Circuit Analysis 1: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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1Formulas and methods

Series & parallel resistors

Resistors in series
Req = R1 + R2 + …
Resistors in parallel
1/Req = 1/R1 + 1/R2 + …
Two in parallel
Req = R1R2 / (R1 + R2)
Ohm's law
V = I·R
Power
P = V·I = I²R = V²/R

Method

  1. Start at the end of the network furthest from the terminals.
  2. Combine resistors in series by adding them, and resistors in parallel with 1/R = 1/R₁ + 1/R₂ + …
  3. Redraw the circuit after each step.
  4. Repeat until one resistor is left between the terminals.

Nodal analysis

Kirchhoff’s current law
ΣIin = ΣIout
Branch current
I = (Va − Vb)/R
Reference node
Vground = 0

Method

  1. Choose a reference node (ground) and label the other node voltages.
  2. Write Kirchhoff’s current law at every unknown node: currents leaving through each branch equal the currents entering.
  3. Express every branch current as (Va − Vb)/R.
  4. Solve the system of equations, then find any current you need.

Mesh analysis

Kirchhoff’s voltage law
ΣV = 0 around every loop
Shared resistor
current = I1 − I2
Current source in a mesh
fixes that mesh current

Method

  1. Draw a clockwise current in every mesh (window) of the circuit.
  2. Apply Kirchhoff’s voltage law around each mesh: the sum of voltage drops is zero.
  3. In a resistor shared by two meshes the current is the difference of the two mesh currents.
  4. Solve the equations; a negative result means the current flows the other way.

Superposition

Superposition
x = x1 + x2 + … (one source at a time)
Switched-off voltage source
short circuit (0 V)
Switched-off current source
open circuit (0 A)
Voltage divider
V2 = V·R2/(R1 + R2)
Current divider
I1 = I·R2/(R1 + R2)

Method

  1. Count the independent sources. You will solve one simpler circuit per source.
  2. Keep one source and switch off the others: a voltage source becomes a short circuit (a wire), a current source becomes an open circuit (a gap).
  3. Solve the simpler circuit for the voltage or current you need, keeping track of its direction.
  4. Repeat for every source, then add the contributions with their signs.

Thévenin & Norton

Thévenin voltage
Vth = Vopen circuit
Thévenin resistance
Rth: all sources off, seen from a–b
Norton current
IN = Vth/Rth
Load current
IL = Vth/(Rth + RL)
Maximum power transfer
RL = Rth, Pmax = Vth²/(4Rth)

Method

  1. Remove the load and mark the two terminals a and b.
  2. Find the open-circuit voltage between a and b: this is Vth.
  3. Switch off all independent sources and find the resistance seen from a–b: this is Rth.
  4. Put the load back on the equivalent circuit: Vth in series with Rth.
  5. For the Norton equivalent, convert: IN = Vth/Rth, with RN = Rth in parallel.

Op-amp circuits

Inverting amplifier
Vout = −(Rf/Rin)·Vin
Non-inverting amplifier
Vout = (1 + Rf/R1)·Vin
Summing amplifier
Vout = −Rf(V1/R1 + V2/R2)
Ideal op-amp
V+ = V−, Iin = 0

RC & RL transients

Time constant
τ = RC | τ = L/R
Response
x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ
Capacitor at DC
open circuit; vC cannot jump
Inductor at DC
short circuit; iL cannot jump
Settled
t ≈ 5τ

Method

  1. Find the initial value just after switching: a capacitor voltage and an inductor current cannot jump.
  2. Find the final value long after switching: a capacitor acts as an open circuit, an inductor as a short circuit.
  3. Find the time constant: τ = RC or τ = L/R, using the resistance seen by the capacitor or inductor.
  4. Use x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ.

2Practice problems

1.Series & parallel resistorsMedium
R1​ = 5 ΩR3​ = 5 Ω+−V1​12 VR2​8 ΩAR4​4 ΩB

Find the equivalent resistance Req seen by the source V1.

V1 = 12 VR1 = 5 ΩR2 = 8 ΩR3 = 5 ΩR4 = 4 Ω
Req = Ω
2.Series & parallel resistorsHard
R1​ = 12 ΩR3​ = 2 ΩR5​ = 25 Ω+−V1​12 VR2​25 ΩAR4​3 ΩBR6​15 Ω

Find the equivalent resistance Req seen by the source V1.

V1 = 12 VR1 = 12 ΩR2 = 25 ΩR3 = 2 ΩR4 = 3 ΩR5 = 25 ΩR6 = 15 Ω
Req = Ω
3.Nodal analysisMedium
R1​ = 12 ΩIR1​R3​ = 3 Ω+−V1​12 VR2​15 ΩAR4​5 ΩBIs​1 A

Use nodal analysis to find the current IR1 through R1 in the direction of the arrow.

V1 = 12 VIs = 1 AR1 = 12 ΩR2 = 15 ΩR3 = 3 ΩR4 = 5 Ω
IR1 = A
4.Nodal analysisHard
R1​ = 20 ΩR3​ = 10 Ω+−V1​9 VR2​3 ΩAR4​8 ΩBIs​?

The current source is adjusted until the node voltage at A is VA = 2.5 V. Use nodal analysis to find the source current Is (direction as shown).

V1 = 9 VR1 = 20 ΩR2 = 3 ΩR3 = 10 ΩR4 = 8 ΩVA = 2.5 V
Is = A
5.Mesh analysisMedium
R1​ = 8 ΩR2​ = 4 Ω+−V1​24 VR3​4 ΩA+−V2​12 VI1​I2​

Use mesh analysis to find the mesh current I1 (clockwise).

V1 = 24 VV2 = 12 VR1 = 8 ΩR2 = 4 ΩR3 = 4 Ω
I1 = A
6.Mesh analysisHard
R1​ = 4 ΩR2​ = 12 ΩR3​ = 20 Ω+−V1​5 VR4​30 ΩAR5​20 ΩB+−V2​12 VI1​I2​I3​

Use mesh analysis to find the mesh current I1 (clockwise).

V1 = 5 VV2 = 12 VR1 = 4 ΩR2 = 12 ΩR3 = 20 ΩR4 = 30 ΩR5 = 20 Ω
I1 = A
7.SuperpositionMedium
R1​ = 20 ΩR3​ = 2 Ω+−V1​15 VR2​12 Ω+−VR2​AR4​6 ΩBIs​2.5 A

Use superposition to find the voltage VR2 across R2 with the polarity shown.

V1 = 15 VIs = 2.5 AR1 = 20 ΩR2 = 12 ΩR3 = 2 ΩR4 = 6 Ω
VR2 = V
8.SuperpositionHard
R1​ = 12 ΩR2​ = 5 ΩIR2​R4​ = 10 Ω+−V1​36 VR3​15 ΩAIs​6 AB−+V2​9 V

Use superposition to find the power PR2 absorbed by R2.

V1 = 36 VIs = 6 AV2 = 9 VR1 = 12 ΩR2 = 5 ΩR3 = 15 ΩR4 = 10 Ω
PR2 = W
9.Thévenin & NortonMedium
R1​ = 12 ΩR3​ = 4 Ω+−V1​15 VR2​3 ΩAab

Find the Norton current IN: the current that flows from a to b if the terminals are shorted.

V1 = 15 VR1 = 12 ΩR2 = 3 ΩR3 = 4 Ω
IN = A
10.Thévenin & NortonHard
R1​ = 8 ΩR3​ = 12 Ω+−V1​36 VR2​6 ΩAIs​4 ABRL​10 ΩIL​

A load RL = 10 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.

V1 = 36 VIs = 4 AR1 = 8 ΩR2 = 6 ΩR3 = 12 ΩRL = 10 Ω
IL = A
11.Op-amp circuitsMedium
−+Vout​Rf​ = 100 kΩR1​ = 10 kΩVin​ = −0.4 V

The op-amp is ideal and runs from ±15 V supplies. Find the output voltage Vout.

Vin = −0.4 VR1 = 10 kΩRf = 100 kΩ
Vout = V
12.Op-amp circuitsHard
−+Vout​R2​ = 1 kΩV1​ = 1.7 VR1​ = 10 kΩV2​ = 0.5 VR3​ = 2 kΩR4​ = 10 kΩ

The op-amp is ideal and runs from ±15 V supplies. The resistor ratios are not matched. Find the output voltage Vout.

V1 = 1.7 VV2 = 0.5 VR1 = 10 kΩR2 = 1 kΩR3 = 2 kΩR4 = 10 kΩ
Vout = V
13.RC & RL transientsMedium
+−Vs​12 Vt = 0R = 33 ΩL22 mHi(t)

The inductor carries no current before the switch closes at t = 0. How long after the switch closes does the current reach 90% of its final value? Find t.

Vs = 12 VR = 33 ΩL = 22 mH
t = ms
14.RC & RL transientsHard
+−Vs​9 Vt = 0R = 10 kΩC10 µF+−vC​vC​(0) = 15 V

Just before the switch closes at t = 0, the capacitor holds 15 V with the polarity shown. How long does it take for vC to reach 11.4 V? Find t.

Vs = 9 VR = 10 kΩC = 10 µFvC(0) = 15 V
t = ms

3Solutions

1.Series & parallel resistors

Answer: Req = 9.235 Ω

Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.

R1​ = 5 ΩR3​ = 5 Ω+−V1​12 VR2​8 ΩAR4​4 ΩB

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R3 + R4 = 5 + 4 = 9 Ω
R2 ∥ 9 = 8·9/(8+9) = 4.235 Ω
Req = R1 + 4.235 = 5 + 4.235 = 9.235 Ω

Answer

Req = 9.235 Ω
2.Series & parallel resistors

Answer: Req = 16.02 Ω

Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.

R1​ = 12 ΩR3​ = 2 ΩR5​ = 25 Ω+−V1​12 VR2​25 ΩAR4​3 ΩBR6​15 Ω

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R5 + R6 = 25 + 15 = 40 Ω
R4 ∥ 40 = 2.791 Ω
R3 + 2.791 = 4.791 Ω
R2 ∥ 4.791 = 4.02 Ω
Req = R1 + 4.02 = 12 + 4.02 = 16.02 Ω

Answer

Req = 16.02 Ω
3.Nodal analysis

Answer: IR1 = 0.5076 A

Nodal analysis: choose a reference node, label the unknown node voltages, write Kirchhoff’s current law at each unknown node using Ohm’s law (I = ΔV/R), and solve the resulting equations.

R1​ = 12 ΩR3​ = 3 Ω+−V1​12 VR2​15 ΩAR4​5 ΩBIs​1 A

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 12 V). Unknowns: VA, VB.

KCL at node A

Currents leaving A through the resistors equal the current pushed into A by sources.

VA/R2 + (VA − V1)/R1 + (VA − VB)/R3 = 0
VA/15 + (VA − 12)/12 + (VA − VB)/3 = 0
0.4833·VA − 0.3333·VB = 1

KCL at node B

Currents leaving B through the resistors equal the current pushed into B by sources.

VB/R4 + (VB − VA)/R3 = Is
VB/5 + (VB − VA)/3 = 1
−0.3333·VA + 0.5333·VB = 1

Solve

Δ = (0.4833)(0.5333) − (−0.3333)(−0.3333) = 0.1467
VA = Δ1/Δ = 0.8667/0.1467 = 5.909 V
VB = Δ2/Δ = 0.8167/0.1467 = 5.568 V

Answer

IR1 = (V1 − VA)/R1 = (12 − 5.909)/12 = 0.5076 A
4.Nodal analysis

Answer: Is = 1.456 A

Nodal analysis: choose a reference node, label the unknown node voltages, write Kirchhoff’s current law at each unknown node using Ohm’s law (I = ΔV/R), and solve the resulting equations.

KCL at node A

VA is known, so this equation gives VB.

(VA − V1)/R1 + VA/R2 + (VA − VB)/R3 = 0
(2.5 − 9)/20 + 2.5/3 = (VB − 2.5)/10
0.5083 = (VB − 2.5)/10 → VB = 7.583 V

KCL at node B

Currents leaving B through the resistors equal the current pushed into B by sources.

Is = (VB − VA)/R3 + VB/R4 = (7.583 − 2.5)/10 + 7.583/8 = 1.456 A

Answer

Is = 1.456 A
5.Mesh analysis

Answer: I1 = 1.8 A

Mesh analysis: give each window of a planar circuit its own loop current, write Kirchhoff’s voltage law around each loop (sum of drops = sum of rises), and solve. Branch currents are the mesh current, or the difference of two mesh currents on a shared branch.

R1​ = 8 ΩR2​ = 4 Ω+−V1​24 VR3​4 ΩA+−V2​12 VI1​I2​

Assign mesh currents

Draw a clockwise current in each window: I1, I2. A resistor shared by two meshes carries the difference of their currents.

KVL around mesh 1

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R1·I1 + R3(I1 − I2) = V1
8·I1 + 4(I1 − I2) = 24
12·I1 − 4·I2 = 24

KVL around mesh 2

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R2·I2 + R3(I2 − I1) = −V2
4·I2 + 4(I2 − I1) = (−12)
−4·I1 + 8·I2 = −12

Solve

Δ = (12)(8) − (−4)(−4) = 80
I1 = Δ1/Δ = 144/80 = 1.8 A
I2 = Δ2/Δ = −48/80 = −0.6 A

Answer

I1 = 1.8 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: I1 = 0.09217 A

Mesh analysis: give each window of a planar circuit its own loop current, write Kirchhoff’s voltage law around each loop (sum of drops = sum of rises), and solve. Branch currents are the mesh current, or the difference of two mesh currents on a shared branch.

R1​ = 4 ΩR2​ = 12 ΩR3​ = 20 Ω+−V1​5 VR4​30 ΩAR5​20 ΩB+−V2​12 VI1​I2​I3​

Assign mesh currents

Draw a clockwise current in each window: I1, I2, I3. A resistor shared by two meshes carries the difference of their currents.

KVL around mesh 1

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R1·I1 + R4(I1 − I2) = V1
4·I1 + 30(I1 − I2) = 5
34·I1 − 30·I2 = 5

KVL around mesh 2

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R2·I2 + R4(I2 − I1) + R5(I2 − I3) = 0
12·I2 + 30(I2 − I1) + 20(I2 − I3) = 0
−30·I1 + 62·I2 − 20·I3 = 0

KVL around mesh 3

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R3·I3 + R5(I3 − I2) = −V2
20·I3 + 20(I3 − I2) = (−12)
−20·I2 + 40·I3 = −12

Solve

Δ = det of the coefficient matrix = 34720
I1 = Δ1/Δ = 3200/34720 = 0.09217 A
I2 = Δ2/Δ = −2160/34720 = −0.06221 A
I3 = Δ3/Δ = −11496/34720 = −0.3311 A

Answer

I1 = 0.09217 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: VR2 = 10.16 V

Superposition: in a linear circuit, any voltage or current is the sum of the responses to each independent source acting alone. To switch a source off, replace an ideal voltage source with a short circuit and an ideal current source with an open circuit. Resistors stay where they are.

R1​ = 20 ΩR3​ = 2 Ω+−V1​15 VR2​12 ΩAR4​6 ΩBOPEN

Case 1 · V1 acting alone

Iₛ becomes an open circuit, so R₃ and R₄ form one series branch from A to ground, in parallel with R₂.

R3+R4 = 2+6 = 8 Ω
R2 ∥ (R3+R4) = 12·8/(12+8) = 4.8 Ω
VA = 15·4.8/(20+4.8) = 2.903 V
VB = VA·R4/(R3+R4) = 2.903·6/8 = 2.177 V
VR2′ = VA − 0 = 2.903 − 0 = 2.903 V
R1​ = 20 ΩR3​ = 2 ΩSHORTR2​12 ΩAR4​6 ΩBIs​2.5 A

Case 2 · Is acting alone

V₁ becomes a short, so R₁ is in parallel with R₂. Iₛ splits at B between R₄ and the path through R₃.

R1 ∥ R2 = 20·12/(20+12) = 7.5 Ω
Req = R4 ∥ (R3+R1∥R2) = 6 ∥ 9.5 = 3.677 Ω
VB = Is·Req = 2.5·3.677 = 9.194 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = 9.194·7.5/9.5 = 7.258 V
VR2″ = VA − 0 = 7.258 − 0 = 7.258 V

Add the contributions

VR2 = VR2′ + VR2″
VR2 = 2.903 + 7.258 = 10.16 V

A negative result means the real polarity is opposite to the one marked.

8.Superposition

Answer: PR2 = 10.24 W

Superposition: in a linear circuit, any voltage or current is the sum of the responses to each independent source acting alone. To switch a source off, replace an ideal voltage source with a short circuit and an ideal current source with an open circuit. Resistors stay where they are.

R1​ = 12 ΩR2​ = 5 ΩR4​ = 10 Ω+−V1​36 VR3​15 ΩAOPENBSHORT

Case 1 · V1 acting alone

V₂ becomes a short and Iₛ an open circuit. Seen from A, R₃ is in parallel with the series pair R₂+R₄.

R3 ∥ (R2+R4) = 15 ∥ 15 = 7.5 Ω
VA = 36·7.5/(12+7.5) = 13.85 V
VB = VA·R4/(R2+R4) = 13.85·10/15 = 9.231 V
IR2′ = (VA − VB)/R2 = (13.85 − 9.231)/5 = 0.9231 A
R1​ = 12 ΩR2​ = 5 ΩR4​ = 10 ΩSHORTR3​15 ΩAIs​6 ABSHORT

Case 2 · Is acting alone

Both voltage sources become shorts. Iₛ splits at B between R₄ (now to ground) and R₂ in series with R₁ ∥ R₃.

R1 ∥ R3 = 6.667 Ω, so the left path is R2+6.667 = 11.67 Ω
Req = R4 ∥ 11.67 = 5.385 Ω
VB = Is·Req = 6·5.385 = 32.31 V
VA = VB·6.667/11.67 = 18.46 V
IR2″ = (VA − VB)/R2 = (18.46 − 32.31)/5 = −2.769 A
R1​ = 12 ΩR2​ = 5 ΩR4​ = 10 ΩSHORTR3​15 ΩAOPENB−+V2​9 V

Case 3 · V2 acting alone

V₁ becomes a short and Iₛ an open circuit. R₁ ∥ R₃ hangs from A, in series with R₂ back to B.

R1 ∥ R3 = 12·15/(12+15) = 6.667 Ω
VB = V2·(R2+R1∥R3)/(R4+R2+R1∥R3) = (−9)·11.67/21.67 = −4.846 V
VA = VB·(R1∥R3)/(R2+R1∥R3) = −4.846·6.667/11.67 = −2.769 V
IR2‴ = (VA − VB)/R2 = (−2.769 − −4.846)/5 = 0.4154 A

Add the current contributions

IR2 = IR2′ + IR2″ + IR2‴ = 0.9231 + (−2.769) + 0.4154 = −1.431 A

Power in the resistor

Power depends on the square of the current, so it does not obey superposition: add the currents first, then square.

PR2 = IR2²·R2 = (−1.431)²·5 = 10.24 W

Answer

PR2 = 10.24 W

Adding the powers of the separate cases would give 43.47 W, which is wrong.

9.Thévenin & Norton

Answer: IN = 0.4688 A

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.

R1​ = 12 ΩR3​ = 4 ΩSHORTR2​3 ΩAabRth​

Find Rth with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R1 ∥ R2 = 12·3/(12+3) = 2.4 Ω
Rth = (R1 ∥ R2) + R3 = 2.4 + 4 = 6.4 Ω
R1​ = 12 ΩR3​ = 4 Ω+−V1​15 VR2​3 ΩAab+−Vth​

Find Vth, the open-circuit voltage

With a–b open, no current flows through R₃, so there is no drop across it and V_b = V_A. R₁ and R₂ form a voltage divider.

Vth = V1·R2/(R1+R2) = 15·3/(12+3) = 3 V
IN​0.4688 ARth​6.4 Ωab

Norton equivalent

A Norton source is the same circuit seen as a current source in parallel with R_th. Its current is the short-circuit current at a–b.

IN = Vth/Rth = 3/6.4 = 0.4688 A
10.Thévenin & Norton

Answer: IL = 3.034 A

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.

R1​ = 8 ΩR3​ = 12 ΩSHORTR2​6 ΩAOPENBabRth​

Find Rth with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R1 ∥ R2 = 8·6/(8+6) = 3.429 Ω
Rth = (R1 ∥ R2) + R3 = 3.429 + 12 = 15.43 Ω
R1​ = 8 ΩR3​ = 12 Ω+−V1​36 VR2​6 ΩAIs​4 ABab+−Vth​

Find Vth, the open-circuit voltage

With a–b open, the only path for Iₛ is through R₃ into node A. Find V_A by superposition of V₁ and Iₛ, then add the drop across R₃.

VA = V1·R2/(R1+R2) + Is·(R1∥R2) = 15.43 + 4·3.429 = 29.14 V
Vth = VB = VA + Is·R3 = 29.14 + 4·12 = 77.14 V
Rth​ = 15.43 Ω+−Vth​77.14 VRL​10 ΩIL​

Connect the load to the Thévenin equivalent

The Thévenin source and R_th in series behave exactly like the original circuit at a–b, so the load sees a simple series loop.

IL = Vth/(Rth + RL) = 77.14/(15.43 + 10) = 3.034 A
11.Op-amp circuits

Answer: Vout = −4.4 V

Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.

Inverting input follows the input

V₋ = V₊ = −0.4 V
I = V₋/R1 = −0.4/10 = −0.04 mA

Output voltage

The current through R₁ also flows through Rf.

Vout = V₋ + I·Rf = −0.4 + (−0.04)·100 = −4.4 V

Answer

Vout = −4.4 V
12.Op-amp circuits

Answer: Vout = 0.2883 V

Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.

Voltage at the + input

V₊ = V2·R4/(R3 + R4) = 0.5·10/(2 + 10) = 0.4167 V

Superpose the two inputs

With V₂ = 0, V₁ sees an inverting amplifier. With V₁ = 0, V₊ is amplified by the non-inverting gain 1 + R₂/R₁.

Vout = −(1/10)·1.7 + (1 + 1/10)·0.4167 = −0.17 + 0.4583 = 0.2883 V

Answer

Vout = 0.2883 V
13.RC & RL transients

Answer: t = 1.535 ms

First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.

Time constant and final current

Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.

τ = L/R = 22 mH / 33 Ω = 0.6667 ms
i(∞) = Vs/R = 12/33 = 0.3636 A

Solve for the time

0.9 = 1 − e−t/τ → t = −τ·ln(1 − 0.9)
t = −0.6667·ln(0.1) = 1.535 ms

Answer

t = 1.535 ms
14.RC & RL transients

Answer: t = 91.63 ms

First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.

Initial value, final value, time constant

The capacitor voltage cannot jump, so it starts at its stored value and heads for the source voltage.

vC(0) = 15 V, vC(∞) = Vs = 9 V
τ = R·C = 10 kΩ · 10 µF = 100 ms
vC(t) = 9 + (15 − 9)·e−t/τ

Solve for the time

11.4 = 9 + (6)·e−t/τ → e−t/τ = (11.4 − 9)/(15 − 9) = 0.4
t = τ·ln((15 − 9)/(11.4 − 9)) = 100·ln(2.5) = 91.63 ms

Answer

t = 91.63 ms

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