Solve the problem.
-Electrical systems are governed by Ohm's law, which states that $V=I R$, where $V=$ voltage, $I$ = current, and $R=$ resistance. If the current in an electrical system is decreasing at a rate of $7 \mathrm{~A} / \mathrm{s}$ while the voltage remains constant at $20 \mathrm{~V}$, at what rate is the resistance increasing when the current is $60 \mathrm{~A}$ ?
A) $\frac{3}{7} \mathrm{ohms} / \mathrm{s}$
B) $\frac{7}{180} \mathrm{ohms} / \mathrm{s}$
C) $\frac{49}{3} \mathrm{ohms} / \mathrm{s}$
D) $\frac{7}{3} \mathrm{ohms} / \mathrm{s}$

Question

Quizplus · Accepted Answer

Using Ohm's law, $V=IR$, we can differentiate both sides with respect to time $t$ to get $\frac{dV}{dt}=I\frac{dR}{dt}+R\frac{dI}{dt}$. Since $V$ is constant, $\frac{dV}{dt}=0$, so we have $I\frac{dR}{dt}+R\frac{dI}{dt}=0$. Solving for $\frac{dR}{dt}$, we get $\frac{dR}{dt}=-\frac{R}{I}\frac{dI}{dt}$. When $I=60 \mathrm{~A}$ and $\frac{dI}{dt}=-7 \mathrm{~A} / \mathrm{s}$, we have $\frac{dR}{dt}=-\frac{20 \mathrm{~V}}{60 \mathrm{~A}}(-7 \mathrm{~A} / \mathrm{s})=\frac{7}{180} \mathrm{ohms} / \mathrm{s}$.

Solve the Problem V=IRV=I RV=IR , Where V=V=V= Voltage III = Current, And R=R=R=

Solve the Problem $V=I R$ , Where $V=$ Voltage $I$ = Current, And $R=$