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

Method

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

Nodal analysis

Kirchhoff’s current law: ΣI_in = ΣI_out
Branch current: I = (V_a − V_b)/R
Reference node: V_ground = 0

Method

Choose a reference node (ground) and label the other node voltages.
Write Kirchhoff’s current law at every unknown node: currents leaving through each branch equal the currents entering.
Express every branch current as (V_a − V_b)/R.
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

Draw a clockwise current in every mesh (window) of the circuit.
Apply Kirchhoff’s voltage law around each mesh: the sum of voltage drops is zero.
In a resistor shared by two meshes the current is the difference of the two mesh currents.
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

Count the independent sources. You will solve one simpler circuit per source.
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).
Solve the simpler circuit for the voltage or current you need, keeping track of its direction.
Repeat for every source, then add the contributions with their signs.

Thévenin & Norton

Thévenin voltage: V_th = V_{open circuit}
Thévenin resistance: R_th: all sources off, seen from a–b
Norton current: I_N = V_th/R_th
Load current: I_L = V_th/(R_th + R_L)
Maximum power transfer: R_L = R_th, P_max = V_th²/(4R_th)

Method

Remove the load and mark the two terminals a and b.
Find the open-circuit voltage between a and b: this is V_th.
Switch off all independent sources and find the resistance seen from a–b: this is R_th.
Put the load back on the equivalent circuit: V_th in series with R_th.
For the Norton equivalent, convert: I_N = V_th/R_th, with R_N = R_th in parallel.

Op-amp circuits

Inverting amplifier: V_out = −(R_f/R_in)·V_in
Non-inverting amplifier: V_out = (1 + R_f/R₁)·V_in
Summing amplifier: V_out = −R_f(V₁/R₁ + V₂/R₂)
Ideal op-amp: V₊ = V₋, I_in = 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; v_C cannot jump
Inductor at DC: short circuit; i_L cannot jump
Settled: t ≈ 5τ

Method

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

2Practice problems

1.Series & parallel resistorsMedium

Find the current I_R1 delivered by the source, which flows through R₁.

V₁ = 6 VR₁ = 10 ΩR₂ = 5 ΩR₃ = 2 ΩR₄ = 12 Ω

I_R1 = A

2.Series & parallel resistorsHard

Find the equivalent resistance R_eq seen by the source V₁.

V₁ = 12 VR₁ = 12 ΩR₂ = 30 ΩR₃ = 6 ΩR₄ = 4 ΩR₅ = 10 ΩR₆ = 15 Ω

R_eq = Ω

3.Nodal analysisMedium

The voltage source Vₛ between A and B has its + terminal at B, so it makes V_B − V_A = Vₛ. Use nodal analysis with a supernode to find the node voltage V_B (measured from ground).

I_s = 4 AV_s = 9 VR₁ = 20 ΩR₂ = 12 Ω

V_B = V

4.Nodal analysisHard

The voltage source Vₛ between A and B has its + terminal at A, so it makes V_A − V_B = Vₛ. Use nodal analysis with a supernode to find the current I_x through the voltage source, from A to B as shown by the arrow.

V₁ = 10 VI_s = 1 AV_s = 15 VR₁ = 10 ΩR₂ = 10 ΩR₃ = 25 Ω

I_x = A

5.Mesh analysisMedium

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

V₁ = 12 VV₂ = 24 VR₁ = 15 ΩR₂ = 12 ΩR₃ = 20 Ω

I₁ = A

6.Mesh analysisHard

Use mesh analysis to find the voltage V_R2 across R₂ with the polarity shown.

V₁ = 12 VV₂ = 6 VR₁ = 2 ΩR₂ = 8 ΩR₃ = 10 ΩR₄ = 25 ΩR₅ = 20 Ω

V_R2 = V

7.SuperpositionMedium

Use superposition to find the power P_R2 absorbed by R₂.

V₁ = 9 VI_s = 2 AR₁ = 8 ΩR₂ = 2 ΩR₃ = 4 ΩR₄ = 10 Ω

P_R2 = W

8.SuperpositionHard

Use superposition to find the voltage V_R3 across R₃ with the polarity shown.

V₁ = 15 VI_s = 5 AV₂ = 9 VR₁ = 12 ΩR₂ = 30 ΩR₃ = 20 ΩR₄ = 15 Ω

V_R3 = V

9.Thévenin & NortonMedium

Find the Thévenin resistance R_th seen from terminals a–b.

V₁ = 18 VR₁ = 5 ΩR₂ = 8 ΩR₃ = 15 Ω

R_th = Ω

10.Thévenin & NortonHard

A load R_L = 2 Ω is connected across a–b. Use a Thévenin equivalent to find the power P_L delivered to the load.

V₁ = 6 VI_s = 5 AR₁ = 15 ΩR₂ = 4 ΩR₃ = 12 ΩR_L = 2 Ω

P_L = W

11.Op-amp circuitsMedium

Choose the feedback resistor R_f that gives this non-inverting amplifier a voltage gain of 2.

R₁ = 10 kΩA_v = 2

R_f = kΩ

12.Op-amp circuitsHard

The integrator uses an ideal op-amp with ±15 V supplies; its output saturates at ±12 V. The capacitor is uncharged when a constant input V_in = 0.6 V is applied at t = 0. How long after t = 0 does the output reach saturation? Find t.

V_in = 0.6 VR = 22 kΩC = 2.2 µFV_sat = ±12 V

t = ms

13.RC & RL transientsMedium

The inductor carries no current before the switch closes at t = 0. Find the current i at t = 2 ms.

V_s = 15 VR = 22 ΩL = 33 mHt = 2 ms

i = A

14.RC & RL transientsHard

Just before the switch closes at t = 0, the capacitor holds 20 V with the polarity shown. Find the capacitor voltage v_C at t = 250 ms.

V_s = 10 VR = 4.7 kΩC = 33 µFv_C(0) = 20 Vt = 250 ms

v_C = V

3Solutions

1.Series & parallel resistors

Answer: I_R1 = 0.4385 A

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.

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 = 2 + 12 = 14 Ω

R₂ ∥ 14 = 5·14/(5+14) = 3.684 Ω

R_eq = R₁ + 3.684 = 10 + 3.684 = 13.68 Ω

Current from the source

The source sees only R_eq, so Ohm’s law gives the current it delivers. That current flows through R₁.

I_R1 = V₁/R_eq = 6/13.68 = 0.4385 A

Answer

I_R1 = 0.4385 A

2.Series & parallel resistors

Answer: R_eq = 19.19 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 = 10 + 15 = 25 Ω

R₄ ∥ 25 = 3.448 Ω

R₃ + 3.448 = 9.448 Ω

R₂ ∥ 9.448 = 7.185 Ω

R_eq = R₁ + 7.185 = 12 + 7.185 = 19.19 Ω

Answer

R_eq = 19.19 Ω

3.Nodal analysis

Answer: V_B = 33.38 V

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.

Form a supernode

The voltage source between A and B carries an unknown current, so KCL cannot be written at A or B separately. Enclose A, B and the source in one supernode: the current entering it equals the current leaving it.

V_B = V_A + V_s = V_A + 9

V_A/R₁ + V_B/R₂ = I_s

Solve

V_A/20 + (V_A + 9)/12 = 4

V_A = (4 − 9/12)/(1/20 + 1/12) = 24.38 V

V_B = 24.38 + 9 = 33.38 V

Answer

V_B = 33.38 V

4.Nodal analysis

Answer: I_x = −1.167 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.

Form a supernode

V_B = V_A − 15

(V_A − V₁)/R₁ + V_A/R₂ + V_B/R₃ = I_s

Solve

(V_A − 10)/10 + V_A/10 + (V_A − 15)/25 = 1

V_A·(1/10 + 1/10 + 1/25) = 1 + 10/10 + 15/25 = 2.6

V_A = 2.6/0.24 = 10.83 V

V_B = −4.167 V

Current through the voltage source

KCL at node A alone: the current arriving through R₁ leaves through R₂ and through the voltage source.

I_R1 = (10 − 10.83)/10 = −0.08333 A

I_x = I_R1 − V_A/R₂ = −0.08333 − 10.83/10 = −1.167 A

Answer

I_x = −1.167 A

A negative result means the real current flows against the arrow.

5.Mesh analysis

Answer: I₁ = −0.1333 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.

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

15·I₁ + 20(I₁ − I₂) = 12

35·I₁ − 20·I₂ = 12

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

12·I₂ + 20(I₂ − I₁) = (−24)

−20·I₁ + 32·I₂ = −24

Solve

Δ = (35)(32) − (−20)(−20) = 720

I₁ = Δ₁/Δ = −96/720 = −0.1333 A

I₂ = Δ₂/Δ = −600/720 = −0.8333 A

Answer

I₁ = −0.1333 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: V_R2 = 3.444 V

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

2·I₁ + 25(I₁ − I₂) = 12

27·I₁ − 25·I₂ = 12

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

8·I₂ + 25(I₂ − I₁) + 20(I₂ − I₃) = 0

−25·I₁ + 53·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

10·I₃ + 20(I₃ − I₂) = (−6)

−20·I₂ + 30·I₃ = −6

Solve

Δ = det of the coefficient matrix = 13380

I₁ = Δ₁/Δ = 11280/13380 = 0.843 A

I₂ = Δ₂/Δ = 5760/13380 = 0.4305 A

I₃ = Δ₃/Δ = 1164/13380 = 0.087 A

Answer

I_R2 = I₂ = 0.4305 A

V_R2 = I_R2·R₂ = 0.4305·8 = 3.444 V

A negative mesh current flows anticlockwise.

7.Superposition

Answer: P_R2 = 6.722 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.

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 = 4+10 = 14 Ω

R₂ ∥ (R₃+R₄) = 2·14/(2+14) = 1.75 Ω

V_A = 9·1.75/(8+1.75) = 1.615 V

V_B = V_A·R₄/(R₃+R₄) = 1.615·10/14 = 1.154 V

I_R2′ = (V_A − 0)/R₂ = (1.615 − 0)/2 = 0.8077 A

Case 2 · I_s 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 = 8·2/(8+2) = 1.6 Ω

R_eq = R₄ ∥ (R₃+R₁∥R₂) = 10 ∥ 5.6 = 3.59 Ω

V_B = I_s·R_eq = 2·3.59 = 7.179 V

V_A = V_B·(R₁∥R₂)/(R₃+R₁∥R₂) = 7.179·1.6/5.6 = 2.051 V

I_R2″ = (V_A − 0)/R₂ = (2.051 − 0)/2 = 1.026 A

Add the current contributions

I_R2 = I_R2′ + I_R2″ = 0.8077 + 1.026 = 1.833 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.

P_R2 = I_R2²·R₂ = 1.833²·2 = 6.722 W

Answer

P_R2 = 6.722 W

Adding the powers of the separate cases would give 3.409 W, which is wrong.

8.Superposition

Answer: V_R3 = −17.46 V

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) = 20 ∥ 45 = 13.85 Ω

V_A = 15·13.85/(12+13.85) = 8.036 V

V_B = V_A·R₄/(R₂+R₄) = 8.036·15/45 = 2.679 V

V_R3′ = 0 − V_A = 0 − 8.036 = −8.036 V

Case 2 · I_s 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 = 7.5 Ω, so the left path is R₂+7.5 = 37.5 Ω

R_eq = R₄ ∥ 37.5 = 10.71 Ω

V_B = I_s·R_eq = 5·10.71 = 53.57 V

V_A = V_B·7.5/37.5 = 10.71 V

V_R3″ = 0 − V_A = 0 − 10.71 = −10.71 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 = 12·20/(12+20) = 7.5 Ω

V_B = V₂·(R₂+R₁∥R₃)/(R₄+R₂+R₁∥R₃) = (−9)·37.5/52.5 = −6.429 V

V_A = V_B·(R₁∥R₃)/(R₂+R₁∥R₃) = −6.429·7.5/37.5 = −1.286 V

V_R3‴ = 0 − V_A = 0 − −1.286 = 1.286 V

Add the contributions

V_R3 = V_R3′ + V_R3″ + V_R3‴

V_R3 = (−8.036) + (−10.71) + 1.286 = −17.46 V

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

9.Thévenin & Norton

Answer: R_th = 18.08 Ω

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

Find R_th 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 = 5·8/(5+8) = 3.077 Ω

R_th = (R₁ ∥ R₂) + R₃ = 3.077 + 15 = 18.08 Ω

10.Thévenin & Norton

Answer: P_L = 40.33 W

Find R_th 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 = 15·4/(15+4) = 3.158 Ω

R_th = (R₁ ∥ R₂) + R₃ = 3.158 + 12 = 15.16 Ω

Find V_th, 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.

V_A = V₁·R₂/(R₁+R₂) + I_s·(R₁∥R₂) = 1.263 + 5·3.158 = 17.05 V

V_th = V_B = V_A + I_s·R₃ = 17.05 + 5·12 = 77.05 V

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.

I_L = V_th/(R_th + R_L) = 77.05/(15.16 + 2) = 4.491 A

P_L = I_L²·R_L = 4.491²·2 = 40.33 W

11.Op-amp circuits

Answer: R_f = 10 kΩ

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.

Gain of a non-inverting amplifier

V₋ = V₊ = V_in, and the current through R₁ also flows through R_f, so V_out = V_in·(1 + R_f/R₁).

A_v = 1 + R_f/R₁

R_f = (2 − 1)·10 = 10 kΩ

Answer

R_f = 10 kΩ

12.Op-amp circuits

Answer: t = 968 ms

Input current and ramp rate

kΩ times µF gives milliseconds.

RC = 22 kΩ · 2.2 µF = 48.4 ms

ramp rate = −V_in/(RC) = −0.6/48.4 = −0.0124 V/ms

Time to saturation

|V_in|·t/(RC) = V_sat → t = V_sat·RC/|V_in| = 12·48.4/0.6 = 968 ms

Answer

t = 968 ms

13.RC & RL transients

Answer: i = 0.5021 A

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 = 33 mH / 22 Ω = 1.5 ms

i(∞) = V_s/R = 15/22 = 0.6818 A

Current at the given time

i(t) = i(∞)·(1 − e^−t/τ) = 0.6818·(1 − e^−2/1.5) = 0.5021 A

Answer

i = 0.5021 A

14.RC & RL transients

Answer: v_C = 12 V

Initial value, final value, time constant

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

v_C(0) = 20 V, v_C(∞) = V_s = 10 V

τ = R·C = 4.7 kΩ · 33 µF = 155.1 ms

v_C(t) = 10 + (20 − 10)·e^−t/τ

Voltage at the given time

v_C(250) = 10 + (10)·e^−250/155.1 = 12 V

Answer

v_C = 12 V

ExamProblems · examproblems.com/courses/circuit-analysis-1

Circuit Analysis 1: cram pack

1Formulas and methods

Series & parallel resistors

Method

Nodal analysis

Method

Mesh analysis

Method

Superposition

Method

Thévenin & Norton

Method

Op-amp circuits

RC & RL transients

Method

2Practice problems

3Solutions

Combine resistors from the far end

Current from the source

Answer

Combine resistors from the far end

Answer

Form a supernode

Solve

Answer

Form a supernode

Solve

Current through the voltage source

Answer

Assign mesh currents

KVL around mesh 1

KVL around mesh 2

Solve

Answer

Assign mesh currents

KVL around mesh 1

KVL around mesh 2

KVL around mesh 3

Solve

Answer

Case 1 · V1 acting alone

Case 2 · Is acting alone

Add the current contributions

Power in the resistor

Answer

Case 1 · V1 acting alone

Case 2 · Is acting alone

Case 3 · V2 acting alone

Add the contributions

Find Rth with every source switched off

Find Rth with every source switched off

Find Vth, the open-circuit voltage

Connect the load to the Thévenin equivalent

Gain of a non-inverting amplifier

Answer

Input current and ramp rate

Time to saturation

Answer

Time constant and final current

Current at the given time

Answer

Initial value, final value, time constant

Voltage at the given time

Answer

Case 1 · V₁ acting alone

Case 2 · I_s acting alone

Case 1 · V₁ acting alone

Case 2 · I_s acting alone

Case 3 · V₂ acting alone

Find R_th with every source switched off

Find R_th with every source switched off

Find V_th, the open-circuit voltage