Chapter 7: Natural and Step Responses of RL and RC Circuits

Open practice · AI allowed

Practice sheet: Chapter 7, Natural and Step Responses of RL and RC Circuits

This sheet is not marked for correctness, and AI tools are allowed. The device-free quiz, at the start of the last class (week 13, session 2, Thu 14 Jan), is built from twins of these questions, so make sure you can do each one on your own.

Two-tier items ask for an answer and a reason. Get both right; on the quiz each earns half the marks.
Predict first: where asked, make your prediction before you calculate.
Confidence: mark each question Sure or Unsure. It is not marked; it shows you what to revise.
Conventions: “a long time” means the circuit has reached dc steady state: an inductor is a short circuit and a capacitor an open circuit. Every response here has the form $x(t) = x_f + (x_0 - x_f)e^{-t/\tau}$, with $\tau = L/R$ or $\tau = RC$, where $R$ is the resistance the inductor or capacitor sees after switching. Check every answer: $x(0)$ must equal the initial value, $x(\infty)$ the final value, and $\tau$ must be positive.
Entering numbers: type each value in the unit shown next to its box (for example, 50 for 50 ms), with a minus sign where needed.

Prefer paper? Open the printable PDF.

Which questions practise which skill

1Find the initial inductor current or capacitor voltage from the circuit before switching S2 S3 S5 S6 S7 S8 S9 S11 S16
2Find the final value from the circuit long after switching S5 S6 S8 S16
3Find $\tau$ from the resistance the element sees, with a test source when there is a dependent source S2 S3 S5 S6 S7 S8 S9 S10 S12 S15 S16
4Write $x(t) = x_f + (x_0 - x_f)\exp(-t/\tau)$ and find other currents and voltages from it S2 S3 S4 S5 S6 S7 S8 S9 S10 S11 S14 S15 S16
5Say which quantities can jump at $t = 0$ and find their values at $t = 0^+$ S1 S2 S5 S7 S15
6Work backwards from a measured response to $\tau$, $R$, $L$ or $C$ S13 S14

S1What cannot jump?Two-tier

A switch moves at $t = 0$ in a circuit with a resistor and an inductor. Which of these must always have the same value just after the switch moves ($t = 0^+$) as just before ($t = 0^-$), in any circuit of this kind?

Answer

a)The inductor’s voltageb)The inductor’s currentc)The resistor’s currentd)The resistor’s voltage

Reason

i)An inductor’s current cannot change instantly: a jump would need an infinite voltage.ii)An inductor’s voltage cannot change instantly, because its current is continuous.iii)A resistor’s current cannot jump, because Ohm’s law fixes it.iv)A resistor’s voltage cannot jump, because the resistance does not change.

Follow-up: name the quantity that cannot jump in a circuit with a capacitor.

a)The capacitor’s currentb)The capacitor’s voltagec)The resistor’s currentd)The resistor’s voltage

S2RL natural response

As in Nilsson Example 7.1. The switch has been closed for a long time and opens at $t = 0$.

a) Find $i_L(0)$.

b) Find the time constant $\tau$.

c) Find $i_L(t)$ for $t \ge 0$ and $v_o(t)$ for $t \ge 0^+$. Each has the form $Ke^{-at}$. Enter $a$:

1/s

c) … and $K$ for $v_o(t)$:

d) Find $v_o(0^-)$.

d) Find $v_o(0^+)$.

d) Why can $v_o$ jump when $i_L$ cannot?

Answer in a sentence, then compare with the explanation.

e) What percentage of the initial stored energy is dissipated in the 12 Ω resistor?

S3RC natural response

As in Nilsson Example 7.3. The switch has been in position x for a long time and moves to position y at $t = 0$.

a) Find $v_C(0)$.

b) Find $\tau$.

c) Find $v_C(t)$ for $t \ge 0$, and $v_o(t)$ and the current $i_o$ down through the 60 kΩ for $t \ge 0^+$. Each has the form $Ke^{-at}$. Enter $a$:

1/s

c) … $K$ for $v_o(t)$:

c) … $K$ for $i_o(t)$:

d) How much energy is dissipated in the 60 kΩ resistor?

S4One time constant laterTwo-tier

In a natural response, what fraction of the initial value remains after one time constant?

Answer

a)63%b)0c)37%d)50%

Reason

i)The time constant is the time the response takes to reach its final value.ii)After one time constant, $x = x_0 e^{-1} = 0.37x_0$.iii)Half of what remains decays in each time constant.iv)After one time constant, $1 - e^{-1} = 63\%$ of $x_0$ remains.

Follow-up: after how many time constants is less than 1% left?

time constants

S5RL step response

As in Nilsson Example 7.5. The switch has been in position a for a long time. At $t = 0$ it moves to position b; it makes contact at b before it breaks contact at a, so the inductor current is never interrupted.

a) Find $i(0)$.

a) Find $i(\infty)$.

a) Find $\tau$.

b) Find $i(t)$ for $t \ge 0$. In the form $i(\infty) + Be^{-t/\tau}$, enter $B$:

c) Find $v(t)$ for $t \ge 0^+$. In the form $Ve^{-t/\tau}$, enter $V$:

c) Why is $v(0^+)$ larger than the 30 V of the source?

Answer in a sentence, then compare with the explanation.

d) When does $v$ equal 30 V?

S6RC step response with a Thévenin equivalent

As in Nilsson Example 7.6. The switch has been in position 1 for a long time and moves to position 2 at $t = 0$.

a) Find $v_o(0)$.

b) For $t \ge 0$, find the Thévenin equivalent that the capacitor sees. $V_{\text{Th}}$, with the same polarity as $v_o$:

b) … $R_{\text{Th}}$:

kΩ

c) Find $v_o(t)$ for $t \ge 0$. First $\tau$:

c) … then $B$ in $v_o(t) = v_o(\infty) + Be^{-t/\tau}$:

d) Find the capacitor current $i_o$, referenced into its + terminal, for $t \ge 0^+$. In the form $Ie^{-t/\tau}$, enter $I$:

S7The general method, RL

As in Nilsson Example 7.7. The switch has been closed for a long time and opens at $t = 0$.

a) Find $i_o(0)$.

b) Find $\tau$.

c) Find $i_o(t)$ for $t \ge 0$. In the form $Ke^{-at}$, enter $a$:

1/s

d) Find $v_o(0^-)$.

d) Find $v_o(0^+)$ and $v_o(t)$ for $t \ge 0^+$. Enter $v_o(0^+)$:

S8The general method, RC from a negative voltage

As in Nilsson Example 7.8. The switch has been in position a for a long time and moves to position b at $t = 0$.

a) Find $v_C(0)$. Watch the polarity.

b) Find $v_C(t)$ for $t \ge 0$. First $v_C(\infty)$:

b) … $\tau$:

b) … $B$ in $v_C(t) = v_C(\infty) + Be^{-t/\tau}$:

c) Find the capacitor current $i$, referenced into its + terminal, for $t \ge 0^+$. In the form $Ie^{-t/\tau}$, enter $I$:

d) When is $v_C$ zero?

S9A dependent source sets $\tau$

As in booklet Problem 7.7. The switch has been closed for a long time and opens at $t = 0$.

a) Find $i_L(0)$.

b) For $t \ge 0$, find the resistance the inductor sees, using a test source.

c) Find $\tau$ and $i_L(t)$ for $t \ge 0$. Enter $\tau$:

d) Find $i_x(t)$ for $t \ge 0^+$. In the form $Ie^{-t/\tau}$, enter $I$:

S10A dependent source in an RC circuit

The 2 µF capacitor has been charged to 30 V. At $t = 0$ the switch closes.

a) Find the resistance the capacitor sees, using a test source.

kΩ

b) Find $\tau$ and $v(t)$ for $t \ge 0$. Enter $\tau$:

c) Find $i(t)$ and $v_x(t)$ for $t \ge 0^+$. In the form $Ke^{-t/\tau}$, enter $K$ for $i(t)$:

c) … $K$ for $v_x(t)$:

S11Spot the errorSpot the error

The worked solution below contains the kind of polarity slip AI chat tools often make. For $t < 0$ a 0.4 µF capacitor has been connected for a long time across the 24 kΩ resistor of a divider: a 36 V source in series with 12 kΩ and 24 kΩ. The source’s + terminal connects to the bottom of the 24 kΩ, so the top of the 24 kΩ is negative. At $t = 0$ the capacitor is switched to a 48 V source in series with 125 kΩ. $v_C$ is positive at the top.

$\tau = (125\text{ k}\Omega)(0.4\ \mu\text{F}) = 50$ ms.
$v_C(0) = 36 \times 24/(12 + 24) = 24$ V.
$v_C(\infty) = 48$ V.
$v_C(t) = 48 + (24 - 48)e^{-20t} = 48 - 24e^{-20t}$ V.
Check: $v_C(0) = 24$ V and $v_C(\infty) = 48$ V, as found.

b) Correct it and find $v_C(t)$. The corrected $v_C(0)$:

b) … $B$ in $v_C(t) = 48 + Be^{-20t}$ V:

c) Line 5 calls itself a check. Why did it not catch the error?

Answer in a sentence, then compare with the explanation.

d) When is $v_C$ zero?

S12Spot the error in a time constantSpot the error

An uncharged 0.4 µF capacitor is connected at $t = 0$ through 5 kΩ to a node N. Node N connects through 30 kΩ to a 60 V source and through 60 kΩ to ground.

$v(0) = 0$, since the capacitor is uncharged.
$v(\infty) = 60 \times 60/(30 + 60) = 40$ V.
$\tau = (5 + 30 + 60\text{ k}\Omega)(0.4\ \mu\text{F}) = 38$ ms.
$v(t) = 40 - 40\exp(-t/0.038)$ V.

a) Which is the first wrong line?

a)Line 1b)Line 2c)Line 3d)Line 4

b) Find the correct $\tau$.

b) … and $v(t)$: enter $a$ in $v(t) = 40 - 40e^{-at}$ V.

1/s

c) What resistance does the capacitor see?

kΩ

c) … and how do you find it?

Answer in a sentence, then compare with the explanation.

S13Work backwards

A capacitor charges through 10 kΩ. Its voltage is 10 V at $t = 0$, rises toward 40 V, and reads 30 V at $t = 2$ ms.

a) Find $\tau$.

b) Find $C$.

µF

c) When does the voltage reach 39 V?

d) The same steps work for an inductor: a current rises from 0 toward 2 A through 20 Ω and reads 1.5 A at $t = 6.93$ ms. Find $\tau$:

d) … and $L$.

S14An artificial pacemakerPredict first

As in Nilsson’s Practical Perspective for this chapter. A source $V_s$ charges a capacitor $C$ through a resistor $R$, starting from 0 V. When $v_C$ reaches $0.75V_s$, a controller discharges the capacitor in a negligible time and sends a pulse to the heart; then charging starts again.

a) Show that each charging takes $t = RC\ln 4$, so the heart rate is $H = 60/(RC\ln 4)$ beats per minute.

Show it on paper, then compare with the explanation.

b) With $C = 2$ µF, what $R$ gives 60 beats per minute?

kΩ

c) With $C = 2$ µF and $R = 300$ kΩ, what is the heart rate?

beats per minute

d) Predict first: what happens to the heart rate if $R$ is doubled?

a)It doubles.b)It halves.c)It does not change.

S15How fast does a relay let go?

A 0.5 H relay coil with 60 Ω winding resistance carries 0.2 A from a 12 V supply. When the driving switch opens at $t = 0$, the coil current continues through a diode (ideal: 0 V when conducting) in series with a 140 Ω resistor, connected across the coil.

a) Find $\tau$ for $t \ge 0$.

b) The relay releases when the current falls below 50 mA. When does it release?

c) What voltage appears across the coil’s terminals at $t = 0^+$? Take it positive at the terminal the supply made positive.

d) Repeat (a) and (b) with the diode alone (no 140 Ω). $\tau$:

d) … and the release time:

d) Why do designers sometimes add the resistor?

Answer in a sentence, then compare with the explanation.

S16Check an answer without solving again

A 0.4 H inductor carries $i(0) = -2$ A. At $t = 0$ it is switched across a 20 V source in series with 5 Ω; $i$ is referenced in the direction the source drives current. Student 1: $i = 4 - 6\exp(-12.5t)$ A. Student 2: $i = 4 - 2\exp(-12.5t)$ A. Use $i(0)$, $i(\infty)$ and $\tau$ to decide which student is right.

Find $i(\infty)$.

Find $\tau$.

Which student is right?

a)Student 1b)Student 2