Electrical circuit diagram with resistors arranged in series and parallel connections powered by a 5V source.

- Calculate the equivalent resistance of R6 and R7 in series: 2 Ω + 2 Ω = 4 Ω.
- This 4 Ω is in parallel with R5 (4 Ω). The equivalent resistance is (4 Ω * 4 Ω) / (4 Ω + 4 Ω) = 2 Ω.
- Add R4 (10 Ω) in series with this 2 Ω: 10 Ω + 2 Ω = 12 Ω.
- R2 (4 Ω) and R3 (8 Ω) are in series: 4 Ω + 8 Ω = 12 Ω.
- This 12 Ω is in parallel with the previous 12 Ω branch: (12 Ω * 12 Ω) / (12 Ω + 12 Ω) = 6 Ω.
- Add R1 (4 Ω) in series with this 6 Ω: 4 Ω + 6 Ω = 10 Ω.
- Total current from the battery: I_total = V / R_total = 5 V / 10 Ω = 0.5 A.
- Voltage drop across R1: V_R1 = I_total * R1 = 0.5 A * 4 Ω = 2 V.
- Voltage at the node after R1: 5 V - 2 V = 3 V.
- Current through the R2-R3 branch: I_branch1 = 3 V / 12 Ω = 0.25 A.
- Current through the R4-(R5||(R6+R7)) branch: I_branch2 = 3 V / 12 Ω = 0.25 A.
- Voltage drop across R4: V_R4 = I_branch2 * R4 = 0.25 A * 10 Ω = 2.5 V.
- Voltage at the node after R4: 3 V - 2.5 V = 0.5 V.
- This 0.5 V is across R5 and also across the series combination of R6 and R7.
- Current through R5: I_R5 = 0.5 V / 4 Ω = 0.125 A.
- Current through R6 and R7: I_R6_R7 = 0.5 V / (2 Ω + 2 Ω) = 0.5 V / 4 Ω = 0.125 A.
- Power dissipated by R6: P_R6 = I_R6_R7² * R6 = (0.125 A)² * 2 Ω = 0.015625 * 2 = 0.03125 W.
- Power dissipated by R7: P_R7 = I_R6_R7² * R7 = (0.125 A)² * 2 Ω = 0.015625 * 2 = 0.03125 W.
- Power dissipated by R5: P_R5 = V² / R5 = (0.5 V)² / 4 Ω = 0.25 / 4 = 0.0625 W.
- Power dissipated by R4: P_R4 = I_branch2² * R4 = (0.25 A)² * 10 Ω = 0.0625 * 10 = 0.625 W.
- Power dissipated by R2: P_R2 = I_branch1² * R2 = (0.25 A)² * 4 Ω = 0.0625 * 4 = 0.25 W.
- Power dissipated by R3: P_R3 = I_branch1² * R3 = (0.25 A)² * 8 Ω = 0.0625 * 8 = 0.5 W.
- Power dissipated by R1: P_R1 = I_total² * R1 = (0.5 A)² * 4 Ω = 0.25 * 4 = 1 W.
- Total power dissipated: 0.03125 W + 0.03125 W + 0.0625 W + 0.625 W + 0.25 W + 0.5 W + 1 W = 2.5 W.
- Total power supplied by battery: P_battery = V * I_total = 5 V * 0.5 A = 2.5 W.