KVL, KCL, and Mesh/Node Analysis
Kirchhoff's Voltage Law (KVL)
\sum_{k} V_k = 0 \quad (\text{around any closed loop})
| Symbol | Description | Unit |
|---|---|---|
| V_k | Voltage across k-th element in loop | V |
Worked example
A loop has a 12 V source, a 4 Ω resistor, and a 2 Ω resistor in series. Find current.
Given: V_s = 12 V, R1 = 4 Ω, R2 = 2 Ω
- KVL: 12 − 4I − 2I = 0
- 6I = 12
- I = 2 A
Answer: I = 2 A
Kirchhoff's Current Law (KCL)
\sum_{k} I_k = 0 \quad (\text{at any node})
| Symbol | Description | Unit |
|---|---|---|
| I_k | Current entering/leaving node k | A |
Worked example
Node A has currents: 3 A entering, 1 A leaving through R1, I2 leaving through R2. Find I2.
Given: I_in = 3 A, I_R1 = 1 A
- KCL: 3 − 1 − I2 = 0
- I2 = 2 A
Answer: I2 = 2 A
Mesh Resistance Matrix (2-mesh)
\begin{bmatrix} R_{11} & -R_{12} \\ -R_{21} & R_{22} \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \end{bmatrix} = \begin{bmatrix} V_1 \\ V_2 \end{bmatrix}
| Symbol | Description | Unit |
|---|---|---|
| R_{11} | Sum of resistances in mesh 1 | Ω |
| R_{22} | Sum of resistances in mesh 2 | Ω |
| R_{12} | Shared resistance between mesh 1 and 2 | Ω |
Worked example
Mesh 1: 10 V source, R1=2 Ω, R3=1 Ω. Mesh 2: 6 V source, R2=3 Ω, R3=1 Ω shared. Find I1, I2.
Given: R11=3, R22=4, R12=1, V1=10, V2=6
- [3, -1; -1, 4][I1; I2] = [10; 6]
- From row1: 3I1 − I2 = 10 ... (i)
- From row2: −I1 + 4I2 = 6 ... (ii)
- From (i): I2 = 3I1 − 10; sub into (ii): −I1 + 4(3I1−10)=6
- −I1 + 12I1 − 40 = 6 → 11I1 = 46 → I1 = 4.18 A
- I2 = 3(4.18)−10 = 2.55 A
Answer: I1 = 4.18 A, I2 = 2.55 A
Thevenin and Norton Equivalents
Thevenin Voltage
V_{th} = V_{oc}
| Symbol | Description | Unit |
|---|---|---|
| V_{th} | Thevenin equivalent voltage | V |
| V_{oc} | Open-circuit voltage at terminals | V |
Worked example
Find Vth at terminals A-B: 20 V source in series with 5 Ω and 10 Ω. Load connected across 10 Ω.
Given: Vs = 20 V, R1 = 5 Ω, R2 = 10 Ω
- With load removed (OC), current through divider: I = 20/(5+10) = 1.333 A
- V_th = V_oc = I × R2 = 1.333 × 10 = 13.33 V
Answer: V_th = 13.33 V
Thevenin Resistance
R_{th} = \frac{V_{oc}}{I_{sc}}
| Symbol | Description | Unit |
|---|---|---|
| R_{th} | Thevenin equivalent resistance | Ω |
| V_{oc} | Open-circuit voltage | V |
| I_{sc} | Short-circuit current | A |
Worked example
Using the same circuit, find I_sc and then R_th.
Given: Vs = 20 V, R1 = 5 Ω, R2 = 10 Ω, V_th = 13.33 V
- When AB shorted: R2 is shorted, so I_sc = Vs/R1 = 20/5 = 4 A
- R_th = V_oc/I_sc = 13.33/4 = 3.33 Ω
- Alternatively: R_th = R1||R2 = (5×10)/(5+10) = 3.33 Ω
Answer: R_th = 3.33 Ω
Norton Current
I_N = I_{sc} = \frac{V_{th}}{R_{th}}
| Symbol | Description | Unit |
|---|---|---|
| I_N | Norton equivalent current | A |
| I_{sc} | Short-circuit current at terminals | A |
Worked example
From the Thevenin equivalent with Vth=13.33 V and Rth=3.33 Ω, find Norton current.
Given: V_th = 13.33 V, R_th = 3.33 Ω
- I_N = V_th / R_th = 13.33 / 3.33 = 4 A
Answer: I_N = 4 A
Maximum Power Transfer
P_{max} = \frac{V_{th}^2}{4 R_{th}}, \quad R_L = R_{th}
| Symbol | Description | Unit |
|---|---|---|
| P_{max} | Maximum power delivered to load | W |
| R_L | Load resistance (equals R_th for max transfer) | Ω |
| V_{th} | Thevenin voltage | V |
Worked example
With Vth = 13.33 V and Rth = 3.33 Ω, find maximum power in load.
Given: V_th = 13.33 V, R_th = 3.33 Ω
- P_max = Vth² / (4 × Rth)
- P_max = (13.33)² / (4 × 3.33)
- P_max = 177.7 / 13.33 = 13.33 W
Answer: P_max = 13.33 W at R_L = 3.33 Ω
AC Circuit Analysis — Impedance and Phasors
Impedance of Series RLC
Z = R + j\left(\omega L - \frac{1}{\omega C}\right)
| Symbol | Description | Unit |
|---|---|---|
| Z | Total impedance | Ω |
| \omega | Angular frequency | rad/s |
| L | Inductance | H |
| C | Capacitance | F |
Worked example
Series RLC: R=10 Ω, L=50 mH, C=100 µF, f=100 Hz. Find Z.
Given: R=10 Ω, L=0.05 H, C=100×10^−6 F, ω=2π×100=628.3 rad/s
- X_L = ωL = 628.3 × 0.05 = 31.42 Ω
- X_C = 1/(ωC) = 1/(628.3 × 100×10^−6) = 15.92 Ω
- X_net = 31.42 − 15.92 = 15.5 Ω
- Z = 10 + j15.5 Ω
- |Z| = √(10² + 15.5²) = √(100+240.25) = √340.25 = 18.45 Ω
Answer: Z = 10 + j15.5 Ω, |Z| = 18.45 Ω
Admittance
Y = \frac{1}{Z} = G + jB
| Symbol | Description | Unit |
|---|---|---|
| Y | Admittance | S |
| G | Conductance | S |
| B | Susceptance | S |
Worked example
Z = 4 + j3 Ω. Find admittance Y.
Given: Z = 4+j3 Ω
- Y = 1/(4+j3) = (4−j3)/[(4+j3)(4−j3)] = (4−j3)/(16+9)
- Y = (4−j3)/25 = 0.16 − j0.12 S
Answer: Y = 0.16 − j0.12 S
Power Factor
pf = \cos\phi = \frac{R}{|Z|}
| Symbol | Description | Unit |
|---|---|---|
| \phi | Phase angle between voltage and current | ° |
| R | Resistance | Ω |
| |Z| | Impedance magnitude | Ω |
Worked example
Z = 10 + j15.5 Ω, |Z| = 18.45 Ω. Find power factor.
Given: R = 10 Ω, |Z| = 18.45 Ω
- pf = R/|Z| = 10/18.45 = 0.542
- φ = cos⁻¹(0.542) = 57.2° lagging
Answer: pf = 0.542 lagging
Resonance
Resonant Frequency
f_0 = \frac{1}{2\pi\sqrt{LC}}
| Symbol | Description | Unit |
|---|---|---|
| f_0 | Resonant frequency | Hz |
| L | Inductance | H |
| C | Capacitance | F |
Worked example
Series resonant circuit: L = 10 mH, C = 10 µF. Find f0.
Given: L = 0.01 H, C = 10×10^−6 F
- f0 = 1 / (2π√(0.01 × 10×10^−6))
- √(10^−7) = 3.162×10^−4
- f0 = 1 / (2π × 3.162×10^−4) = 1 / (1.987×10^−3) = 503.3 Hz
Answer: f0 = 503.3 Hz
Quality Factor (Series)
Q = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}
| Symbol | Description | Unit |
|---|---|---|
| Q | Quality factor | |
| \omega_0 | Resonant angular frequency | rad/s |
| R | Series resistance | Ω |
Worked example
Series RLC: R=5 Ω, L=10 mH, C=10 µF. Find Q at resonance.
Given: R=5 Ω, L=0.01 H, C=10×10^−6 F
- √(L/C) = √(0.01/10×10^−6) = √1000 = 31.62
- Q = 31.62 / 5 = 6.32
Answer: Q = 6.32
Bandwidth
BW = \frac{f_0}{Q} = \frac{R}{2\pi L}
| Symbol | Description | Unit |
|---|---|---|
| BW | 3 dB bandwidth | Hz |
| f_0 | Resonant frequency | Hz |
| Q | Quality factor |
Worked example
f0 = 503.3 Hz, Q = 6.32. Find bandwidth.
Given: f0 = 503.3 Hz, Q = 6.32
- BW = f0/Q = 503.3/6.32 = 79.6 Hz
Answer: BW = 79.6 Hz
Half-Power Frequencies
f_{1,2} = f_0 \mp \frac{BW}{2}
| Symbol | Description | Unit |
|---|---|---|
| f_1 | Lower half-power frequency | Hz |
| f_2 | Upper half-power frequency | Hz |
Worked example
f0 = 503.3 Hz, BW = 79.6 Hz. Find f1 and f2.
Given: f0 = 503.3 Hz, BW = 79.6 Hz
- f1 = 503.3 − 79.6/2 = 503.3 − 39.8 = 463.5 Hz
- f2 = 503.3 + 39.8 = 543.1 Hz
Answer: f1 = 463.5 Hz, f2 = 543.1 Hz
Two-Port Network Parameters
Z-Parameters (Open Circuit)
\begin{bmatrix} V_1 \\ V_2 \end{bmatrix} = \begin{bmatrix} z_{11} & z_{12} \\ z_{21} & z_{22} \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \end{bmatrix}
| Symbol | Description | Unit |
|---|---|---|
| z_{11} | Input impedance with output open | Ω |
| z_{21} | Forward transfer impedance | Ω |
Worked example
T-network: Z1=10 Ω, Z2=10 Ω, Z3=20 Ω. Find z11 and z21.
Given: Z1=10 Ω, Z2=10 Ω, Z3=20 Ω (shunt)
- z11 = Z1 + Z3 = 10 + 20 = 30 Ω
- z21 = Z3 = 20 Ω
Answer: z11 = 30 Ω, z21 = 20 Ω
Y-Parameters (Short Circuit)
\begin{bmatrix} I_1 \\ I_2 \end{bmatrix} = \begin{bmatrix} y_{11} & y_{12} \\ y_{21} & y_{22} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \end{bmatrix}
| Symbol | Description | Unit |
|---|---|---|
| y_{11} | Input admittance with output shorted | S |
| y_{21} | Forward transfer admittance | S |
Worked example
Pi-network: Ya=0.1 S, Yb=0.1 S, Yc=0.05 S. Find y11.
Given: Ya=0.1 S, Yb=0.1 S, Yc=0.05 S (series)
- y11 = Ya + Yc = 0.1 + 0.05 = 0.15 S
- y21 = −Yc = −0.05 S
Answer: y11 = 0.15 S, y21 = −0.05 S
ABCD Transmission Parameters
\begin{bmatrix} V_1 \\ I_1 \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} V_2 \\ -I_2 \end{bmatrix}
| Symbol | Description | Unit |
|---|---|---|
| A | Voltage ratio (OC output) | |
| B | Transfer impedance (SC output) | Ω |
| C | Transfer admittance (OC output) | S |
| D | Current ratio (SC output) |
Worked example
Series impedance Z=j10 Ω (L-section). Find ABCD parameters.
Given: Series element Z = j10 Ω, no shunt element
- A = 1, B = Z = j10 Ω
- C = 0, D = 1 (for series element only)
Answer: A=1, B=j10 Ω, C=0, D=1
Transient Response (RL, RC, RLC)
RL Circuit Time Constant
\tau = \frac{L}{R}
| Symbol | Description | Unit |
|---|---|---|
| \tau | Time constant | s |
| L | Inductance | H |
| R | Resistance | Ω |
Worked example
RL series circuit: R=100 Ω, L=50 mH. Find time constant and current at t=τ when 10 V applied.
Given: R=100 Ω, L=0.05 H, V=10 V
- τ = L/R = 0.05/100 = 0.5 ms
- I_final = V/R = 10/100 = 0.1 A
- i(τ) = I_final(1 − e^−1) = 0.1×0.632 = 63.2 mA
Answer: τ = 0.5 ms; i(τ) = 63.2 mA
RC Circuit Time Constant
\tau = RC
| Symbol | Description | Unit |
|---|---|---|
| \tau | Time constant | s |
| R | Resistance | Ω |
| C | Capacitance | F |
Worked example
RC circuit: R=10 kΩ, C=22 µF. Find τ and capacitor voltage at t=2τ for 5 V step input.
Given: R=10×10^3 Ω, C=22×10^−6 F, V=5 V
- τ = RC = 10000 × 22×10^−6 = 0.22 s
- v_C(2τ) = 5(1 − e^−2) = 5×(1 − 0.1353) = 5×0.8647 = 4.32 V
Answer: τ = 0.22 s; v_C(2τ) = 4.32 V
RLC Damping Coefficient
\alpha = \frac{R}{2L}, \quad \omega_0 = \frac{1}{\sqrt{LC}}
| Symbol | Description | Unit |
|---|---|---|
| \alpha | Damping coefficient | Np/s |
| \omega_0 | Undamped natural frequency | rad/s |
Worked example
Series RLC: R=10 Ω, L=0.1 H, C=100 µF. Find α, ω0 and determine response type.
Given: R=10 Ω, L=0.1 H, C=100×10^−6 F
- α = R/(2L) = 10/(2×0.1) = 50 Np/s
- ω0 = 1/√(0.1×100×10^−6) = 1/√(10^−5) = 316.2 rad/s
- Since α=50 < ω0=316.2 → underdamped response
Answer: α=50 Np/s, ω0=316.2 rad/s, underdamped
Network Theorems — Superposition and Reciprocity
Superposition Principle
V_{total} = V_{source1-only} + V_{source2-only} + \cdots
| Symbol | Description | Unit |
|---|---|---|
| V_{total} | Response due to all sources | V |
Worked example
Two sources: 10 V and 6 V in a bridge. Resistor R=4 Ω carries current I1=2 A due to 10 V alone and I2=−0.5 A due to 6 V alone. Find total current.
Given: I1 = 2 A, I2 = −0.5 A
- I_total = I1 + I2 = 2 + (−0.5) = 1.5 A
Answer: I_total = 1.5 A
Reciprocity Theorem
\frac{V_1}{I_2}\bigg|_{V_2=0} = \frac{V_2}{I_1}\bigg|_{V_1=0} = z_{12} = z_{21}
| Symbol | Description | Unit |
|---|---|---|
| z_{12} | Transfer impedance port 1→2 | Ω |
| z_{21} | Transfer impedance port 2→1 | Ω |
Worked example
A passive network with V1=20 V produces I2=4 A. Verify reciprocity by finding I1 when V2=20 V.
Given: V1=20 V gives I2=4 A
- z21 = V1/I2 = 20/4 = 5 Ω
- By reciprocity z12 = z21 = 5 Ω
- When V2 = 20 V: I1 = V2/z12 = 20/5 = 4 A
Answer: I1 = 4 A (confirms reciprocity)
Quick reference
| Formula | Expression |
|---|---|
| KVL | \sum V_k = 0 |
| KCL | \sum I_k = 0 |
| Thevenin Voltage | V_{th} = V_{oc} |
| Thevenin Resistance | R_{th} = V_{oc}/I_{sc} |
| Norton Current | I_N = V_{th}/R_{th} |
| Max Power Transfer | P_{max} = V_{th}^2/(4R_{th}) |
| Series RLC Impedance | Z = R + j(\omega L - 1/\omega C) |
| Resonant Frequency | f_0 = 1/(2\pi\sqrt{LC}) |
| Quality Factor | Q = (1/R)\sqrt{L/C} |
| Bandwidth | BW = f_0/Q |
| RL Time Constant | \tau = L/R |
| RC Time Constant | \tau = RC |
| Damping Coefficient | \alpha = R/(2L) |
| Power Factor | pf = R/|Z| |
| Admittance | Y = 1/Z |
Exam tips
- GATE network analysis questions heavily test Thevenin equivalents — always deactivate independent sources (short V-sources, open I-sources) when finding Rth.
- For resonance problems, examiners often ask for Q, BW, and half-power frequencies together — memorise the trio f0, Q, and BW = f0/Q.
- Two-port parameter conversion questions appear in ESE; know the conversion between Z, Y, and ABCD parameters using the standard tables.
- Transient response type (over/under/critically damped) depends on α vs ω0 — the comparison α² vs ω0² is the deciding condition, always check this first.
- Mesh analysis matrix setup errors (wrong sign on mutual resistance) are the most common mistake in university papers — R12 is always subtracted.
- Maximum power transfer efficiency is only 50% — examiners frequently ask this as a follow-up to the condition R_L = R_th.