A parallel circuit has a 179.7 V battery and 3 resistors, R1 = 68.6
Ω
, R2 = 123.6
Ω
, and R3 = 366.9
Ω
. How much power is develop in the circuit?

Question

A parallel circuit has a 179.7 V battery and 3 resistors, R1 = 68.6 
Ω
 , R2 = 123.6 
Ω
 , and R3 = 366.9 
Ω
 .  How much power is develop in the circuit?

BlackTom AI · Accepted Answer

We are given a parallel circuit with a supply voltage of 179.7 V and three resistors in parallel: R1 = 68.6 Ω, R2 = 123.6 Ω, and R3 = 366.9 Ω.

First, compute the equivalent resistance for resistors in parallel using 1/R_eq = 1/R1 + 1/R2 + 1/R3.
- 1/R1 = 1/68.6 ≈ 0.01458
- 1/R2 = 1/123.6 ≈ 0.008088
- 1/R3 = 1/366.9 ≈ 0.002724
Sum = 0.01458 + 0.008088 + 0.002724 ≈ 0.025392
Therefore, R_eq ≈ 1 / 0.025392 ≈ 39.38 Ω.

Next, use the power formula for a load with a given voltage: P = V^2 / R_eq.
- V^2 = (179.7)^2 ≈ (180 − 0.3)^2 = 32400 − 108 + 0.09 ≈ 32292.09
- P ≈ 32292.09 / 39.38 ≈ 820.0 W (approximately).

To verify consistency, note that lowering the equivalent resistance increases power for a fixed voltage, and the calculated R_eq around 39.38 Ω with V ≈ 180 V yields a power on the order of several hundred watts, specifically about 8.2×10^2 W.

Key checks:
- Correct parallel formula was used for R_eq.
- The arithmetic for each reciprocal and the final division aligns with typical unit analysis (voltage in volts, resistance in ohms, power in watts).
- Units: volts squared divided by ohms gives watts, which matches the desired quantity.

If you compare to common roundings, small differences in rounding intermediate steps (e.g., 0.01458 vs 0.0146) will only slightly shift the final result, keeping it near 820 W.

A parallel circuit has a 179.7 V battery and 3 resistors, R1 = 68.6 Ω , R2 = 123.6 Ω , and R3 = 366.9 Ω . How much power is develop in the circuit? 简答题

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