Power dissipated by resistors in series versus in parallel for fixed voltage

The most straightforward way to reason about this doesn't require much math.

The power delivered by the voltage source to either pair of resistors is inversely proportion to their combined resistance, i.e, if the combined resistance is greater, the power delivered is smaller.

$$p_R = \dfrac{V^2}{R}$$

Now, recall that:

• the series combination of two resistances is always greater than either individual resistance

• the parallel combination of two resistances is always less than than resistance of either individual resistance

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For example, suppose that both resistors have the same value of resistance $R$.

Now, if the two resistors are connected in series, the equivalent resistance is $R_{EQ}=2R$.

But, if the two resistors are connected in parallel, the equivalent resistance is $R_{EQ}=\dfrac{R}{2}$.

Thus, the power for the series combination is:

$$p_{series} = \dfrac{1}{2}\dfrac{V^2}{R} $$

Whilst the power for the parallel combination is:

$$p_{parallel} = 2\dfrac{V^2}{R}$$

In this case, the parallel combination dissipates 4 times the power of the series combination.

In general, if the power consumed would depend on the circuit structure. But for a simple case, such as two resistors connected in series versus the same resistors connected in parallel (with identical voltage sources in both), the power dissipated in the parallel combination will be greater.

Power depends on voltage across the circuit and resistance of the circuit

\begin{equation} P = \frac{V^2}{R};\\ P_{series} = \frac{V^2}{(R_1+R_2)};\\ P_{parallel} = \frac{V^2}{(R_1^{-1}+R_2^{-1})^{-1}}=\frac{V^2}{\frac{R_1R_2}{R_1+R_2}}=\frac{V^2(R_1+R_2)}{R_1R_2};\\ \frac{P_{series}}{P_{parallel}} = \frac{R_1R_2}{(R_1+R_2)^2} \end{equation}

Since $R_1$ and $R_2$ are always positive, $R_1R_2 < (R_1+R_2)^2$ i.e. $P_{series} < P_{parallel}$