The p.d. across a $3 \Omega$ resistor is $6 \mathrm{~V}$. The current flowing in the resistor will be:(a) $\frac{1}{2} \mathrm{~A}$(b) $1 \mathrm{~A}$(c) $2 \mathrm{~A}$(d) $6 \mathrm{~A}$
Given:
Potential difference, V=6V
Resistance, R=3 ohms
To find:
Current, I
Solution:
According to the ohm's law:
$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$
$\mathrm{I}=\frac{\mathrm{6}}{\mathrm{3}}$
= 2 A
Answer is option (c)
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