Physics · Thermal Properties Of Matter · NEET
They are in PARALLEL. Picture two rods glued lengthwise, like two lanes of a road running together. Heat enters both left faces at temperature T1 and leaves both right faces at temperature T2. Both rods have the SAME length L and the SAME temperature difference (T1 - T2). Whenever the temperature difference is shared, the arrangement is parallel. (In series, rods are joined end to end, share the same heat current, and the drop splits between them.)
In parallel the two heat currents add. Total current H = H1 + H2 = K1*A*(dT)/L + K2*A*(dT)/L. We treat the pair as one rod of area (A + A) = 2A: H = Keff*(2A)*(dT)/L. Setting them equal, Keff*2A = K1*A + K2*A, so Keff = (K1 + K2)/2. This only gives the simple average when both cross-sections are equal.
Use the area-weighted formula Keff = (K1*A1 + K2*A2)/(A1 + A2). The (K1 + K2)/2 result is just the special case where A1 = A2. Always check the figure: if areas differ, weight each K by its own area, not by 1/2.
SAME for both: the length L and the temperature difference (T1 - T2). DIFFERENT: the heat current through each rod, because K1 and K2 differ, so the rod with higher K carries more heat. This is the mirror image of series, where the current is the same but the temperature drop differs.
Thermal conductance (K*A/L) plays the role of electrical conductance (1/R). In parallel, conductances ADD, so K*A values add. That is why the combined conductivity is a weighted average that goes UP, never down. If you instead added thermal resistances you would be doing the series rule by mistake.
Two rods A and B of different materials are welded together side by side as shown in the figure. Their thermal conductivities are K1 and K2. The effective thermal conductivity of the composite rod (for heat flowing from the left face at T1 to the right face at T2) will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
When they are placed side by side so both share the same length and the same two end temperatures. Heat flows through both at the same time, and the two heat currents add.
Keff = (K1 + K2)/2. The conductivities average because the combined rod has double the cross-sectional area while length and temperature difference stay the same.
More. Adding a second rod adds another path, so total heat current increases. The effective conductivity always lies between K1 and K2 (never below the smaller one).
Keff = (K1*A1 + K2*A2)/(A1 + A2). Use this when the two rods have different cross-sectional areas; it reduces to (K1 + K2)/2 when A1 = A2.
End-to-end joined rods share the same heat current = series (thermal resistances add). Side-by-side welded rods share the same temperature difference = parallel (K*A values add).
