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.
Each branch spends power
Each branch resistor turns electrical energy into heat. Branch power can be checked from current squared times resistance.
P branch = I 2 R P_{\text{branch}} = I^{2}R P branch = I 2 R Branch power in a divider The same branch currents now mark where power is spent as heat. 12 V A 9 A 2 ohm 4 ohm 9 A 6 A 3 A + - 12 V + - 12 V heat heat
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.
R a R b I a I b P a P b P total 2 ohm 2 ohm 6 A 6 A 72 W 72 W 144 W 2 ohm 4 ohm 6 A 3 A 72 W 36 W 108 W 4 ohm 12 ohm 3 A 1 A 36 W 12 W 48 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} R a 2 ohm 2 ohm 4 ohm R b 2 ohm 4 ohm 12 ohm I a 6 A 6 A 3 A I b 6 A 3 A 1 A P a 72 W 72 W 36 W P b 72 W 36 W 12 W P total 144 W 108 W 48 W Branch power in a divider The diagram shows the middle power-budget row. 12 V A 9 A 2 ohm 4 ohm 9 A 6 A 3 A + - 12 V + - 12 V heat heat
Top branch power
For the shown case, the top branch has the larger current and therefore the larger power.
P a = I a 2 R a = ( 6 A ) 2 ⋅ 2 ohm = 72 W P_{\text{a}} = I_{\text{a}}^{2}R_{\text{a}} = \left(6\ \text{A}\right)^{2}\cdot 2\ \text{ohm} = 72\ \text{W} P a = I a 2 R a = ( 6 A ) 2 ⋅ 2 ohm = 72 W
Bottom branch power
The bottom branch has less current, so it spends fewer watts.
P b = I b 2 R b = ( 3 A ) 2 ⋅ 4 ohm = 36 W P_{\text{b}} = I_{\text{b}}^{2}R_{\text{b}} = \left(3\ \text{A}\right)^{2}\cdot 4\ \text{ohm} = 36\ \text{W} P b = I b 2 R b = ( 3 A ) 2 ⋅ 4 ohm = 36 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.
P source = 12 V ⋅ 9 A = 108 W = P a + P b = 72 W + 36 W = 108 W P_{\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} P source = 12 V ⋅ 9 A = 108 W = P a + P b = 72 W + 36 W = 108 W Branch power in a divider Heat labels mark the two loads whose powers add. 12 V A 9 A 2 ohm 4 ohm 9 A 6 A 3 A + - 12 V + - 12 V heat heat