- #1

doktorwho

- 181

- 6

## Homework Statement

When the switch is open, the voltage is ##U_{ab}##. When the switch is closed the calculated max current is ##I_{max}##. Calculate the resistance of ##R_p## when its power is max and the current of ##I_g##

##R1=100,R3=200,R4=400##,##0<Rp<100## and ##U_{ab}=2V##,##I_{max}=10mA##

This is the problem we had today in school and i have solved the second part of it but am not clear about the first part. I will post the problem and the solution and hope you can help answer what's bugging me.

## Homework Equations

3. The Attempt at a Solution [/B]

The solution: The equivalent simplified system:

##E_t=U_{ab}=2V## {This part i don't get. Why is it U_{ab} when we also have the thevenin resistor so the total voltage doesn't equal ##U_{ab}## right?}

##I_{max}=\frac{E_t}{R_t+R_p}=\frac{E_t}{R_t}##

##R_t=200##

##R_t=\frac{R1R3}{R1+R3} + \frac{R2R4}{R2+R4}=200##

##R_2=200##

##U_{ab}=I_gR_t##

##I_g=I_{max}=10mA## {This should be because if ##I_g## was higher than the resistor could not handle it? Could have we stated this from the start then?}

since ##R_{pmax}<R_t## then ##R_p=R_{pmax}=100## {How did they deduce that ##R_{pmax}## is smaller than ##R_t##. Cant see they derived it anywhere? What am i missing?