Discussion
A transistor dissipates \( 2 \ \mathrm{W} \) with a junction-to-case resistance of \( 3 \ \mathrm{^\circ C/W} \) and a case at \( 60 \ \mathrm{^\circ C} \). Its junction reaches
Answer: C
Adding the rise across the junction-to-case path:
\[ T_{j} = 60 + (2 \times 3) = 66 \ \mathrm{^\circ C} \]
giving 66 degrees. The value 63 uses one watt and 126 doubles the case temperature. Thermal problems are all the same series circuit with temperature in place of voltage and power in place of current, which makes them easy once the analogy is accepted.
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