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

Answer:

$-0.5\times 10^{-4} A/s$

Explanation:

We are given that

$\frac{dV}{dt}=-0.01 V/s$

R=600 ohms

I=0.04 A

$\frac{dR}{dt}=0.5ohm/s$

$V=IR$

$\frac{dV}{dt}=\frac{\partial V}{dI}\frac{dI}{dt}+\frac{\partial V}{dR}\frac{dR}{dt}$

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

Substitute the values

$-0.01=600\times \frac{dI}{dt}+0.04\times 0.5$

$-0.01-0.04\times 0.5=600\frac{dI}{dt}$

$-0.03=600\frac{dI}{dt}$

$\frac{dI}{dt}=\frac{-0.03}{600}=-0.5\times 10^{-4}A/s$