Very elementary doubt about power dissipation in DC circuits. Please see!

Suppose there are two circuits as described below:

a) A 12 V Battery to which connected are a 10Ω and 20Ω resistors in series thereby completing the circuit.
b) A 12 V Battery to which connected are a 10Ω and a 100Ω resistors in series thereby completing the circuit.

Both of the above circuits will consume power and thus depleting the battery.

My question is whether the power spent by the battery will be equal in both cases or not? If not then which one of the two cases will cause the battery to run out faster?
I am asking this by the perspective of the battery and its runtime. The confusion arose in my mind due to reading things like "the remaining power is dissipated by the resistor" and stuff like that.

Please respond

The battery is producing power not spending it in both cases but all batteries have an internal resistance which dissipates some power and the amount depends on the current you draw from the battery so 'a)' loses slightly more power than 'b)'.

In terms of battery life, that depends upon it's capacity and how much current you draw from it. The current drawn in case 'a' is (I = V/R) 12/(10+20) = 0.4A and in case 'b' it is 12/(10+100) = 0.109A so with four times the current consumed in circuit 'a' the battery would be depleted faster.

Brian.

So, in other words, if I have a DC circuit which consumes a certain amount of power from a battery thereby depleting it and now, for any arbitrary reason, I want the battery to last longer and thus reduce the amount of power drawn from the battery so as to prolong its runtime (disregard the inability of the circuit to function due to changes in voltage and/or current across it), all I have to do is put a resistance in series with the circuit? Is that right?

As Betwixt showed in his examples, it is not the power, but the product of current and time that determines how much charge is removed from the battery.

If you have a certain (electronic) circuit or product, you can't just add resistance in series as very likely the circuit/product will stop operating.

The capacity of a battery is expressed in Ah or mAh. You can see it as the product of time and current. If you have a 50Ah battery, it will be completely depleted when you draw 1A during 50h, or for example 10A during 5h. Please note that this is a simplification as actual capacity of a battery depends on the current drawn from it and for example temperature.

Batteries such as a car battery don't want to be depleted completely as this will reduce the life time, or even destroys it. Normally you don't fully deplete any battery system during its normal life time.

If you have a certain (electronic) circuit or product, you can't just add resistance in series as very likely the circuit/product will stop operating.

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I already stated in my reply that for the sake of the discussion here, it's not one of the concerns whether or not the circuit will continue or stop operating. I simply ask whether or not the power usage will vary.

And how is it that power doesn't affect the capacity? If the capacity is a function of current and time then current is a function of voltage and power as well. It is power which when multiplied by time which gives the energy spent (or used). If not so then why do we get electricity bill based on kWH units?

When you buy a battery, they state the nominal voltage and the Ah rating, not energy or power. See below the formulas to relate Ah product to Energy and power.
t should be taken in s (seconds).

1 Ah = 3600 A*s, 1mAh = 3.6 A*s

(instantaneous) power = U*I (or V*I if you like),
Energy = integral(U*I*dt), when voltage is constant: Energy = U*I*t

So when you put more batteries of same capacity in series, you have more stored energy because of the higher voltage. When you put the same batteries in parallel, you have also more energy, but now because of the higher overall battery capacity (expressed in Ah).

WimRFP said:

When you buy a battery, they state the nominal voltage and the Ah rating, not energy or power. See below the formulas to relate Ah product to Energy and power.
t should be taken in s (seconds).

1 Ah = 3600 A*s, 1mAh = 3.6 A*s

(instantaneous) power = U*I (or V*I if you like),
Energy = integral(U*I*dt), when voltage is constant: Energy = U*I*t

So when you put more batteries of same capacity in series, you have more stored energy because of the higher voltage. When you put the same batteries in parallel, you have also more energy, but now because of the higher overall battery capacity (expressed in Ah).

Click to expand...

That's exactly what I meant. Thanks for clarifying though. But, i still didn't get my answer (or did i?) :S