Verified Solution • 2025–26 Session moderate

QPages 198-203: Q76: HC Verma: Concepts of Physics - Volume 2 Electric Current in Conductors Pages 198-203: Q76 Solution: A capacitance C charged to a potential difference V is…

⚠️ Common Exam Traps & Conceptual Pitfalls

When integrating $i^2 R$ for heat, students often forget that squaring the exponential decay term $e^{-t/RC}$ results in $e^{-2t/RC}$, effectively doubling the decay rate. This leads to the correct upper limit evaluation of $e^{-2}$ rather than the erroneous $e^{-1}$, which is a critical distinction for accurate energy calculation.

A capacitance C charged to a potential difference V is discharged by connecting its plates through a resistance R. Find the heat dissipated in one time constant after the connections are made. Do this by calculating

\int i^2R \ dt

and also by finding the decrease in the energy stored in the capacitor.

Step-by-Step Verified Solution

**Answer:**

H = \\frac{1}{2}CV^2(1 - e^{-2}) = 0.432CV^2

. **Step-by-Step Explanation:**

i = \\frac{V}{R}e^{-t/RC}

.

H = \\int_0^{\\tau} i^2R dt = \\frac{V^2}{R}\\int_0^{\\tau} e^{-2t/RC}dt = \\frac{CV^2}{2}(1 - e^{-2})

[citation:2].

All Questions in this Chapter

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