# Contradicting relation between total power consumed and power consumed by each individual resistor in series and parallel configuration

In a given series configuration of n resistances of equal resistance r, I derived the relation bewteen total power and individual power consumption as follows:

P(Total) = I^2 * R

R= nr (equivalent resistance)

P(total) = n*I^2*r = n*p(individual) {p individual = I^2*r}

If the configuration is parallel:

P (total) = V^2/R = V^2 * n/r =n* p(individual) {p individual = V^2/r}

But this seems to imply that power consumed is the same for both.

Can someone point out where I went wrong?

Thanks

## 2 Answers

They only look like they're consuming the same power because you're implicitly using a different power source in each case.

For example, let's first look at a constant-voltage source, which maintains a constant voltage drop $$V$$ across whatever load it's connected to, and allows the current to vary. Let $$I_s$$ be the current drawn by the series configuration, and let $$I_p$$ be the current drawn by the parallel configuration. The series configuration has an equivalent resistance of $$nr$$, so $$I_s=\frac{V}{nr}$$. In contrast, the parallel configuration has an equivalent resistance of $$r/n$$, so $$I_p=\frac{nV}{r}$$.

This means that the power consumed by the series configuration is $$P_s=I_s^2(nr)=\frac{V^2}{nr}$$, and the power consumed by the parallel configuration is $$P_p=I^2_p(\frac{r}{n})=\frac{nV^2}{r}$$. The power consumed by the parallel configuration is higher for a constant-voltage source.

In contrast, a constant-current source maintains a constant current $$I$$ through whatever load it's connected to, and allows the voltage drop to vary. Let $$V_s$$ be the voltage drop across the series configuration and $$V_p$$ be the voltage drop across the parallel configuration. We see, as before, that $$V_s=Inr$$ and $$V_p=\frac{Ir}{n}$$.

This means that the power consumed by the series configuration is $$P_s=\frac{V_s^2}{nr}=I^2nr$$ and the power consumed by the parallel configuration is $$P_p=\frac{nV_p^2}{r}=\frac{I^2r}{n}$$. The power consumed by the series configuration is higher for a constant-current source.

So which configuration consumes more power depends on what kind of power supply it's connected to. In your question, you compare the power consumed by series resistors attached to a constant-voltage source with the power consumed by parallel resistors attached to a constant-current source, which really aren't the same situation.

Yes, I can point it out.

You went wrong in line seven of your question.

In both the parallel and serial case the total power consumed is indeed n times the power consumed by an individual resistor.

However, in the parallel case, each resistor takes the full voltage V. The power consumed by each resistor is VV/r so the total power is nVV/r.

If the resistors are in series the `voltage drops gradually over each of them, so an individual resistor sees only V/n volts. The power consumed by each resistor is VV/nnr. The total power consumed is nVV/nnr or VV/nr.

The power differs by a factor of nn.

Best wishes.