NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 7.1. A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle? Solution : (a) $I_{rms} = \frac{v_{rms}}{R} = \frac{220}{100} = 2.20 A$ (b) $Net power = V_{rms} \times I_{rms} = 220 \times 2.20$ = 484 W 7.2 (a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current? Solution: (a) $V_{rms} = \frac{V_{0}}{\sqrt{2}} =\frac{300}{\sqrt{2}} = 212.1 V$ (b) $I_{rms} = \frac{I_{0}}{\sqrt{2}}$ $I_{0} = I_{rms} \sqrt{2} = 10 \sqrt{2} = 14.1A$ 7.3 A 44 mH inductor is connected to 220 V , 50 Hz ac supply. Determine the rms value of the current in the circuit. Solution: Here , Reactance $X_{L} = 2 \pi\nu L = 2\pi \times 50 \times 44 \times 10^{-3}$ $\therefore I_{rms} = \frac{V_{rms}}{X_{L}} = \frac{220}{2\p...