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A circuit has a current of 2 A. If the resistance in the circuit decreases to one-fourth of its original amount while the voltage remains constant, what will be the resulting current?

0.5 A

2 A

4 A

8 A

Verified Answer :

I =V/R

initally, R = V/2

then, new R= V/8

so, current will be multiple of 4

equals to 8 A

Answer: The resulting current 8 A.

Explanation:

Current , I = 2A

Voltage applied = V

Resistance offered by the circuit = R

V= IR (Ohm's law)

\(V=2 A\times R\)

Now,the resistance in the circuit decreases to one-fourth of its original amount while the voltage remains constant

The new current I' will be given as:

\(V=I'\times \frac{1}{4}R\)

\(2 A\times R=I'\times \frac{1}{4}R\)

\(I'=8 A\)

The resulting current will be 8 A when resistance in the circuit decreases to one-fourth of its original amount while the voltage kept constant.

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