Let the resistance of an electrical component | vedclass

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Question

(C) According to Ohm's law, the relationship between potential difference $(V)$, current $(I)$, and resistance $(R)$ is given by:$V = IR$ or $I = \fra

Vedclass · Accepted Answer

The current becomes half.

Vedclass · Answer

(C) According to Ohm's law, the relationship between potential difference $(V)$, current $(I)$, and resistance $(R)$ is given by:$V = IR$ or $I = \frac{V}{R}$Given that the resistance $(R)$ remains constant, the current is directly proportional to the potential difference $(I \propto V)$.If the potential difference is reduced to half of its former value, let the new potential difference be $V' = \frac{V}{2}$.Substituting this into the formula for the new current $(I')$:$I' = \frac{V'}{R} = \frac{\frac{V}{2}}{R} = \frac{1}{2} \left( \frac{V}{R} \right) = \frac{I}{2}$Therefore, the current flowing through the electrical component is reduced to half of its original value.

$I$ (amperes)	$0.5$	$1.0$	$2.0$	$3.0$	$4.0$
$V$ (volts)	$1.6$	$3.4$	$6.7$	$10.2$	$13.2$

Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

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