MCQ
In the circuit shown below : The potential difference across the $3 \Omega $ resistor is :
  • A
    $\frac{1}{9}\text{V}$
  • B
    $\frac{1}{2}\text{V}$
  • $1V$
  • D
    $2V$

Answer

Correct option: C.
$1V$
The resistors of $1 \Omega , 2 \Omega $ and $3 \Omega $ are connected in series. Therefore, the net resistance,
$\text{R} = \text{R}_1 + \text{R}_2 +\text{R}_3$
$\text{R}=1\Omega+2\Omega+3\Omega=6\Omega$
Current in the circuit will be,
$\text{I}=\frac{\text{V}}{\text{R}}$
or $\text{I}=\frac{2}{6}=\frac{1}{3}\text{A}$
Current $=\frac{1}{3}\text{A}$
Therefore, the voltage across the $3I$ resistor,
$V = IR$
or $\text{V}=\frac{1}{3}\times3=1\text{V}$

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