Question

The voltage V in a circuit that satisfies the law V = IR is slowly dropping as the battery wears out. At the same time, the resistance R is increasing as the resistor heats up. Use the equation dV dt = ∂V ∂I dI dt + ∂V ∂R dR dt to find how the current I is changing at the instant when R = 600 ohms, I = 0.04 amps, dR/dt = 0.5 ohms/sec, and dV /dt = −0.01 volts/sec. Hint: We need to find the rate of change of I, with respect to time. Find the partial derivatives of V with respect to I and R, then substitute into the equation for dV /dt.

Answer 1

`-0.5* 10^(-4) A/s`

Step-by-step explanation:

We are given that

`(dV)/(dt)=-0.01 V/s`

R=600 ohms

I=0.04 A

`(dR)/(dt)=0.5ohm/s`

`V=IR`

`(dV)/(dt)=(\partial V)/(dI)(dI)/(dt)+(\partial V)/(dR)(dR)/(dt)`

`(dV)/(dt)=R(dI)/(dt)+I(dR)/(dt)`

Substitute the values

`-0.01=600* (dI)/(dt)+0.04* 0.5`

`-0.01-0.04* 0.5=600(dI)/(dt)`

`-0.03=600(dI)/(dt)`

`(dI)/(dt)=(-0.03)/(600)=-0.5* 10^(-4)A/s`