04Electric Circuits
Apply Ohm's law and Kirchhoff's rules to analyze series, parallel, and combination DC circuits; calculate current, voltage, resistance, and power.
Electric circuits power every device you use. Circuit analysis is a fundamental skill for physics, engineering, and electronics.
How do charges flow through a circuit, and what rules govern the distribution of current, voltage, and power in series and parallel networks?
An electric circuit is a closed conducting path through which charge flows. Current (I) is the rate of charge flow: I = Q/t. The driving force is a potential difference (voltage), and every conductor opposes flow through its resistance (R). Ohm's law, V = IR, links these three quantities for ohmic materials. Resistance depends on material and geometry: R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area. Circuits can be wired in series (one path) or parallel (multiple paths), each with distinct rules for combining resistance, current, and voltage. Kirchhoff's two laws provide a systematic framework for any network. Power dissipated in a resistor is P = IV = I²R = V²/R. Real batteries have internal resistance r, so the terminal voltage is Vt = EMF − Ir. Finally, RC circuits charge and discharge exponentially with time constant τ = RC.
Series vs. Parallel Circuits
Series Circuit
- Resistance: Rtotal = R₁ + R₂ + … + Rₙ (sum of all)
- Current: Same through every element — I is identical everywhere
- Voltage: Splits across resistors — V₁ + V₂ + … = Vsupply
- If one element fails (open), the entire circuit breaks
- Adding more resistors increases total resistance
Parallel Circuit
- Resistance: 1/Rtotal = 1/R₁ + 1/R₂ + … (reciprocal sum)
- Voltage: Same across every branch — V is identical everywhere
- Current: Splits among branches — I₁ + I₂ + … = Itotal
- If one branch fails, others continue to operate
- Adding more branches decreases total resistance
Kirchhoff's Laws
Kirchhoff's Voltage Law (KVL)
The algebraic sum of all voltages around any closed loop equals zero:
ΣV = 0Based on conservation of energy — a charge returning to its starting point gains and loses the same total energy. EMF sources add voltage; resistors drop voltage.
Kirchhoff's Current Law (KCL)
The total current entering a junction equals the total current leaving it:
ΣIin = ΣIoutBased on conservation of charge — charge cannot accumulate at a node. Apply at every junction in a complex network.
Key Equations
I = Q / tAmperes (A); Q in coulombs, t in secondsV = IRVolts = Amperes × OhmsR = ρL / Aρ = resistivity (Ω·m), L = length, A = areaR_series = R₁ + R₂ + … + RₙTotal > any individual R1/R_parallel = 1/R₁ + 1/R₂ + …Total < smallest individual RP = IV = I²R = V²/RWatts (W); energy per secondV_t = EMF − Irr = internal resistance; drops under loadτ = RCSeconds; V_C = V₀(1 − e^(−t/τ)) chargingWorked Examples
Three resistors R₁ = 10 Ω, R₂ = 20 Ω, and R₃ = 30 Ω are connected in series to a 12 V battery. Find the total resistance, the current in the circuit, and the voltage across each resistor.
Identify configuration: series — resistances add directly.
R_total = R₁ + R₂ + R₃ = 10 + 20 + 30 = 60 Ω.
Apply Ohm's law for the whole circuit: I = V / R_total = 12 / 60 = 0.2 A.
In series, the same current (0.2 A) flows through every resistor.
Voltage across R₁: V₁ = IR₁ = 0.2 × 10 = 2 V.
Voltage across R₂: V₂ = IR₂ = 0.2 × 20 = 4 V.
Voltage across R₃: V₃ = IR₃ = 0.2 × 30 = 6 V.
Check (KVL): V₁ + V₂ + V₃ = 2 + 4 + 6 = 12 V ✓
Two resistors R₁ = 6 Ω and R₂ = 12 Ω are connected in parallel across a 12 V supply. Find the total resistance, total current, and the current through each branch.
Identify configuration: parallel — use reciprocal formula.
1/R_total = 1/R₁ + 1/R₂ = 1/6 + 1/12 = 2/12 + 1/12 = 3/12.
R_total = 12/3 = 4 Ω.
In parallel, the same voltage (12 V) appears across each branch.
Total current: I_total = V / R_total = 12 / 4 = 3 A.
Current through R₁: I₁ = V / R₁ = 12 / 6 = 2 A.
Current through R₂: I₂ = V / R₂ = 12 / 12 = 1 A.
Check (KCL): I₁ + I₂ = 2 + 1 = 3 A = I_total ✓
A toaster has a resistance of 24 Ω and is plugged into a 120 V outlet. Find the current drawn, the power consumed, and the energy used in 10 minutes.
Find current using Ohm's law: I = V / R = 120 / 24 = 5 A.
Find power: P = IV = 5 × 120 = 600 W.
Alternatively: P = V²/R = (120)²/24 = 14,400/24 = 600 W ✓
Convert time: t = 10 min = 10 × 60 = 600 s.
Energy: E = Pt = 600 × 600 = 360,000 J = 360 kJ.
A battery has an EMF of 9 V and an internal resistance r = 0.5 Ω. It is connected to an external resistor R = 8.5 Ω. Find the current in the circuit and the terminal voltage of the battery.
The total resistance in the circuit is R_total = R + r = 8.5 + 0.5 = 9 Ω.
Apply Ohm's law to the full loop: I = EMF / R_total = 9 / 9 = 1 A.
Terminal voltage: V_t = EMF − Ir = 9 − (1)(0.5) = 9 − 0.5 = 8.5 V.
Alternatively: V_t = IR (voltage across external resistor) = 1 × 8.5 = 8.5 V ✓
Note: the internal resistance 'steals' 0.5 V from the EMF.
An RC circuit has a capacitor C = 100 μF and a resistor R = 10 kΩ. Find the time constant τ. If the supply voltage is 10 V, what voltage has the capacitor reached after one time constant?
Convert units: C = 100 × 10⁻⁶ F = 10⁻⁴ F; R = 10 × 10³ Ω = 10⁴ Ω.
Time constant: τ = RC = 10⁴ × 10⁻⁴ = 1 s.
Charging equation: V_C(t) = V₀(1 − e^(−t/τ)).
At t = τ = 1 s: V_C = 10 × (1 − e^(−1)) = 10 × (1 − 0.368) = 10 × 0.632 = 6.32 V.
This is 63.2% of the supply voltage — a universal result at t = τ.
After 5τ (5 s), the capacitor is considered fully charged (≈ 99.3% of V₀).
Guided Practice
Four resistors of 5 Ω, 10 Ω, 15 Ω, and 20 Ω are connected in series to a 25 V battery. Find the total resistance and the current flowing through the circuit.
Hint: In series, simply add all resistances: R_total = R₁ + R₂ + R₃ + R₄. Then use I = V / R_total.
Three resistors of 4 Ω, 6 Ω, and 12 Ω are connected in parallel. Calculate the equivalent resistance of the combination.
Hint: Use 1/R_total = 1/4 + 1/6 + 1/12. Find a common denominator (12), add the fractions, then take the reciprocal.
A 60 W light bulb operates on a 120 V supply. Find its resistance and the current it draws.
Hint: Use P = V²/R to find R first: R = V²/P. Then use I = V/R or I = P/V.
A battery with EMF = 12 V and internal resistance r = 1 Ω drives a current of 2 A through an external circuit. Find the terminal voltage and the external resistance.
Hint: Terminal voltage: V_t = EMF − Ir. Then find external resistance using R = V_t / I.
An RC circuit has R = 5 kΩ and C = 200 μF. How long does it take for the capacitor to charge to approximately 86.5% of the supply voltage?
Hint: 86.5% ≈ (1 − e⁻²), which corresponds to t = 2τ. First find τ = RC, then multiply by 2.
Key Vocabulary
Electric Current
The rate of flow of electric charge past a point in a circuit: I = Q/t. Measured in amperes (A), where 1 A = 1 C/s. Conventional current flows from positive to negative terminal.
Example: A current of 2 A means 2 coulombs of charge pass a point every second.
Ohm's Law
For an ohmic conductor, the current through it is directly proportional to the voltage across it and inversely proportional to its resistance: V = IR. Holds when temperature is constant.
Example: A 12 Ω resistor with 6 V across it carries a current of I = 6/12 = 0.5 A.
Resistance
The opposition a material offers to the flow of electric current, measured in ohms (Ω). Depends on material resistivity ρ, length L, and cross-sectional area A: R = ρL/A.
Example: A longer, thinner wire has greater resistance than a short, thick wire of the same material.
Series Circuit
A circuit in which components are connected end-to-end along a single path. The same current flows through all components; voltages add up to the supply voltage.
Example: Three resistors of 10 Ω, 20 Ω, and 30 Ω in series give R_total = 60 Ω.
Parallel Circuit
A circuit in which components are connected across the same two nodes, providing multiple current paths. The same voltage appears across all branches; currents add up to the total.
Example: Two 6 Ω resistors in parallel give R_total = 3 Ω — less than either individual resistor.
Kirchhoff's Voltage Law
The algebraic sum of all potential differences (EMFs and voltage drops) around any closed loop in a circuit equals zero: ΣV = 0. Reflects conservation of energy.
Example: In a loop with a 9 V battery and resistors dropping 3 V and 6 V: 9 − 3 − 6 = 0 ✓
Kirchhoff's Current Law
The total current entering any junction (node) in a circuit equals the total current leaving it: ΣI_in = ΣI_out. Reflects conservation of charge.
Example: If 5 A enters a junction and splits into two branches, the branches must carry currents that sum to 5 A.
Power
The rate at which electrical energy is converted to other forms (heat, light, etc.): P = IV = I²R = V²/R. Measured in watts (W). Energy = Power × time.
Example: A 100 Ω resistor carrying 2 A dissipates P = I²R = 4 × 100 = 400 W.
Quick Check — Multiple Choice
Interactive Practice — 5 Questions
Three resistors of 2 Ω, 4 Ω, and 6 Ω are connected in series. What is the total resistance?
Two identical 8 Ω resistors are connected in parallel. What is the equivalent resistance?
A resistor dissipates 50 W when a current of 5 A flows through it. What is its resistance?
Which statement about a parallel circuit is correct?
A battery has EMF = 6 V and internal resistance r = 0.5 Ω. When a 5.5 Ω external resistor is connected, the terminal voltage is:
Independent Practice
A 9.0 V battery is connected to a 45 Ω resistor. Calculate (a) the current through the resistor and (b) the power dissipated.
Three resistors (10 Ω, 20 Ω, 30 Ω) are connected in series to a 12 V source. Find the total resistance, current, and voltage across each resistor.
Three resistors (6 Ω, 12 Ω, 4 Ω) are connected in parallel to a 24 V source. Find the equivalent resistance and the current through each resistor.
A capacitor with capacitance 50 μF is connected to a 12 V battery. Find (a) the charge stored and (b) the energy stored.
★ In a circuit, a 12 V battery (internal resistance 0.5 Ω) is connected to two resistors: R₁ = 3 Ω in series with the parallel combination of R₂ = 6 Ω and R₃ = 12 Ω. (a) Find the equivalent external resistance. (b) Find the terminal voltage of the battery. (c) Find the current through each of R₂ and R₃. (d) Find the power delivered to R₁.
ChallengeCommon Mistakes
Adding resistances in parallel like series resistors (R_total = R₁ + R₂)
Parallel: 1/R_total = 1/R₁ + 1/R₂. For two resistors: R_total = R₁R₂/(R₁+R₂). The total is always LESS than the smallest individual resistor
Applying KVL by adding all voltage drops without regard to sign
KVL: going around a loop, voltage RISES (through a battery from − to +) are positive; voltage DROPS (through a resistor in the direction of current) are negative. ΣV = 0
Thinking current is 'used up' as it flows through resistors
Current (charge per second) is conserved at every junction (KCL). Resistors dissipate energy (power = I²R), not current
Forgetting that in an RC circuit, the capacitor is fully charged (I = 0) at t → ∞
RC charging: V_C = V₀(1 − e^(−t/τ)), I = (V₀/R)e^(−t/τ). At t = τ, capacitor is 63.2% charged; at t = 5τ, it is 99.3% charged (essentially full)
Math Tips
Two-resistor parallel shortcut: R_total = R₁R₂/(R₁+R₂). Three or more: use 1/R_total = 1/R₁ + 1/R₂ + 1/R₃. Always verify R_total < smallest R
Power triangle: P = IV = I²R = V²/R. Choose the form based on what you know. If you know I and R, use P = I²R. If you know V and R, use P = V²/R
KCL node method: label unknown currents, write ΣI_in = ΣI_out at each node, write KVL for each independent loop. You need as many equations as unknowns
RC time constant τ = RC (seconds). After each τ, the remaining charge/voltage decreases by factor e ≈ 2.718. After 5τ, the circuit is effectively at steady state