I=E/R
so, the new I will be
E/(R/2) = 2E/R
if the resistance is halved and the voltage is not changed, what will happen to the current in resistive circuit?
2 answers
V=IR
When V=constant,
V=(2/2)*IR
=>V=(2*I)*(R/2)
So, double current flow when R is halved.
Now,
P=VI
=>(2*P)=V*(2*I)
So, power double if V=constant & R is halved.
When V=constant,
V=(2/2)*IR
=>V=(2*I)*(R/2)
So, double current flow when R is halved.
Now,
P=VI
=>(2*P)=V*(2*I)
So, power double if V=constant & R is halved.