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Do two resistors in parallel dissipate more power for a fixed applied voltage compared to the same two resistors in series?

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    $\begingroup$ Here's an idea: find net resistance and hence dissipated power in each case. $\endgroup$
    – pho
    Commented Oct 23, 2013 at 23:27
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/81992/2451 $\endgroup$
    – Qmechanic ♦
    Commented Oct 24, 2013 at 7:47

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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

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.

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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.

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    $\begingroup$ This is a statement rather than an explanation.... $\endgroup$
    – Bernhard
    Commented Oct 30, 2013 at 22:11
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    $\begingroup$ No offense, but the OP did not ask for an explanation. He merely asked a generic question "Is A sometimes greater than B?" If the question provided me with even the most remote of hints that the OP had some trouble with a concept, or maybe he/she was stuck at some point, I would have gladly provided a satisfactory explanation. $\endgroup$
    – Sam29
    Commented Oct 30, 2013 at 22:37
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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. $P_{series} < P_{parallel}$

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for same voltage supply, the power consumed by two resistances in series connection is less in compare to power consumed by same resistances in parallel connection. Therefore we can say that - P(series) < p(parallel)

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in P=VI so when connect the the batteries in parallel then voltage is improving and its current increases because of the less of the resistance ana powrer is directle proportional to both thats why power is also increases....

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They should because light bulbs connected in parallel definitely give you more energy per unit time :). It is because in parallel you have to circuits...for example, if you have a simple circuit with one bulb, and if you have total of 4 A of current, what will the current be if you add one more, in parallel? If you remember, sum of currents in two branches must always give a total current. So, you would say, ok, its 2 A per branch, but you would be wrong...now, you have total of 8 A, 4 A in each branch :) its cool, you are increasing your current by adding new branches.Electrons it those additional wires are effected by the battery in the same way as in the first wire, and you generate same current, provided you have equal bulbs. This current now pours into a meeting point of all branches and rushes into just ONE wire and since copper or any other material wants to have same density of its electrons, it must compensate for this...buy moving away at the same rate at which electrons are rushing in. So, each wire takes the same amount as one and in general you take away batteries energy faster, ie, you generate more power, yeah.. :) But, question that has been bugging me is when you have a series why should all the energy be dissipated in that bulb? Someone said its because the potential difference, it has to calm to its zero value when particle travels from neg to positive terminal, but that does not mean it should lose its all energy in dissipation, right? It could hit positive terminal and then come to zero. Why should it all be lost in a bulb? Please, comment.

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  • $\begingroup$ The total voltage is always dropped across the total load. In other words it don't matter if you measure at the load or the battery because it will be the same because you are measuring the difference in potential caused by the load. $\endgroup$
    – user77331
    Commented Apr 9, 2015 at 10:10
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but $$Heat=I^2*R*t$$ so, $$Heat_{\parallel}= \frac{I^2*R}{2t}$$ $$Heat series=i^2* 2rt=4 *Heat_\parallel$$ therefore heat produced should be greater in a series circuit.

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