Verified Solution Science Electricity

A circuit consists of three resistors: R1, R2, and R3. R1 = 5 Ω is connected in series with the parallel combination of R2 = 10 Ω and R3 = 10 Ω. If this circuit is connected to a 12V battery, what is the power dissipated across R1?

5 views 1 helpful Updated Jun 30, 2026
Solution ✔ Verified
  • A7.2 W
  • B12.8 W
  • C20 W
  • D30 W
Explanation

First, calculate the equivalent resistance of R2 and R3 in parallel: R_p = (R2 × R3) / (R2 + R3) = (10 × 10) / (10 + 10) = 100 / 20 = 5 Ω. The total equivalent resistance of the circuit is R_eq = R1 + R_p = 5 Ω + 5 Ω = 10 Ω. The total current flowing through the circuit is I = V / R_eq = 12V / 10 Ω = 1.2 A. Since R1 is in series with the parallel combination, the current flowing through R1 is also 1.2 A. The power dissipated across R1 is P1 = I² × R1 = (1.2 A)² × 5 Ω = 1.44 × 5 = 7.2 W.

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