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QPages 218-219: Q12: HC Verma: Concepts of Physics - Volume 2 Thermal and Chemical Effects of Electric Current Pages 218-219: Q12 Solution: The resistor shown in the figure is dipped into a calorimeter…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently take the total circuit power $P=V^2/R_{\text{eq}}$ and assume it all heats the water, overlooking that only the immersed resistor dissipates heat. First find the current through that resistor (using current division) and then use $Q=I^2R_{\text{water}}t$ to compute the temperature rise.

The 2.0 Ω2.0 \ \Omega resistor shown in the figure is dipped into a calorimeter containing water. The heat capacity of the calorimeter together with water is 2000 J K−12000 \text{ J K}^{-1}. a) If the circuit is active for 15 minutes, what would be the rise in the temperature of the water? b) Suppose the 6.0 Ω6.0 \ \Omega resistor gets burnt. What would be the rise in the temperature of the water in the next 15 minutes?

Step-by-Step Verified Solution

**Answer:** (a) ΔT=2.916\Delta T = 2.916 K (b) ΔT=3.6\Delta T = 3.6 K. **Step-by-Step Explanation:** (a) Effective resistance: Reff=1+(6∥2)=1+1.5=2.5 ΩR_{eff} = 1 + (6 \parallel 2) = 1 + 1.5 = 2.5\ \Omega. Current i=6/2.5=2.4i = 6/2.5 = 2.4 A. Current through 2 �: i′=2.4×6/8=1.8i' = 2.4 \times 6/8 = 1.8 A. Heat = 1.82×2×15×60=58321.8^2 \times 2 \times 15 \times 60 = 5832 J. ΔT=5832/2000=2.916\Delta T = 5832/2000 = 2.916 K. (b) When 6 � burns: Reff=1+2=3 ΩR_{eff} = 1 + 2 = 3\ \Omega. i=2i = 2 A. Heat = 22×2×900=72002^2 \times 2 \times 900 = 7200 J. ΔT=7200/2000=3.6\Delta T = 7200/2000 = 3.6 K [citation:9].

All Questions in this Chapter

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