MOSFET

Suppose the transistor acts as a switch.

If so, the voltage applied to the control point will be: $\frac{V_{in}*r}{r+r}=5(V)$ , which is above the threshold voltage.

Therefore, the transistor must act as a resistor.

The current through the transistor is: $I=\frac{V_{in}}{R+R_{on}}$ .

The power released on the transistor is: $W=I^2R_{on}=0.444(J)$ .

Basically, once we see a problem like this, our first instinct should be to get rid of the "variables", which basically means, in this context, that we should determine what the MOSFET will do.

Let us first presume that the MOSFET will act as a switch, and we will try to derive a contradiction. Whence, we know that the voltage would be $\frac{(V-in) \times r}{2r} = 5V$ . Now, the problem statement says that:

switches (open circuit) if the voltage at the control is below the threshold

Since the threshold is $2V$ and clearly $5V > 2V$ , we have the desired contradiction.

Now, we have successfully determined that the MOSFET will act as a resistor. Now recall that the power $(W)$ released on a resistor is $I^2 \times R_{on}$ (*), where I is the current. So we need to determine $I$ .

Recall once again that the current across equivalent resistor $R_t$ is $I = \frac{V}{R_t}$ . However in the problem, our desired $R_t$ is split into R and MOSFET arranged in series, recall that in such a scenario, $R_t = R_{on} + R$ .

Subbing everything back into (*) gives:

$W = 0.444$

Since the control voltage $V_{CP} = \frac {V_{in}r}{r+r} = \frac {10}{2} = 5V > V_t (=2V)$ , the MOSFET is on and acts as a $100\Omega$ resistor.

When the MOSFET is on, the current through it,

$I = \frac {V_{in}}{R+R_{on}} = \frac {10}{50+100} = 0.066666667 A$

The power released on the MOSFET,

$P = I^2R_{on} = 0.066666667^2 \times 100 = \boxed {0.444}W$ .