Find Max Power Dissipation in Circuits with Variables

[stunner5000pt]

In the circuit, E, R1 and R2 have constant values. However R has a variable value. FInd an epxression for R that results in the max power dissipation in that resistor.
so do i do this
$$i_{1} = i + i_{2}$$
$$E - i_{1} R_{1} - i_{2} R_{2} = 0$$
$$E - i_{1} R_{1} - iR = 0$$
$$i_{2} R_{2} - iR = 0$$
not too sure about the last one...
and the power dissipation in that resitor is given by $$P = i^2 R$$ But to find the max power dissipation do i plug that expression for i back into the above expression and then differentiate OR sub i into this
$$\frac{dP}{dR} = 2iR + i^2$$
i used a sample value for i and got different values for the derivative. Which is correct?
Please help!

 

[BobG]

You should use your Thevenin equivalent and Norton equivalent for this one - unless they're making you do this using differentiation and limits at least once just to make sure you truly appreciate Thevenin and Norton equivalents.

 

[quicknote]

To find the maximum power dissipation in the resistor, we need to maximize the power equation, P = i^2R. This can be done by taking the derivative of P with respect to R and setting it equal to 0. So, we have:

\frac{dP}{dR} = 2iR + i^2 = 0

Solving for R, we get:

R = -\frac{i}{2}

Substituting this value back into the original power equation, we get:

P = i^2 \left(-\frac{i}{2}\right) = -\frac{i^3}{2}

To find the maximum power dissipation, we need to find the maximum value of i. Since E, R1, and R2 are constant, the current i will be maximum when R is minimum. Therefore, to maximize power dissipation, we need to have R = 0. This means that the maximum power dissipation will occur when the resistor is short-circuited.

In conclusion, to find the maximum power dissipation in the circuit, we need to short-circuit the variable resistor R. This will result in the expression for R that results in the maximum power dissipation being R = 0.