Verified Solution • 2025–26 Session moderate

QPages 330-331: Q11: HC Verma: Concepts of Physics - Volume 2 Alternating Current Pages 330-331: Q11 Solution: A resistor of resistance is connected to an AC source . Find…

A resistor of resistance 100 Ω100 \ \Omega is connected to an AC source ε=12 V)sin⁡250π s−1)t\varepsilon = 12 \text{ V}) \sin250 \pi \text{ s}^{-1})t. Find the energy dissipated as heat during t=0t = 0 to t=1.0 mst = 1.0 \text{ ms}.

Step-by-Step Verified Solution

**Answer:** Energy ≈1.4×10−3 J\approx 1.4 \times 10^{-3} \, \text{J} or as calculated. **Step-by-Step Explanation:** Erms=122 VE_{rms} = \frac{12}{\sqrt{2}} \, \text{V}. Pavg=Erms2R=144/2100=0.72 WP_{avg} = \frac{E_{rms}^2}{R} = \frac{144/2}{100} = 0.72 \, \text{W}. Energy in 1 ms=0.72×10−3 J1 \, \text{ms} = 0.72 \times 10^{-3} \, \text{J} [citation:9].

All Questions in this Chapter

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