HC Verma: Concepts of Physics - Volume 2 Alternating Current Pages 330-331: Q7 Solution: The current in a discharging LR circuit is given by , where is…

QPages 330-331: Q7: HC Verma: Concepts of Physics - Volume 2 Alternating Current Pages 330-331: Q7 Solution: The current in a discharging LR circuit is given by , where is…

⚠️ Common Exam Traps & Conceptual Pitfalls

When computing $I_{ ext{rms}}=sqrt{rac{1}{T}int_0^T i^2,dt}$ for a decaying exponential, the integral $int_0^T e^{-2t/ au}dt$ yields $( au/2)(1-e^{-2T/ au})$. The factor $ au$ cancels with the $1/T$ prefactor, so no extra $ au$ remains in the final expression. Forgetting this cancellation leads to an incorrect $ au$ in the denominator.

The current in a discharging LR circuit is given by

i = i_0 e^{-t/\tau}

, where

\tau

is the time constant of the circuit. Calculate the rms current for the period

t = 0

t = \tau

Step-by-Step Verified Solution

**Answer:**

I_{rms} = i_0 \sqrt{\frac{1 - e^{-2}}{2\tau}}

or similar. **Step-by-Step Explanation:**

I_{rms}^2 = \frac{1}{\tau} \int_0^\tau i_0^2 e^{-2t/\tau} dt = \frac{i_0^2}{\tau} \left[ \frac{-\tau}{2} e^{-2t/\tau} \right]_0^\tau = \frac{i_0^2}{2} (1 - e^{-2})

. Thus,

I_{rms} = i_0 \sqrt{\frac{1 - e^{-2}}{2}}

[citation:9].

QPages 330-331: Q7: HC Verma: Concepts of Physics - Volume 2 Alternating Current Pages 330-331: Q7 Solution: The current in a discharging LR circuit is given by , where is…

Step-by-Step Verified Solution

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