Question

The voltage V in a simple circuit is related to the current I and the resistance R by Ohm's Law: V=IR. An illustration of a simple square circuit. Parts of the surface are labeled V for voltage, I for current, and R for resistance, but values for these quantities are not indicated. Find the rate of change dI/dt of the current I if R=600 Ω, I=0.04 A, dR/dt=−0.5 Ω/s, and dV/dt=0.04 V/s.

Answer 1

0.0001 A/s

Step-by-step explanation:

Since V = IR,

dV/dt = d(IR)/dt = IdR/dt + RdI/dt using product rule

dV/dt = IdR/dt + RdI/dt

dI/dt = (dV/dt - IdR/dt)/R which is the rate of change of current

dV/dt = 0.04 V/s, dR/dt = -0.5 Ω/s, I = 0.04 A and R = 600 Ω

Substituting these values into dI/dt, we have

dI/dt = [0.04 V/s - (0.04 A × -0.5 Ω/s)]/600 Ω

= (0.04 V/s + 0.02 V/s)/600 Ω

= 0.06 V/s/600 Ω

= 0.0001 A/s

Answer 2

dI/dt = 0.0004 A/s

Step-by-step explanation:

R = 600 ohms

I = 0.04 A

dR/dt = -0.5 ohms/s

dV/dt = 0.04 V/s

From Ohm's law V = IR

Taking the derivative of both sides with respect to t using product rule

dV/dt = I dR/dt + R dI/dt

0.04 = -( 0.04*0.5) + (600) dI/dt

0.04 + 0.2 = 600 dI/dt

0.24 = 600 dI/dt

dI/dt = 0.24/600

dI/dt = 0.0004 A/s