How can Kirchoff's first law be applied to electrical circuits?

[nameVoid]

According to Kirchoff's first law for electrical circuits V=RI+L(dI/dt) where constants V, R and L denote the electromotive force, the resistance, and the inductance, respectively, and I denotes the current at time t. If the electromotive force is terminated at time t=0 and if the current is I(0) at the instant of removal, prove that I=I(0)e^(-Rt/L)

Homework Equations

The Attempt at a Solution

$$v=RI+L(\frac{dI}{dT})$$
$$dt=\frac{LdI}{v-RI}$$
$$t+b=Lln|v-RI|(-R)$$
$$e^{\frac{t+b}{-LR}}=v-RI$$
$$t-> v-> 0 , I=I(0)$$
$$e^{\frac{b}{-LR}}=-RI(0)$$
$$-e^{\frac{t}{-LR}}(RI(0))=v-RI$$
$$I=\frac{e^{\frac{t}{-LR}}RI(0)+v}{R}$$
im afraid to do anymore work on this but looks good to me so far

 

[Mark44]

According to Kirchoff's first law for electrical circuits V=RI+L(dI/dt) where constants V, R and L denote the electromotive force, the resistance, and the inductance, respectively, and I denotes the current at time t. If the electromotive force is terminated at time t=0 and if the current is I(0) at the instant of removal, prove that I=I(0)e^(-Rt/L)

Homework Equations

The Attempt at a Solution

$$v=RI+L(\frac{dI}{dT})$$
$$dt=\frac{LdI}{v-RI}$$

OK up to here, but I think instead of (-R) you want (-1/R) in the next line. This comes from the substitution you did but don't show. In any case, you can check your answer by differentiating it and substituting back into the original differential equation.

Also, pay closer attention to your variables. You started with T and changed to t. Since this is time, t is a better choice. If you run into differential equations that involve time and temperature (usually represented by t and T, respectively), you'll run into trouble.

$$t+b=Lln|v-RI|(-R)$$
$$e^{\frac{t+b}{-LR}}=v-RI$$
$$t-> v-> 0 , I=I(0)$$
$$e^{\frac{b}{-LR}}=-RI(0)$$
$$-e^{\frac{t}{-LR}}(RI(0))=v-RI$$
$$I=\frac{e^{\frac{t}{-LR}}RI(0)+v}{R}$$
im afraid to do anymore work on this but looks good to me so far