Question

In the circuit shown below:

Ώ

The potential difference across the 3 Ω resistor is:

(a) $\frac{1}{9}$ V
(b) $\frac{1}{2}$ V
(c) 1 V
(d) 2 V

Answer 1

(c) 1V
The resistors of 1 $\mathrm{\Omega }$, 2 $\mathrm{\Omega }$ and 3 $\mathrm{\Omega }$ are connected in series. Therefore, the net resistance,
R =R₁ +R₂ +R₃
R=1 $\mathrm{\Omega }$ + 2 $\mathrm{\Omega }$ + 3 $\mathrm{\Omega }$ = 6 $\mathrm{\Omega }$
Current in the circuit will be,
I = V/R
or I = 2/6 = 1/3 A
Current = $\frac{1}{3}$A
Therefore, the voltage across the 3 Ώ resistor,
V = IR
or V= $\frac{1}{3}$ x 3 = 1 V