Circuit Theory (Electrical) — Overview

Circuit theory is the grammar of electrical engineering — every device from a mobile charger to a power grid is governed by the same few laws about voltage, current, and energy that you are about to master completely.

Foundational Quantities

Definition: Charge (Q) — the fundamental electrical quantity. Unit: Coulomb (C). One electron carries 1.6 × 10⁻¹⁹ C.

Definition: Current (I) — rate of flow of charge. I = Q/t. Unit: Ampere (A). 1 A = 1 C/s. Conventional current flows from + to −; electron flow is opposite.

Definition: Voltage / Potential Difference (V) — energy per unit charge. V = W/Q. Unit: Volt (V). It is the "pressure" that drives current through a circuit.

Definition: Resistance (R) — opposition to current flow. Unit: Ohm (Ω). Depends on material, length, cross-section: R = ρL/A.

Ohm's Law

Ohm's Law: V = IR — the voltage across a conductor is proportional to the current flowing through it, provided temperature is constant.

The three derived forms:

• V = IR (find voltage)
• I = V/R (find current)
• R = V/I (find resistance)

Power:
P = VI = I²R = V²/R. Unit: Watt (W). Energy = P × t = VIt. Unit: Joule (J) or Watt-hour.

Real-world example: A 60 W bulb connected to a 240 V supply draws I = P/V = 60/240 = 0.25 A. Its resistance = V/I = 240/0.25 = 960 Ω.

Common misconception: Ohm's Law is not universal — it only holds for ohmic (linear) resistors. Semiconductor diodes, transistors, and bulb filaments at varying temperatures are non-ohmic devices.

Kirchhoff's Laws

These two laws are the backbone for analysing any circuit, however complex.

KCL — Kirchhoff's Current Law: The algebraic sum of currents at any node (junction) is zero. Equivalently: sum of currents entering a node = sum of currents leaving it. Based on conservation of charge.

KVL — Kirchhoff's Voltage Law: The algebraic sum of all voltages around any closed loop is zero. Based on conservation of energy. (What goes up in potential must come back down.)

Question: In a series circuit, a 12 V battery drives current through R₁ = 4 Ω and R₂ = 8 Ω. Find the current and voltage across each resistor.
Solution:
Step 1: Total resistance = 4 + 8 = 12 Ω (series).
Step 2: I = V/R = 12/12 = 1 A (same current everywhere in series — KCL).
Step 3: V₁ = IR₁ = 1 × 4 = 4 V; V₂ = 1 × 8 = 8 V.
Conclusion: I = 1 A; V₁ = 4 V; V₂ = 8 V. Check: 4 + 8 = 12 V ✓ (KVL satisfied).

Resistors — Series and Parallel

Series combination:

• R_total = R₁ + R₂ + R₃ + ...
• Same current through all; voltages add up.
• Total resistance always greater than the largest individual resistor.

Parallel combination:

• 1/R_total = 1/R₁ + 1/R₂ + ...
• For two resistors: R_total = (R₁ × R₂)/(R₁ + R₂)
• Same voltage across all; currents add up.
• Total resistance always less than the smallest individual resistor.

Why it matters: Household wiring uses parallel connections so each appliance gets the full mains voltage (220 V) and can be switched on/off independently.

Capacitors

Definition: Capacitance (C) — ability to store charge per unit voltage. C = Q/V. Unit: Farad (F). In practice, microfarad (µF) and picofarad (pF) are common.

• Parallel combination: C_total = C₁ + C₂ + ... (capacitors add in parallel — same as resistors in series).
• Series combination: 1/C_total = 1/C₁ + 1/C₂ + ... (capacitors in series behave like resistors in parallel).
• Energy stored: E = ½CV².
• Capacitors block DC (after charging) but pass AC — used in filtering, timing, and coupling circuits.

Inductors

Definition: Inductance (L) — property of a coil by which it opposes a change in current. V = L(di/dt). Unit: Henry (H).

• Series: L_total = L₁ + L₂ + ...
• Parallel: 1/L_total = 1/L₁ + 1/L₂ + ...
• Energy stored: E = ½LI².
• Inductors oppose AC (especially high frequency) but pass DC freely — opposite behaviour to capacitors.

Memory trick: Resistors and inductors behave identically in series/parallel. Capacitors behave the opposite way.

Network Theorems

These theorems let you simplify any network to a two-terminal equivalent:

Thevenin's Theorem: Any linear two-terminal network can be replaced by a single voltage source V_Th (open-circuit voltage at the terminals) in series with a resistance R_Th (the equivalent resistance seen from the terminals with all independent sources set to zero — voltage sources short-circuited, current sources open-circuited).

Norton's Theorem: Dual of Thevenin's — the same network can be replaced by a current source I_N (= V_Th / R_Th) in parallel with R_N (= R_Th).

Superposition Theorem: In a linear circuit with multiple sources, the response (voltage or current) at any element is the sum of the responses due to each source acting alone (others set to zero). Valid only for linear elements.

Maximum Power Transfer Theorem: Maximum power is delivered to a load resistance R_L when R_L = R_Th (the source's Thevenin equivalent resistance). At this condition, power in load = V_Th²/(4R_Th).

Question: A Thevenin equivalent has V_Th = 20 V and R_Th = 5 Ω. What load gives maximum power, and what is that power?
Solution:
Step 1: R_L = R_Th = 5 Ω.
Step 2: P_max = V_Th² / (4 × R_Th) = 400 / 20 = 20 W.
Conclusion: Load of 5 Ω receives maximum power of 20 W.

AC Circuits

Alternating current has voltage and current varying sinusoidally:
v(t) = V_max sin(ωt + φ), where ω = 2πf (angular frequency in rad/s) and φ is the phase angle.

RMS (Root Mean Square) values:

• V_RMS = V_max / √2 ≈ 0.707 V_max
• I_RMS = I_max / √2
• India's mains supply: 230 V RMS, 50 Hz.

Reactance:

• Inductive reactance: X_L = ωL = 2πfL. Increases with frequency.
• Capacitive reactance: X_C = 1/(ωC) = 1/(2πfC). Decreases with frequency.

Impedance (Z): The AC equivalent of resistance — combines R, X_L, X_C:
Z = √(R² + (X_L − X_C)²)

Phase angle: φ = arctan((X_L − X_C)/R). If X_L > X_C, circuit is inductive (current lags voltage). If X_C > X_L, circuit is capacitive (current leads voltage). Mnemonic: CIVIL — in a Capacitor, I leads V; V leads I in an inductor (L).

Power factor: cos φ. Unity (1.0) means pure resistive — all supplied power is consumed. Low power factor means wasted reactive power — a problem in industrial motors. Utilities charge penalties for low power factor.

Resonance in series RLC:
At resonance, X_L = X_C, so Z = R (minimum). Current is maximum.
Resonant frequency: f_r = 1 / (2π√(LC))

Question: Find the resonant frequency for L = 10 mH and C = 100 µF.
Solution:
Step 1: f_r = 1 / (2π × √(0.01 × 100×10⁻⁶))
Step 2: LC = 0.01 × 10⁻⁴ = 10⁻⁶. √(LC) = 10⁻³.
Step 3: f_r = 1 / (2π × 10⁻³) = 1000 / (2π) ≈ 159 Hz.
Conclusion: Resonant frequency ≈ 159 Hz.

Three-Phase AC Systems

Three-phase is used for power generation, transmission, and large motors because it is more efficient than single-phase.

Star (Y) connection:

• Line voltage V_L = √3 × Phase voltage V_ph
• Line current = Phase current

Delta (Δ) connection:

• Line voltage = Phase voltage
• Line current I_L = √3 × Phase current I_ph

Three-phase power:
P = √3 × V_L × I_L × cos φ

Real-world example: A 415 V (line), 50 Hz, 3-phase supply powers a factory motor. The motor draws 10 A line current at power factor 0.8. Active power = √3 × 415 × 10 × 0.8 ≈ 5,750 W ≈ 5.75 kW.

RRB JE Focus Areas

The RRB Junior Engineer paper tests:

• Setting up and solving simple resistive circuits using Kirchhoff's laws.
• Calculating resonant frequency of LC/RLC circuits.
• Maximum power transfer condition.
• RMS/average values and their relationships.
• Power factor and three-phase power calculations.

Circuit Theory is the single highest-yield technical chapter for RRB JE (Electrical) CBT-2 — expect roughly 12–18 questions built on Ohm's law, Kirchhoff's laws, network theorems and AC fundamentals. It is also the foundation for Machines and Power Systems, so mastering it pays twice.

Must-know relations

For a resistor, Ohm's law gives V = IR. Kirchhoff's Current Law (KCL): the algebraic sum of currents at a node is zero. Kirchhoff's Voltage Law (KVL): the algebraic sum of EMFs and voltage drops around any closed loop is zero.

Quantity	Formula
Series resistance	R = R₁ + R₂ + R₃
Two resistors in parallel	R = R₁R₂ / (R₁ + R₂)
Power	P = VI = I²R = V²/R
Resistivity	R = ρL / A
RMS of sine	0.707 × Vₘ
Half-cycle average	0.637 × Vₘ = 2Vₘ/π
Form factor (sine)	1.11
Peak factor (sine)	1.414
Series RLC resonance	f₀ = 1 / (2π√(LC))

Energy stored: capacitor = ½CV², inductor = ½LI². Time constants: RC circuit τ = RC, RL circuit τ = L/R.

Exam Tricks & Tips

• 🎯 Two-resistor parallel = product ÷ sum — memorise R₁R₂/(R₁+R₂); equal resistors in parallel just halve.
• 🎯 Max power transfer: load gets maximum power when R_L = R_th (source/Thevenin resistance); efficiency then is only 50%.
• 🎯 At series resonance X_L = X_C, impedance is minimum (= R), current is maximum, power factor = unity.
• 🎯 RMS = 0.707 Vₘ and average = 0.637 Vₘ apply only to a pure sine; a full-cycle average of a symmetrical AC wave is zero.
• 🎯 Thevenin: replace any two-terminal linear network by V_th in series with R_th; find R_th by shorting voltage sources and opening current sources.
• 🎯 Balanced star–delta: R_delta = 3 × R_star.
• ❌ Common mistake: adding capacitors in series like resistors in series — capacitors in series add as reciprocals (1/C = 1/C₁ + 1/C₂); they add directly only in parallel.

Expected exam pattern

Numerical single-answer questions: solve a resistive network, apply KVL/KCL, compute RMS/average/form factor, find resonant frequency, or apply Thevenin/Norton/superposition/max-power-transfer. Values are chosen for clean mental arithmetic.

Quick recap

V = IR anchors everything. KCL at nodes, KVL around loops. Series adds, parallel is product-over-sum. Learn RMS (0.707), average (0.637), form factor 1.11, peak factor 1.414, and resonance f₀ = 1/(2π√(LC)). Network theorems (Thevenin, Norton, superposition, max power transfer) are near-guaranteed marks.

Cover the answer, recall, then check. 12 cards on core circuit-theory facts for RRB JE (Electrical) CBT-2.

Q1. State Ohm's law.
A1. V = IR — voltage equals current times resistance (at constant temperature).

Q2. State KCL and KVL.
A2. KCL: algebraic sum of currents at a node = 0. KVL: algebraic sum of voltages (EMFs and drops) around any closed loop = 0.

Q3. Equivalent of two resistors R₁ and R₂ in parallel?
A3. R = R₁R₂ / (R₁ + R₂). Equal resistors in parallel → half of one.

Q4. Three ways to write electrical power.
A4. P = VI = I²R = V²/R.

Q5. RMS and half-cycle average of a sine wave of peak Vₘ?
A5. RMS = 0.707 Vₘ; average (half cycle) = 0.637 Vₘ = 2Vₘ/π.

Q6. Form factor and peak factor of a sine wave?
A6. Form factor = RMS/average = 1.11; peak (crest) factor = peak/RMS = 1.414.

Q7. Resonant frequency of a series RLC circuit?
A7. f₀ = 1 / (2π√(LC)). At resonance X_L = X_C, impedance minimum, current maximum, pf = 1.

Q8. Condition for maximum power transfer to a load?
A8. R_L = R_th (Thevenin/source resistance). Efficiency at this point is 50%.

Q9. Energy stored in a capacitor and in an inductor?
A9. Capacitor: ½CV². Inductor: ½LI².

Q10. Time constant of an RC and an RL circuit?
A10. RC circuit: τ = RC. RL circuit: τ = L/R.

Q11. How does resistance depend on dimensions and material?
A11. R = ρL/A — proportional to resistivity ρ and length L, inversely proportional to cross-sectional area A.

Q12. Balanced star-to-delta resistance relation?
A12. R_delta = 3 × R_star (each delta resistor is three times the equivalent star resistor).

Worked Example

Problem: In a circuit, two 3 Ω resistors are connected in parallel, and this combination is in series with another 3 Ω resistor across a 9 V battery. Find the total current drawn from the battery.

Solution:
Step 1 — Parallel pair: two equal 3 Ω in parallel = 3 ÷ 2 = 1.5 Ω.
Step 2 — Add the series resistor: R_total = 1.5 + 3 = 4.5 Ω.
Step 3 — Total current from the 9 V source: I = V ÷ R_total = 9 ÷ 4.5 = 2 A.

Answer: Total current = 2 A.