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 = 24 VR₁ = 2 ΩR₂ = 6 ΩR₃ = 10 ΩR₄ = 12 Ω

I_R1 = A

2.Series & parallel resistorsHard

Find the current I_R2 through R₂.

V₁ = 36 VR₁ = 15 ΩR₂ = 2 ΩR₃ = 8 ΩR₄ = 30 ΩR₅ = 10 ΩR₆ = 8 Ω

I_R2 = A

3.Nodal analysisMedium

Three branches are connected between node A and ground, as shown. Use nodal analysis to find the current I₂ in branch 2, flowing up into node A in the direction of the arrow.

V₁ = 9 VR₁ = 20 ΩV₂ = 24 VR₂ = 20 ΩR₃ = 4 Ω

I₂ = A

4.Nodal analysisHard

The current source is adjusted until the node voltage at A is V_A = 5 V. Use nodal analysis to find the source current I_s (direction as shown).

V₁ = 12 VR₁ = 20 ΩR₂ = 3 ΩR₃ = 8 ΩR₄ = 2 ΩV_A = 5 V

I_s = A

5.Mesh analysisMedium

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

V₁ = 12 VV₂ = 15 VR₁ = 8 ΩR₂ = 20 ΩR₃ = 5 Ω

V_R2 = V

6.Mesh analysisHard

Use mesh analysis to find the current I_R5 through R₅ in the direction of the arrow.

V₁ = 15 VV₂ = 30 VR₁ = 15 ΩR₂ = 10 ΩR₃ = 30 ΩR₄ = 25 ΩR₅ = 6 Ω

I_R5 = A

7.SuperpositionMedium

Find the value of the current source I_s (direction as shown) that makes the current I_R3 through R₃ equal to zero.

V₁ = 9 VR₁ = 5 ΩR₂ = 20 ΩR₃ = 4 ΩR₄ = 15 Ω

I_s = A

8.SuperpositionHard

A linear resistive circuit contains three independent sources: V₁, V₂ and I_s. The current I through a load is measured three times. With all sources at their normal values, I = 18 mA. With V₁ doubled and the other two unchanged, I = 20 mA. With the direction of I_s reversed and the other two unchanged, I = 3 mA. Find the load current I when V₂ is switched off and V₁ and I_s have their normal values.

normal: I = 18 mA2·V₁: I = 20 mA−I_s: I = 3 mA

I = mA

9.Thévenin & NortonMedium

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

V₁ = 9 VR₁ = 6 ΩR₂ = 15 ΩR₃ = 20 Ω

I_N = A

10.Thévenin & NortonHard

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

V₁ = 12 VI_s = 2 AR₁ = 6 ΩR₂ = 20 ΩR₃ = 25 Ω

I_N = A

11.Op-amp circuitsMedium

Choose the feedback resistor R_f that gives this non-inverting amplifier a voltage gain of 15.

R₁ = 4.7 kΩA_v = 15

R_f = kΩ

12.Op-amp circuitsHard

The integrator uses an ideal op-amp with ±15 V supplies; its output saturates at ±13 V. The capacitor is uncharged when a constant input V_in = 1.4 V is applied at t = 0. How long after t = 0 does the output reach saturation? Find t.

V_in = 1.4 VR = 47 kΩC = 0.22 µFV_sat = ±13 V

t = ms

13.RC & RL transientsMedium

The inductor carries a steady current of 1 A. At t = 0 the switch closes the loop through R and the source is removed, so the inductor current decays through R. Find the current i at t = 2 ms.

i(0) = 1 AR = 100 ΩL = 150 mHt = 2 ms

i = A

14.RC & RL transientsHard

Just before the switch closes at t = 0, the capacitor holds 12 V with the polarity shown. Find the capacitor voltage v_C at t = 75 ms.

V_s = 6 VR = 2.2 kΩC = 68 µFv_C(0) = 12 Vt = 75 ms

v_C = V

3Solutions

1.Series & parallel resistors

Answer: I_R1 = 3.574 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 = 10 + 12 = 22 Ω

R₂ ∥ 22 = 6·22/(6+22) = 4.714 Ω

R_eq = R₁ + 4.714 = 2 + 4.714 = 6.714 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 = 24/6.714 = 3.574 A

Answer

I_R1 = 3.574 A

2.Series & parallel resistors

Answer: I_R2 = 1.94 A

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 + 8 = 18 Ω

R₄ ∥ 18 = 11.25 Ω

R₃ + 11.25 = 19.25 Ω

R₂ ∥ 19.25 = 1.812 Ω

R_eq = R₁ + 1.812 = 15 + 1.812 = 16.81 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 = 36/16.81 = 2.141 A

Current through R₂

Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.

V_A = V₁ − I_R1·R₁ = 36 − 2.141·15 = 3.88 V

I_R2 = V_A/R₂ = 3.88/2 = 1.94 A

Answer

I_R2 = 1.94 A

3.Nodal analysis

Answer: I₂ = −1.093 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

Take the bottom wire as ground. Each branch carries (V_A − V_k)/R_k away from A, where V_k is the voltage of its source (zero for a plain resistor). The currents leaving A add up to zero.

(V_A − 9)/20 + (V_A − (−24))/20 + V_A/4 = 0

Solve for V_A

V_A = (9/20 + (−24)/20 + 0/4) / (1/20 + 1/20 + 1/4) = −0.75/0.35 = −2.143 V

Branch current

The arrow points up through the branch, into node A.

I₂ = ((−24) − V_A)/R₂ = ((−24) − (−2.143))/20 = −1.093 A

Answer

I₂ = −1.093 A

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

4.Nodal analysis

Answer: I_s = 9.083 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

V_A is known, so this equation gives V_B.

(V_A − V₁)/R₁ + V_A/R₂ + (V_A − V_B)/R₃ = 0

(5 − 12)/20 + 5/3 = (V_B − 5)/8

1.317 = (V_B − 5)/8 → V_B = 15.53 V

KCL at node B

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

I_s = (V_B − V_A)/R₃ + V_B/R₄ = (15.53 − 5)/8 + 15.53/2 = 9.083 A

Answer

I_s = 9.083 A

5.Mesh analysis

Answer: V_R2 = −9 V

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

8·I₁ + 5(I₁ − I₂) = 12

13·I₁ − 5·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

20·I₂ + 5(I₂ − I₁) = (−15)

−5·I₁ + 25·I₂ = −15

Solve

Δ = (13)(25) − (−5)(−5) = 300

I₁ = Δ₁/Δ = 225/300 = 0.75 A

I₂ = Δ₂/Δ = −135/300 = −0.45 A

Answer

I_R2 = I₂ = −0.45 A

V_R2 = I_R2·R₂ = −0.45·20 = −9 V

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: I_R5 = 0.9829 A

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

15·I₁ + 25(I₁ − I₂) = 15

40·I₁ − 25·I₂ = 15

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

10·I₂ + 25(I₂ − I₁) + 6(I₂ − I₃) = 0

−25·I₁ + 41·I₂ − 6·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

30·I₃ + 6(I₃ − I₂) = (−30)

−6·I₂ + 36·I₃ = −30

Solve

Δ = det of the coefficient matrix = 35100

I₁ = Δ₁/Δ = 17100/35100 = 0.4872 A

I₂ = Δ₂/Δ = 6300/35100 = 0.1795 A

I₃ = Δ₃/Δ = −28200/35100 = −0.8034 A

Answer

I_R5 = I₂ − I₃ = 0.9829 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: I_s = 0.48 A

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+15 = 19 Ω

R₂ ∥ (R₃+R₄) = 20·19/(20+19) = 9.744 Ω

V_A = 9·9.744/(5+9.744) = 5.948 V

V_B = V_A·R₄/(R₃+R₄) = 5.948·15/19 = 4.696 V

I_R3′ = (V_A − V_B)/R₃ = (5.948 − 4.696)/4 = 0.313 A

Case 2 · I_s acting alone

V₁ becomes a short, so R₁ ∥ R₂ hangs from A to ground. Iₛ splits at B between R₄ and the path R₃ + R₁∥R₂; the part through R₃ flows from B to A, against the arrow.

R₁ ∥ R₂ = 4 Ω

I_R3″ = −I_s·R₄/(R₄ + R₃ + R₁∥R₂) = −I_s·15/23 = −0.6522·I_s

Set the total to zero

I_R3′ + I_R3″ = 0.313 − 0.6522·I_s = 0

I_s = 0.313/0.6522 = 0.48 A

Answer

I_s = 0.48 A

8.Superposition

Answer: I = 9.5 mA

Write the measurements with contributions

Each contribution is proportional to its own source.

I₁ + I₂ + I₃ = 18

2·I₁ + I₂ + I₃ = 20

I₁ + I₂ − I₃ = 3

Solve for the contributions

I₁ = 20 − 18 = 2 mA

I₃ = (18 − 3)/2 = 7.5 mA

I₂ = 18 − 2 − 7.5 = 8.5 mA

Switch V₂ off

I = I₁ + I₃ = 2 + 7.5 = 9.5 mA

Answer

I = 9.5 mA

9.Thévenin & Norton

Answer: I_N = 0.2647 A

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 = 6·15/(6+15) = 4.286 Ω

R_th = (R₁ ∥ R₂) + R₃ = 4.286 + 20 = 24.29 Ω

Find V_th, 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.

V_th = V₁·R₂/(R₁+R₂) = 9·15/(6+15) = 6.429 V

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.

I_N = V_th/R_th = 6.429/24.29 = 0.2647 A

10.Thévenin & Norton

Answer: I_N = 2.312 A

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 = 6·20/(6+20) = 4.615 Ω

R_th = (R₁ ∥ R₂) + R₃ = 4.615 + 25 = 29.62 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) = 9.231 + 2·4.615 = 18.46 V

V_th = V_B = V_A + I_s·R₃ = 18.46 + 2·25 = 68.46 V

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.

I_N = V_th/R_th = 68.46/29.62 = 2.312 A

11.Op-amp circuits

Answer: R_f = 65.8 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 = (15 − 1)·4.7 = 65.8 kΩ

Answer

R_f = 65.8 kΩ

12.Op-amp circuits

Answer: t = 96.01 ms

Input current and ramp rate

kΩ times µF gives milliseconds.

RC = 47 kΩ · 0.22 µF = 10.34 ms

ramp rate = −V_in/(RC) = −1.4/10.34 = −0.1354 V/ms

Time to saturation

|V_in|·t/(RC) = V_sat → t = V_sat·RC/|V_in| = 13·10.34/1.4 = 96.01 ms

Answer

t = 96.01 ms

13.RC & RL transients

Answer: i = 0.2636 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

τ = L/R = 150 mH / 100 Ω = 1.5 ms

Decay of the current

i(t) = 1·e^−2/1.5 = 0.2636 A

Answer

i = 0.2636 A

14.RC & RL transients

Answer: v_C = 9.634 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) = 12 V, v_C(∞) = V_s = 6 V

τ = R·C = 2.2 kΩ · 68 µF = 149.6 ms

v_C(t) = 6 + (12 − 6)·e^−t/τ

Voltage at the given time

v_C(75) = 6 + (6)·e^−75/149.6 = 9.634 V

Answer

v_C = 9.634 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

Current from the source

Current through R₂

Answer

KCL at node A

Solve for VA

Branch current

Answer

KCL at node A

KCL at node B

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

Set the total to zero

Answer

Write the measurements with contributions

Solve for the contributions

Switch V₂ off

Answer

Find Rth with every source switched off

Find Vth, the open-circuit voltage

Norton equivalent

Find Rth with every source switched off

Find Vth, the open-circuit voltage

Norton equivalent

Gain of a non-inverting amplifier

Answer

Input current and ramp rate

Time to saturation

Answer

Time constant

Decay of the current

Answer

Initial value, final value, time constant

Voltage at the given time

Answer

Solve for V_A

Case 1 · V₁ acting alone

Case 2 · I_s acting alone

Find R_th with every source switched off

Find V_th, the open-circuit voltage

Find R_th with every source switched off

Find V_th, the open-circuit voltage