Question

According to Ohm's Law, in an electrical circuit the voltage V (in volts) is related to the current I (in amps) and the resistance R (in ohms) by V = TR. Sup pose that when the current is 5 amps and the resistance is 6 ohms, the current is increasing by 0.3 amps/minute. What will be the rate of change of the resistance at that moment if the voltage remains unchanged? Make sure your answer has units! .

Answer 1

Answer:

The resistance is decreasing at a rate of 0.36 ohms/minute.

Step-by-step explanation:

The mathematical form of the Ohm's law is given by :

V = IR ...(1)

Where V is voltage, I is current and R is resistance

Given,

I = 5 A

R = 6 ohms

dI/dt = 0.3 A/min

Differentiate equation (1) wrt t:

$\dfrac{dV}{dt}=I\dfrac{dR}{dt}+R\dfrac{dI}{dt}$

When V is constant, dV/dt = 0

$0=5\times \dfrac{dR}{dt}+6\times 0.3\ A/s\\\\1.8=-5\times \dfrac{dR}{dt}\\\\\dfrac{dR}{dt}=-0.36\ \Omega/min$

So, the resistance is decreasing at a rate of 0.36 ohms/minute.