Textbook Example: find the load resistance $R_L$ such that the power at the load is $500\times10^6~\text{watt}$.

By Voltage Divider, voltage of load is:

$$V_L = \frac{R_T}{R_L + R_T}V_s\tag{1}$$

And the Power over load is:

$$P_L = \frac{V_S^2}{R_L}\tag{2}$$

Substitute (1) into (2)

$$P_L = \bigg(\frac{R_L V_S}{R_L+R_T}\bigg)^2 \frac{1}{R_L}\tag{3}$$

algebraically rearranging:

$$R_L^2 + \bigg(2R_T - \frac{V_S^2}{P_L}\bigg)R_L + R_T^2 = 0\tag{4}$$

Substituting $R_T = 14~ \Omega$ and $V_s = 500\times 10^6~V$

$$R_L^2 - 152 R_L + 196 = 0\tag{5}$$

When I solve this using quadratic formula i get two positive values for $R_L$:

$R_L = 76 \pm 75~\Omega$$

$R_L = 151~\Omega$ or $R_L = 1~ \Omega$

How do I know which is the correct value of $R_L$ to use?

Is it possible for both values to be correct? how would you know which is valid or that both are valid?

The only reason i was asking because the book i'm looking at fudged one of the values to have a negative resistance, to avoid the problem... but the calculation really comes out with both the resistances positive.



  • 1
    $\begingroup$ I just did the calculation and obtained your answer. Since both of the values are valid (the question has given no further information), I think it would be necessary to give both. This is fine! $\endgroup$
    – 13509
    Jan 6, 2020 at 17:07

2 Answers 2


The problem has indeed two solutions: $150.7\Omega $ and $1.3 \Omega$. Or to be exact $76\Omega \pm \sqrt{5584}\Omega$

Think about it: At the two extreme values of $R_L = 0, R_L = \infty$ the power through the load is zero. You get the maximum load at the power for $R_L = R_T$ which comes out to be about 3.21 GW.

That means as you vary the load resistance from $R_L = 0, R_L = \infty$ your load power will start at $0W$, go all the way up to $3.21 GW$ and come back to $0W$ again. Along the way you will hit $500 MW$ twice.

Both are correct solutions. From a practical stand point, the $1.3 \Omega$ doesn't make much sense since you are burning up 10 times more power on the grid then on the load so things are likely to get very hot hot & crispy.

  • $\begingroup$ This kind of extremal analysis is very useful for developing intuition about what is possible. And while the two options are equally physically valid, the quetion's author is presumbly hoping that the students will apply a little common sense to the problem--it's a valuable skill. $\endgroup$ Jan 6, 2020 at 23:36

The book is sloppy.

It should say:

$$b = 2R_T - \frac{V_S^2}{P_L} =2(14\,\Omega)-\frac{(300{\,\rm kV})^2}{500{\,\rm MW}}=28\,\Omega-180\,\Omega=-152\,\Omega$$

So they left some units out, and then made a factor of 10 error in the answer. Hence: your disagreement.

I'd also argue the approach is muddled, as there is no reason to invoke a voltage divider (I suppose it is part of the lesson, but still).

The physicist orient approach avoids rote formulae and focuses on what can be determined directly from the statement of the problem. It says: The power in $R_L$ is

$$ P = I^2R_L$$

and the current in the circuit is:

$$ I = \frac V R = \frac V {R_T + R_L}$$

Combining those gives the quadratic equation in $R_L$.

There is a solution with a 1.3 Ohm load, which dissipates 58 GW into the line, or 50 KW per foot. I don't think that is a good answer (for an engineer).

Also: Shouldn't a negative solution for $R_L$ raise a red flag? What does it mean? Is it for current flowing backwards? No... the power has to be positive, and $I^2$ is positive, so a negative resistance must indicate and error.

  • $\begingroup$ I kind of like it to be negative..because it makes it easier to throw away that root as not being a solution to the problem... what throws me off is that the 1 ohm is a solution and its also an inconsistent solution that doesn't drop 500 Mwatts of power over the load...Its interesting, because they normally don't let you see the ugly problems like that.. usually they are toy problems that are always nice with one negative root that you throw away...which makes me paranoid that I didn't understand something fundamental about electricity because both roots are positive... $\endgroup$
    – pico
    Jan 6, 2020 at 18:14
  • $\begingroup$ but 1.3 Ohm does drop 500 MW over the load. $\endgroup$
    – JEB
    Jan 6, 2020 at 18:32
  • $\begingroup$ Consider $R_L = 1~\Omega$, then by voltage divider equation: $V_L = \frac{1}{1+14}V_s=20~kV$...then calculating power over the load: $P_L = \frac{V_L^2}{R_L}=\frac{(20~kV)^2}{1}=400~\text{MWatt}$....however, $R_L = 151~\Omega$...$V_L=\frac{151}{151+14}=275~kV$..then calculating power over load $P_L = \frac{V_L^2}{R_L} =\frac{(275\times10^3)^2}{151}=500~\text{Mwatt}$... the only thing i could figure for way 1 ohm is only 400 Mwatt instead of 500 Mwatt is rounding error... or a mistake some where in calculation... $\endgroup$
    – pico
    Jan 7, 2020 at 3:46
  • $\begingroup$ I agree with Hilmar's reasoning... which makes it even more puzzling... $\endgroup$
    – pico
    Jan 7, 2020 at 3:52
  • $\begingroup$ i think its just really sensitive to rounding error in the 1 ohm range... because i can get it to overshoot 500 Mwatts if i use 1.3 ohms.... i noticed the book author truncated 14 ohm down from 14.3 ohms as well.. $\endgroup$
    – pico
    Jan 7, 2020 at 4:01

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