MCQ
The p.d. across a $ 3Ω$ resistor is 6V. The current flowing in the resistor will be:
- A$\frac{1}{2}\text{A}$
- B1A.
- C2A.
- D6A.
Explanation:
If the p.d. across a $3\Omega$ resistor is 6V, the current flowing in the resistor will be 2A as current (I) is given by the equation, $\text{I}=\frac{\text{V} (\text{Voltage})}{\text{R}(\text{Resistance})}.$
or $\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=6\ \frac{\text{V}}{3}\ Ω$
$\text{I} = 2\text{A.}$
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