On a real voltage source of \(10V\) with internal resistance of \(100\Omega\), a resistor is connected. The power on the resistor is \(200mW\). Find the resistance.

I know Ohm's law, Kirchhoff's laws, Power formulas
\(P=I^{2}R\)
\(P=\frac{E^{2}}{R}\)
\(P=EI\)

I tried finding R or I from \(R=\frac{P}{I^{2}}\)


This may be an easy example but I don't know I can't find R.
I'm solving harder problems, but can't find this one.
Thanks in advance.

 

The source terminal voltage

V=10-100*I

The load power

P=0.2=V*I

V=0.2/I=10-100*I

and so on to solve for I from a quadratic equation & finally R=P/I^2=0.2/I^2

There is an unexpected outcome by the way......

 

yeah, I tried this
but from the quadratic equation there are two outcomes
I1 and I2,
0.028 and 0.072
Im really confused.

 

Simply you have two proper solution to your problem.

 

@Jony130
wow how can that be true.
never thought there can be two solutions...
I solved problems like this one before, but from the quadratic equation there was only one outcome.
for example I remember this one...

\(6I^{2}-24I-24=0\)
this equation has only one solution, which is 2.

anyway thanks Jony and t_n_k

 

It is not difficult to see that there must be two solutions to this problem. The maximum power will be transferred into the load when it is made equal to the generator resistance, in this case 100Ω. This results from the Maximum Power Transfer Theorem:

http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

The load voltage at maximum power will therefore be half the open-circuit value, i.e. 5V, so that the current will be 5V/100Ω = 50mA, and the output power will be 5V*50mA = 250mW.

The load power will also reduce if its resistance is varied either up or down from 100Ω, so we can expect to obtain 200mW for two current levels, one somewhat greater than 50mA and another somewhat less.

 

Plotting V-I characteristic against the P=0.2W curve shows there are two valid conditions for 0.2W load power.

 

\(6I^{2}-24I-24=0\)
this equation has only one solution, which is 2.

anyway thanks Jony and t_n_k

Just as an aside, this equation has two solutions, neither of which is 2.

 

Just as an aside, this equation has two solutions, neither of which is 2.

yep, you're right.
+24 instead of -24, the "c" element.

\(6I^{2}-24I+24=0\)