If a resistance is halved in value and the potential difference across it is tripped,calculate the ratio of the new current to the old one
the voltage is also tripled, so the current ratio is 6:1
Apologies for the oversight. You are correct. Since the potential difference across the resistance is tripped (increased by a factor of 3), the correct ratio of the new current to the old one would indeed be 6:1.
Thank you for pointing out the error.
Assuming the old current is I_old, the new current is I_new, the old resistance is R_old, and the new resistance is R_new.
According to Ohm's Law, V = IR, where V is the potential difference.
For the old resistance:
V = I_old * R_old
For the new resistance:
V = I_new * R_new
As given in the question, the resistance is halved and the potential difference is tripled. So, we have:
R_new = R_old / 2
V = 3V
From the above equations, we can write:
I_old * R_old = I_new * R_new
I_old * R_old = I_new * (R_old / 2)
Rearranging the above equation to solve for I_new, we get:
I_new = 2 * I_old
Therefore, the ratio of the new current to the old one is 2:1.