Homework problem

A MOSFET has a power dissipation of $P_{d} = 10\text{W}$. The MOSFET is mounted on a heatsink. There is an isolation pad between the MOSFET case and the heatsink. We are given the thermal resistances: -

Junction to case: $R_{th,jc}= 1.7 \: \frac{°\text{C}}{\text{W}}$

Isolation pad: $R_{th,cs}= 1.3 \: \frac{°\text{C}}{\text{W}}$

Heatsink to ambient air: $R_{th,sa}= 2.0 \: \frac{°\text{C}}{\text{W}}$

The ambient temperature around the MOSFET is 25°C. What is the temperature of the MOSFET case?

My attempt

The scenario must look something like this: -

Here I have represented the thermal resistances as resistors, the dissipated power as a current source, and the temperatures as potentials. From this I calculate the following: -

$$P_d = \frac{T_c-T_a}{R_{th,cs}+R_{th,sa}} \Leftrightarrow$$$$T_c = P_d(R_{th,cs}+R_{th,sa})+T_a = (10\cdot 3.3)°C + 25°C = 58°C$$

So the case must be 58°C.

Is my thought process/calculations correct or have I misunderstood something? Moreover, are calculations like these useful/used when designing circuits or are they only good for "back of the envelope" calculations?