Answer:
I' = 40 mA
Explanation:
Given that,
The current in the wires of a circuit is 60.0 mA.
If the voltage impressed across the ends of the circuit were doubled and the resistance were tripled such that,
V' = 2V and R'=3R
The relation between current, voltage and resistance is given by :
V = IR
Let I' be the new current. So,
V'=I'R'
⇒ [tex]I'=\dfrac{V'}{R'}\\\\I'=\dfrac{2V}{3R}\\\\I'=\dfrac{2}{3}\times \dfrac{V}{R}\\\\I'=\dfrac{2}{3}\times I\\\\I'=\dfrac{2}{3}\times 60\\\\I'=40\ mA[/tex]
So, the new current would be 40 mA.
