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

Explanation:

According to ohm's law, V = IR where;

V is the supply voltage (in volts)

I is the current supplied (in amperes)

R is the resistance (in ohms)

Initially, I = 2A, hence V = 2*R;

V = 2R 1

If the resistance in the circuit decreases to one-fourth of its original amount while the voltage remains constant, then the new resistance will be expressed as R₂ = 1/4 R and V₂ = V

Substituting the new conditions into the ohms law formula, we will have;

V₂ = I₂R₂

I₂ is the resulting new current

V = I₂ (1/4 R)

V = I₂ * R/4

V = I₂R/4 2

Substituting equation 1 into 2 we will have;

2R = I₂R/4

2 = I₂/4

cross multiply

I₂ = 4*2

I₂ = 8A

Hence the resulting new current is 8A

its c i just checked it