What is wrong with this derivation of the power dissipated in a series circuit?

Question

Suppose for a circuit of two bulbs joined in series. From Kirchoff's Current Law, the current through both bulbs is same, say $I$. Then, $$P_{\text{net}}=I^2R_{\text{s}}$$ $$P_1=I^2R_1$$ $$P_2=I^2R_2$$

Also, in a series circuit, $$R_{\text{s}}=R_1+R_2$$

Multiplying through by $I^2$: $$I^2R_{\text{s}}=I^2R_1+I^2R_2$$

From the above equations, $$\boxed{P_{\text{net}}=P_1+P_2}$$

But in my book, it says that in series the power is given by: $$\dfrac{1}{P_{\text{net}}}=\dfrac{1}{P_1}+\dfrac{1}{P_2}$$

So, what did I do wrong?

Answer 1

What you've done is correct, assuming that $P_1$ and $P_2$ are the powers dissipated in the bulbs in their series configuration, and is what I'd call a consistency check. You're showing that the addition rule for resistances in series is consistent with energy conservation: the total energy dissipated per second is the sum of the energies dissipated per second in the individual bulbs.

What I suspect, though, is that $P_1$ and $P_2$ do not mean what you're taking them to mean, but are supposed to be the powers if each bulb were separately connected to the same constant voltage source.

In that case one does get the answer in your book, but only by assuming that the resistances of the bulbs are constant (independently of the applied pd). This is not the case, because filament lamps are seriously non-ohmic!

Answer 2

When resistors are in series, the equivalent resistance of the circuit goes up, and the current through the circuit goes down. This means that you can't assume that the current that would exist through either resistor alone is also the current that will exist in both resistors when they are in series. To work your problem, use Ohm's law to calculate a new current, then calculate power through the circuit using the equivalent resistance.

Answer 3

The context in your book is: Given a constant voltage difference across the circuit what is the power dissipated.

Let us calculate: $$P_1 = V^2/R_1$$ $$P_2 = V^2/R_2$$ $$P_{total} = V^2/(R_1 + R_2)$$ Rearrange them and you'll get the result.

Answer 4

Assuming that the resistance of the bulbs is independent of the current passing through them.

When the bulbs are connected to a supply of voltage $V$ then the power dissipated in each of the bulbs is $P_1 = \dfrac{V^2}{R_1}$ and $P_2 = \dfrac{V^2}{R_2}$.

With the bulbs in series they do not have a voltage of $V$ across each of them and the total resistance is $R_1+R_2$

So now the power dissipated in the bulbs $P = \dfrac{V^2}{R_1+R_2}$

All you now need to do relate $P$ to $P_1$ and $P_2$.