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QPages 218-219: Q2: HC Verma: Concepts of Physics - Volume 2 Thermal and Chemical Effects of Electric Current Pages 218-219: Q2 Solution: A coil of resistance is connected across a battery of emf 6.0…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often write the electrical power as $P=V/R$ instead of the correct $P=V^2/R$ (or $I^2R$) after applying Ohm's law; this underestimates the heat generated and yields a time that is too large. Always substitute $I=V/R$ into $P=VI$ to get $P=V^2/R$ before equating $Q=Pt$.

A coil of resistance 100 Ω100 \ \Omega is connected across a battery of emf 6.0 V. Assume that the heat developed in the coil is used to raise its temperature. If the heat capacity of the coil is 4.0 J K−14.0 \text{ J K}^{-1}, how long will it take to raise the temperature of the coil by 15�C?

Step-by-Step Verified Solution

**Answer:** t=166.67t = 166.67 s ≈\approx 2.8 min. **Step-by-Step Explanation:** R=100 ΩR = 100\ \Omega, ε=6\varepsilon = 6 V, heat capacity =4.0= 4.0 J/K, ΔT=15�C\Delta T = 15�\text{C}. Heat required: Q=4.0×15=60Q = 4.0 \times 15 = 60 J. Heat produced: Q=ε2RtQ = \frac{\varepsilon^2}{R} t.
60=36100t⇒t=600036=166.67 s60 = \frac{36}{100} t \Rightarrow t = \frac{6000}{36} = 166.67\ \text{s}

All Questions in this Chapter

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