Question

Let the resistance of an electric component remains 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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Answer 1

Answer:

We know that according to Ohm’s law

V = IR

where

V= potential difference

I= Current

R = Resistance

We can also modify the equation as

I=V/R ——- (i)

Now given that the potential difference across the two ends of the component decreases to half

∴ So let the new potential difference be Vʹ=V/2

Resistance remains constant across the electrical component

So the new current is drawn through the electrical component is Iʹ = Vʹ/R

= (V/2)/R {Substituting Vʹ=V/2 in the above equation}

= (1/2) (V/R)

= (1/2) I = I/2

Therefore, the amount of current flowing through the electrical component is reduced by half.

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Answer 2

$\green{\tt Case\: I } \\ \\ \tt Resistance = {R}_{1} \\ \tt Potential Difference = {V}_{1} \\ \tt Current = {I}_{1} \\ \\ \green{ \tt Case\:II} \\ \\ \tt Resistance = {R}_{2} \\ \tt Potential Difference = {V}_{2} \\ \tt Current = {I}_{2} \\ \\ \tt According\:to\:the\: question \\ \\ \tt {R}_{1} = {R}_{2} \\ {V}_{1} = \frac{{V}_{2}}{2} \\ \\ \tt We\: know \: that \\ \\ \tt According\: to \: ohms \: law \\ \\ \tt V\: \alpha \:I \\ \\ \tt {V}_{1} \: \alpha \: {I}_{1} \\ \\ \tt {V}_{2} \: \alpha \: {I}_{2} \\ \\ \tt \frac{{V}_{1}}{{V}_{2}} = \frac{{I}_{1}}{{I}_{2}} \\ \\ \tt \frac {{I}_{1}}{{I}_{2}} = \frac{{V}_{1}}{\frac{{V}_{2}}{2}} \\ \\ \frac{{I}_{1}}{{I}_{2}} = 2 \\ \\ \bold{\orange{\tt {I}_{2} = \frac{{I}_{1}}{2}}} \\ \\ \boxed{\bold{\underline{\red{\tt So\: current\:will\:be\:half\:of\:initial\:current}}}}$