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I am trying to compare watts exiting the battery vs watts being consumed by the bulb.

I have set up a test circuit shown below.

Test circuit

Should not A-B watts match C-D watts??

Sorry for such a basic question...I'm learning here...

William Mingin
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    If A-B watts did match C-D watts, then that would mean that $R_1$ was 0 Ω, or that the voltage across the resistor is 0 V and/or that the current through the resistor is 0 A. – Harry Svensson Oct 30 '18 at 10:59
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    And this is theoretical. In a real circuit, everything has resistance - even the wires. So there is power loss in every single piece. – rdtsc Oct 30 '18 at 12:12

2 Answers2

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You should also include the power dissipation from the resistor in there.

$$ P_{battery} = -(P_{resistor} + P_{lamp}) $$

Sven B
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What you are ignoring is the power consumed by \$R_1\$.

The product of AB (2.07W) is the total power supplied to the circuit, while the product of CD (0.48W) is the power consumed by the lamp.

$$P_{R_1} = (V_A - V_C)\ \times \ I = (6.15V - 1.434V) \times 0.335A = 1.58W$$

1.58W + 0.48W is 2.06W which is the power the battery is supplying to the circuit.

Harry Svensson
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StainlessSteelRat
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