Question

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

Answer 1

I =V/R

initally, R = V/2

then, new R= V/8

so, current will be multiple of 4

equals to 8 A

Answer 2

Answer: The resulting current 8 A.

Step-by-step explanation:

Current , I = 2A

Voltage applied = V

Resistance offered by the circuit = R

V= IR (Ohm's law)

`V=2 A* 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'* (1)/(4)R`

`2 A* R=I'* (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.