hence, the current flowing through the resistor R1 (4 ohms) is 1A.