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Asked in RS Academy GUJCET booklet · Heating and power in resistors
Given: 10 Ω in parallel with 4 Ω and 5 Ω in series; the heat in the 10 Ω resistor is 10 cal s⁻¹.
Idea: the two branches have the same potential difference V across them, so compare the heating rates through H=I²R with I=V/(R_branch).
10 Ω branch: I₁=V/(10), so H₁₀=(V²)/(100)×10=(V²)/(10).
Since H₁₀=10, V² is 100 in these units, and the whole 9 Ω branch gives (V²)/9=(100)/9≈11 cal s⁻¹.
4 Ω+5 Ω branch: I₂=V/9, so H₄=(V²)/(81)×4.
(H₄)/(H₁₀)=(4/81)/(1/10)=(40)/(81)≈0.494.
H₄≈0.494×10=4.94 cal s⁻¹, that is about 5 cal s⁻¹.
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