Match List - I with List - II. Choose the correct answer from the options given below :List - I List - II (A) Electrical resistance (I) °C/W (B) Thermal resistance (II) W (C) Time constant (III) RC (D) Rate of heat transfer (IV) V/A
(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
Every pairing follows from reading the units: (A)-(IV), (B)-(I), (C)-(III), (D)-(II), option 4.
| Quantity | Units | Why |
|---|---|---|
| Electrical resistance | V/A | Ohm's law, \(R=V/I\) |
| Thermal resistance | °C/W | Temperature rise per watt dissipated |
| Time constant | RC | Ohms times farads gives seconds |
| Rate of heat transfer | W | Energy per unit time is power |
The thermal-electrical analogy runs right through the list, and seeing it makes every pairing obvious at once:
\(V\leftrightarrow\Delta T,\qquad I\leftrightarrow P,\qquad R_{elec}\leftrightarrow R_{th}\)
Ohm's law \(V=IR\) becomes
\(\Delta T=P\,R_{th}\)
so thermal resistance carries degrees per watt exactly as electrical resistance carries volts per amp. Heat flow plays the part of current, and temperature difference the part of voltage.
Why this matters in practice. A power device's junction temperature is found by treating the heat path as a chain of resistances in series:
\(T_{j}=T_{a}+P\left(R_{th,jc}+R_{th,cs}+R_{th,sa}\right)\)
— junction to case, case to sink, sink to ambient. A transistor dissipating 10 W with a total of 5 °C/W runs 50 °C above ambient, and the whole heat-sink calculation is a resistor-network problem in disguise.
The time constant completes the analogy. Just as \(\tau=RC\) in an electrical circuit, a thermal system has \(\tau=R_{th}C_{th}\) where thermal capacitance is the mass times specific heat — which is why a large heat sink warms up slowly and can absorb short bursts of dissipation without its temperature rising much.
A check on the answer : only options 3 and 4 pair electrical resistance with V/A, and only option 4 then gives thermal resistance the °C/W it must have.
Hence, the correct code is (A)-(IV), (B)-(I), (C)-(III), (D)-(II).
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