10.2 Kirchhoff’s laws

Learning objectives

$ΣI_in = ΣI_out or, with signed currents, ΣI = 0$

At a junction in a steady circuit, the total current entering equals the total current leaving. Since I = ΔQ/Δt, any difference sustained for time Δt would leave net charge ΔQ = (ΣI_in − ΣI_out)Δt accumulating at the junction. A steady node does not accumulate charge, so the totals are equal.

Draw an arrow for every branch and choose one sign convention before writing the equation. You may sum entering and leaving currents separately, or assign one direction positive and use an algebraic sum. A negative solved value reverses the assumed arrow.

At a node, 2.0 A and 0.50 A enter while 1.5 A and I leave. First law gives 2.0 + 0.50 = 1.5 + I, so I = 1.0 A leaving. In 5.0 s that branch carries Q = It = 5.0 C.

Current is not used up at a junction. Charge can divide between branches, but the rate of charge arrival and departure must balance in steady operation.

$Σ(emf) = Σ(p.d. drops) or, algebraically around a closed loop, ΣV = 0$

After one complete circuit loop, a charge returns to its starting point with the same energy per unit charge. Energy gained per coulomb in sources therefore equals energy transferred per coulomb in circuit components, including internal resistance when present.

Traversal	Potential change
through a source from − to +	+ε rise
through a source from + to −	−ε fall
through a resistor in the direction of conventional current	−IR drop
through a resistor opposite to conventional current	+IR rise

Traversing a loop in the current direction through a 12 V source and resistive drops of 4 V and 8 V gives +12 − 4 − 8 = 0. For each coulomb, 12 J is supplied and 12 J is transferred; no net energy is gained after the closed loop.

The law does not say every component has the same p.d. Signs are set by the chosen traversal and polarity; reversing the loop reverses every term but leaves the physical equation equivalent.

Let resistors R₁, R₂, …, Rₙ be in one unbranched series path. Let I be the current and V the total p.d. across the combination. The equivalent resistance R_T is defined by V = IR_T.

With no junction between the resistors, Kirchhoff's first law gives the same current I through every resistor.

$V = V₁ + V₂ + ··· + Vₙ$

$IR_T = IR₁ + IR₂ + ··· + IRₙ = I(R₁ + R₂ + ··· + Rₙ)$

$For I ≠ 0: R_T = R₁ + R₂ + ··· + Rₙ$

The derivation uses common current and additive p.d.s; it does not assume equal p.d.s. It applies to components in one series path, not to a network containing a branch.

$R_T = R₁ + R₂ + ··· and I = V/R_T$

Confirm there is one unbranched current path, add the resistances, use the total applied p.d. to find the common current, then use V_i = IR_i for each resistor. Check that the individual p.d.s sum to the supply p.d.

A 12 V supply is connected to 2.0 Ω, 3.0 Ω and 5.0 Ω in series. R_T = 10.0 Ω and I = 12/10.0 = 1.2 A. The drops are 2.4 V, 3.6 V and 6.0 V; they sum to 12.0 V.

$For series resistors: V₁/V₂ = R₁/R₂$

Series p.d.s are equal only when the resistances are equal. Adding another positive series resistance increases R_T and, for a fixed supply p.d., decreases the circuit current.

Let R₁, R₂, …, Rₙ connect between the same two nodes. Let V be the p.d. across the combination, I the total current and R_T the equivalent resistance, so I = V/R_T.

Kirchhoff's second law applied to loops through different branches shows that every branch has the same p.d. V.

$I = I₁ + I₂ + ··· + Iₙ$

$V/R_T = V/R₁ + V/R₂ + ··· + V/Rₙ$

$For V ≠ 0: 1/R_T = 1/R₁ + 1/R₂ + ··· + 1/Rₙ$

Parallel branches share p.d., not necessarily current. The reciprocal formula follows because branch currents add; directly adding the resistance values would describe a series path instead.

$1/R_T = Σ(1/R_i) and, for exactly two branches, R_T = R₁R₂/(R₁ + R₂)$

Confirm the resistors connect between the same two nodes. Add conductances 1/R_i, then invert the complete sum. The product-over-sum shortcut is valid for exactly two resistors; for three or more, use the general reciprocal relation or reduce in stages.

For 6.0 Ω, 3.0 Ω and 2.0 Ω in parallel, 1/R_T = 1/6 + 1/3 + 1/2 = 1.0 Ω⁻¹, so R_T = 1.0 Ω. Across 6.0 V, the branch currents are 1.0 A, 2.0 A and 3.0 A, summing to 6.0 A = V/R_T.

Check	Required result
magnitude	R_T is less than the smallest positive branch resistance
identical n resistors R	R_T = R/n
current	V/R_T equals the sum of V/R_i

Do not report the reciprocal sum as R_T: after calculating 1/R_T, invert it. Adding another conducting parallel branch lowers the equivalent resistance because it provides another route for current.

Name every branch current and choose arrows; write independent junction equations; choose independent closed loops and assign voltage signs consistently; replace resistor p.d.s by IR; solve the simultaneous equations; then check every junction and loop. Include internal resistance as a series resistor inside its source loop when relevant.

Example network: a 12 V ideal source and 2.0 Ω series resistor feed node A. From A, currents I₄ and I₆ return to the source through separate 4.0 Ω and 6.0 Ω branches. Let total current I reach A through the 2.0 Ω resistor.

$junction A: I = I₄ + I₆ 4 Ω loop: 12 = 2I + 4I₄ 6 Ω loop: 12 = 2I + 6I₆$

The two branch equations give I₄ = (12 − 2I)/4 and I₆ = (12 − 2I)/6. Substitute into I = I₄ + I₆: I = (12 − 2I)(1/4 + 1/6), so I = 30/11 = 2.73 A. Then I₄ = 18/11 = 1.64 A and I₆ = 12/11 = 1.09 A.

Check	Substitution	Result
junction	18/11 + 12/11	30/11 A = I
4 Ω loop	2(30/11) + 4(18/11)	12 V
6 Ω loop	2(30/11) + 6(12/11)	12 V

Do not write a single current through every branch. If a solved current is negative, retain the magnitude and reverse its assumed arrow; the algebra has identified the physical direction rather than failed.

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—10.3 Potential dividersRead

10.2 Kirchhoff’s laws

Learning objectives

Kirchhoff's first law is charge conservation at a junction

Kirchhoff's second law is energy conservation around a loop

Derive the series-resistance formula from Kirchhoff's laws

Use series resistance and recover each voltage drop

Derive the parallel-resistance formula from both Kirchhoff laws

Use parallel resistance with reciprocal and physical checks

Solve a branched circuit with junction and loop equations

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