DC Electric Circuits

Time Constant Calculations

52 questions By Tony R. Kuphaldt

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Question 49 of 52

∫f(x) dx Calculus alert!

Differential equations may be used to model the charging behavior of an L/R circuit. Take, for instance, this simple L/R circuit:

We may develop a loop equation based on Kirchhoff’s Voltage Law, knowing that the voltage of the power source is constant (40 volts), and that the voltage drops across the inductor and resistor are V_L = L[dI/dt] and V_R = IR, respectively:

40 − IR − L dI
dt = 0

Show that the specific solution to this differential equation, assuming an initial condition of I = 0 at t = 0, is as follows:

I = 0.8(1 − e^−25t)

Reveal answer

40 − IR − L dI
dt = 0

40 − IR = L dI
dt

40 − IR
L = dI
dt

dt
L = dI
40 − IR

⌠ ⌡ 1
L dt = ⌠ ⌡ 1
40 − IR dI

Substitution: u = 40 − IR ; du
dI = −R ; dI = − 1
R du

1
L ⌠ ⌡ dt = − 1
R ⌠ ⌡ 1
u du

t
L K₁ = − 1
R | lnu |

− tR
L K₂ = | lnu |

e^{−[tR/L] K₂} = | u |

K₃e^−[tR/L] = u

K₃e^−[tR/L] = 40 − IR

IR = 40 − K₃e^−[tR/L]

I = 40
R − K₄e^−[tR/L]

Given the initial condition of zero current (I = 0) at time zero (t = 0), the constant of integration must be equal to [40/R] in our specific solution:

I = 40
R − 40
R e^−[tR/L]

I = 40
R (1 − e^−[tR/L])

Substituting the given component values into this specific solution gives us the final equation:

I = 0.8(1 − e^−25t)

Notes:
L/R time constant circuits are an excellent example of how to apply simple differential equations. In this case, we see that the differential equation is first-order, with separable variables, making it comparatively easy to solve.

It should also be evident to students that any initial condition for current may be set into the general solution (by changing the value of the constant).

Question 50 of 52

Assume that the switch in this circuit is toggled (switched positions) once every 5 seconds, beginning in the “up” (charge) position at time t = 0, and that the capacitor begins in a fully discharged state at that time. Determine the capacitor voltage at each switch toggle:


Time	Switch motion	V_C (volts)

0 s	discharge → charge	0 volts

5 s	charge → discharge

10 s	discharge → charge

15 s	charge → discharge

20 s	discharge → charge

25 s	charge → discharge

Reveal answer

Question 51 of 52

This passive integrator circuit is powered by a square-wave voltage source (oscillating between 0 volts and 5 volts at a frequency of 2 kHz). Determine the output voltage (v_out) of the integrator at each instant in time where the square wave transitions (goes from 0 to 5 volts, or from 5 to 0 volts), assuming that the capacitor begins in a fully discharged state at the first transition (from 0 volts to 5 volts):


Transition	v_out

#1 (0 → 5 volts)	0 volts

#2 (5 → 0 volts)

#3 (0 → 5 volts)

#4 (5 → 0 volts)

#5 (0 → 5 volts)

#6 (5 → 0 volts)

#7 (0 → 5 volts)

#8 (5 → 0 volts)

Reveal answer

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P
pthg3 May 10, 2021

Maybe this will help someone else. The general formulas for V(t) and I(t) in question 25 (and the x(t) versions in question s 23 and 14) contain typos (or maybe hypertext coding glitches). They should actually be V(t) = (Vf-Vo)(1-e^(-t/𝛕)) + Vo, I(t) = (If-Io)(1-e^(-t/𝛕)) + Io in question 25. Those are correct in the PDF download version. In questions 23 and 24 the equations are x = xinitial + ( xfinal − xinitial ) ( 1 − e[(−t)/(τ)] ).

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- eetechzac November 15, 2021
  
  This has been updated. Thanks!
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