I have connected 2 cells one to one bulb another to 2 bulbs ,which would finish faster (it is a series circuit)?

I have thought of two solutions:-

Method 1: The cell connected with one bulb would finish faster as it offers less resistance and the charges would move faster in the circuit and take less time to reach the cell's other terminal and the charges would be used up faster.

Problem: Cell doesn't work by giving charges so it should not be the case or is it?

Method 2: The cell with 2 bulbs would finish faster because they require more voltage and current and electric energy so it should finish faster.

Problem: A cell maintains a constant voltage so only current and resistance can change but not the Voltage and so there would be no difference in the potential difference or electric energy.

I think the method 1 is correct but method 2 looks more logical and is answered by common sense but still I think method 1 is correct anyways both have their own problems so which one is correct and how can we resolve the problem with the correct method.


2 Answers 2


It depends how you connect the bulbs.

In series, the resistances add, so the total resistance is 2R.

In parallel, the total resistance is $(\frac{1}{R} + \frac{1}{R})^{-1} = R/2$

Since $P = \frac{V^{2}}{R}$, lower resistance drains the battery faster.

So 2 parallel bulbs drains the cell first (as according to your common sense). Then 1 single bulb. Then in last place, 2 bulbs in series.

To appeal more to common sense:

Parallel circuit: more copies of a circuit (same voltage difference) draining from the same battery should drain the battery faster. Turning on more lightbulbs in your house uses energy faster (because they are in parallel).

Series circuit: a large enough number of bulbs in series with very large total resistance would be like the resistance of an open circuit (resistance of air) with no current. This obviously drains the battery slower, like a battery sitting in your shelf.

  • $\begingroup$ It is about series circuit but still I would like to see it in more detail, why two bulbs in parallel would drain the battery faster. $\endgroup$
    – Mohd Saad
    Sep 27, 2021 at 5:22
  • $\begingroup$ In parallel, you can imagine you have two copies of the same circuit draining the same amount of power. Each circuit has the same voltage difference, and the same resistance, so it also has the same current, so the two copies have the same power. So the power drain should be 2x. $\endgroup$
    – Alwin
    Sep 27, 2021 at 7:04
  • $\begingroup$ This makes sense. $\endgroup$
    – Mohd Saad
    Sep 27, 2021 at 7:51

1 is correct. Given battery potential $V_b$, load resistance $R_l$, battery internal resistance $R_b$

The power dissipated by the circuit composed of a battery and a load resistance is (from $P = I^2R, I = V/R$)

$P = V_b^2/(R_l + R_b)$

As you can see, increasing $R_l$ results in less power dissipation.

  • $\begingroup$ @g s yes it looks correct and can you tell me what this equation is called, I have seen it for the first time so that I can know more about it. $\endgroup$
    – Mohd Saad
    Sep 27, 2021 at 4:17
  • 1
    $\begingroup$ This is Ohm's Law, with some algebraic manipulation. $\endgroup$
    – g s
    Sep 27, 2021 at 4:18
  • $\begingroup$ @g s No I am asking about how you got the equation for power dissipation i.e. $P=I^2R$ $\endgroup$
    – Mohd Saad
    Sep 27, 2021 at 4:22
  • 1
    $\begingroup$ That one isn't usually given a name. It's the power dissipated by a resistor. If we were to name it, it's a generalization of of Joule's Law ($dQ = I^2Rdt$), where Q is heat, not charge. $\endgroup$
    – g s
    Sep 27, 2021 at 4:32
  • $\begingroup$ @g s Okay, I this looks good. $\endgroup$
    – Mohd Saad
    Sep 27, 2021 at 4:38

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