Stem-Master
Physics / Chapter 30: Induction and Inductance / Section 30-8

RL Circuits — Question 6

Section 30-8 · Physics pages 4902–4903

Question 6 A parallel resistor creates an immediate source current beside a delayed RL branch An ideal voltage step V is applied across two parallel branches: a resistor R1, and a series combination R2 with an initially unenergized inductor L. All resistances and L are positive. V R2 L i R1 Schematic; labels, not drawing lengths, specify the data. (a) Find total source current as a function of time. (b) Find its initial and final values and explain which current can jump.
Show solutionHide solution
Question 6 - Solution A parallel resistor creates an immediate source current beside a delayed RL branch V R2 L i R1 (a) Method and calculation i1 = V/R1, i2 = V R2 (1 −e−tR2/L). is(t) = V R1 + V R2 (1 −e−tR2/L) . (b) Interpretation and further calculation is(0+) = V/R1, is(∞) = V/R1 + V/R2 . i2(0+) = i2(0−) = 0; i1 jumps when the ideal source voltage steps. Physical check Inductor current continuity does not require the total source current to be continuous when a separate resistive branch exists.

Question 6

A parallel resistor creates an immediate source current beside a delayed RL branch

An ideal voltage step V is applied across two parallel branches: a resistor R1, and a series combination R2 with an initially unenergized inductor L. All resistances and L are positive.

Diagram shown in the original SVG worksheet.

  1. Find total source current as a function of time.

  2. Find its initial and final values and explain which current can jump.

Question 6 - Solution

A parallel resistor creates an immediate source current beside a delayed RL branch

Diagram shown in the original SVG worksheet.

(a) Method and calculation

i1=V/R1,i2=VR2(1−e−tR2/L).i_1=V/R_1,\quad i_2=\frac V{R_2}(1-e^{-tR_2/L}).is(t)=VR1+VR2(1−e−tR2/L).\boxed{i_s(t)=\frac V{R_1}+\frac V{R_2}(1-e^{-tR_2/L})}.

(b) Interpretation and further calculation

is(0+)=V/R1,is(∞)=V/R1+V/R2.\boxed{i_s(0^+)=V/R_1,\quad i_s(\infty)=V/R_1+V/R_2}.i2(0+)=i2(0−)=0;i1 jumps when the ideal source voltage steps.i_2(0^+)=i_2(0^-)=0;\quad i_1\text{ jumps when the ideal source voltage steps}.

Physical check

Inductor current continuity does not require the total source current to be continuous when a separate resistive branch exists.

← Section questionsChapter contents