Mixed Circuit Audits

Mixed Power Balance Audit

branch heating equals source power

A mixed-power audit checks that resistor heating in the branches matches the power supplied by the source. The balance closes exactly only for the declared branch currents, resistances, and ideal steady DC network; real circuits have resistor tolerance, lead resistance, meter loading, supply variation, self-heating, parasitic paths, temperature drift, and voltage/current/power measurement uncertainty.

Power balance closes the mixed-current story by checking each branch load and the source total. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Each branch spends power

Each branch resistor turns electrical energy into heat. Branch power can be checked from current squared times resistance.

Pbranch=I2RP_{\text{branch}} = I^{2}R
Branch power in a dividerThe same branch currents now mark where power is spent as heat.12 VA9 A2 ohm4 ohm9 A6 A3 A+-12 V+-12 Vheatheat

Scan three divider power budgets

The power rows reuse the current-divider rows. Each branch power is computed from that branch's current and resistance, then the branch powers add.

RaRbIaIbPaPbPtotal2 ohm2 ohm6 A6 A72 W72 W144 W2 ohm4 ohm6 A3 A72 W36 W108 W4 ohm12 ohm3 A1 A36 W12 W48 W\begin{array}{c|c|c|c|c|c|c}R_{\text{a}}&R_{\text{b}}&I_{\text{a}}&I_{\text{b}}&P_{\text{a}}&P_{\text{b}}&P_{\text{total}}\\2\ \text{ohm}&2\ \text{ohm}&6\ \text{A}&6\ \text{A}&72\ \text{W}&72\ \text{W}&144\ \text{W}\\2\ \text{ohm}&4\ \text{ohm}&6\ \text{A}&3\ \text{A}&72\ \text{W}&36\ \text{W}&108\ \text{W}\\4\ \text{ohm}&12\ \text{ohm}&3\ \text{A}&1\ \text{A}&36\ \text{W}&12\ \text{W}&48\ \text{W}\\\end{array}
Branch power in a dividerThe diagram shows the middle power-budget row.12 VA9 A2 ohm4 ohm9 A6 A3 A+-12 V+-12 Vheatheat

Top branch power

For the shown case, the top branch has the larger current and therefore the larger power.

Pa=Ia2Ra=(6 A)2⋅2 ohm=72 WP_{\text{a}} = I_{\text{a}}^{2}R_{\text{a}} = \left(6\ \text{A}\right)^{2}\cdot 2\ \text{ohm} = 72\ \text{W}

Bottom branch power

The bottom branch has less current, so it spends fewer watts.

Pb=Ib2Rb=(3 A)2⋅4 ohm=36 WP_{\text{b}} = I_{\text{b}}^{2}R_{\text{b}} = \left(3\ \text{A}\right)^{2}\cdot 4\ \text{ohm} = 36\ \text{W}

Branch powers add to source power

The branch powers are 72 watts and 36 watts. Together they equal the source power delivered by the same source voltage and total current.

Psource=12 V⋅9 A=108 W=Pa+Pb=72 W+36 W=108 WP_{\text{source}} = 12\ \text{V}\cdot 9\ \text{A} = 108\ \text{W} = P_{\text{a}} + P_{\text{b}} = 72\ \text{W} + 36\ \text{W} = \hl{108}\ \text{W}
Branch power in a dividerHeat labels mark the two loads whose powers add.12 VA9 A2 ohm4 ohm9 A6 A3 A+-12 V+-12 Vheatheat