RL Circuits — Question 1

Question 1

A voltage step fixes both the final RL current and its initial slope

An initially unenergized series circuit with L=.300 H and R=6.00 ohms is connected to an ideal 12.0-V source at t=0.

1. Solve for current and give the time constant.

2. Find the initial current slope and current after one time constant.

Question 1 - Solution

A voltage step fixes both the final RL current and its initial slope

(a) Method and calculation

$$L\dot i+Ri=V,\quad i(0)=0\Rightarrow\boxed{i(t)=\frac VR(1-e^{-t/\tau}),\quad\tau=L/R=.0500\,s}.$$$$i_\infty=12.0/6.00=2.00\,A.$$

(b) Interpretation and further calculation

$$\boxed{\dot i(0^+)=V/L=40.0\,A/s}.$$$$\boxed{i(\tau)=2(1-e^{-1})=1.26424\,A}.$$

Physical check

The inductor initially takes the entire source voltage; at steady current its ideal voltage is zero.