To stay within the end-of-load rating for a 2000-watt circuit, how should two 1000-watt loads be connected?

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Multiple Choice

To stay within the end-of-load rating for a 2000-watt circuit, how should two 1000-watt loads be connected?

Explanation:
When staying under a circuit’s end-of-load rating, the key idea is how connection method changes the total power drawn from the supply. Two 1000-watt devices wired in series across a 120-volt circuit share the same current, and the circuit’s voltage is split between them. Each 1000-watt device at 120 V has a resistance of about 14.4 ohms (R = V^2 / P). In series, the total resistance is about 28.8 ohms, so the current is 120 V divided by 28.8 ohms, which is roughly 4.17 A. The power each device dissipates is P = I^2 R ≈ (4.17 A)^2 × 14.4 Ω ≈ 250 W. The total drawn from the circuit is about 500 W, well under the 2000 W limit. If the two loads were connected in parallel, each would still draw about 1000 W at 120 V, totaling the full 2000 W, which sits at the circuit’s limit and leaves little room for voltage drops or inefficiencies.

When staying under a circuit’s end-of-load rating, the key idea is how connection method changes the total power drawn from the supply. Two 1000-watt devices wired in series across a 120-volt circuit share the same current, and the circuit’s voltage is split between them.

Each 1000-watt device at 120 V has a resistance of about 14.4 ohms (R = V^2 / P). In series, the total resistance is about 28.8 ohms, so the current is 120 V divided by 28.8 ohms, which is roughly 4.17 A. The power each device dissipates is P = I^2 R ≈ (4.17 A)^2 × 14.4 Ω ≈ 250 W. The total drawn from the circuit is about 500 W, well under the 2000 W limit.

If the two loads were connected in parallel, each would still draw about 1000 W at 120 V, totaling the full 2000 W, which sits at the circuit’s limit and leaves little room for voltage drops or inefficiencies.

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