Param Himalaya - परम हिमालय

8.1 Figure 8.5 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.  (a) Calculate the capacitance and the rate of change of potential difference between the plates.  (b) Obtain the displacement current across the plates. c) Is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor? Explain  Solution :  (a)Using  $$C = \frac{\epsilon_{0} A}{d}$$ $$C = \frac{8.854 \times 10^{-12} \times \pi r^{2}}{d}$$ $$C = \frac{8.854 \times 10^{-12} \times 3.14 \times (12\times 10^{-2})^{2}}{5\times 10^{-2}}$$ $$C = 800.7 \times 10^{-14}$$ $$C = 8 pF$$ Now using  $$V = \frac{Q}{C}$$ $$\frac{dV}{dt} = \frac{1}{C}\frac{dQ}{dt}$$ $$or \frac{dV}{dt} = \frac{I}{C} = \frac{0.15}{8 \times 10^{-12}}$$ $$\frac{dV}{dt}= 1.875 \times 10^{10}VS^{-1}$$ (b) Displacement Current  $$I_{d} = \...

NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current - Param Himalaya  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}}...

Ncert Solution class 12 Physics chapter 6 Electromagnetic induction : 6.1 Predict the direction of induced current in the situation described by the following Fig. 2(a) to (f). Ans : (a)  The given figure shows the South Pole of a bar magnet moving towards a closed loop. The direction of the induced current in a closed loop is given by Lenz's law. Using Lenz's rule, the direction of the induced current will be such that the face of the loop towards the approaching South Pole behaves as South Pole. Because the current induced in solenoid will oppose the approach of magnet and therefore the current at the nearer face should flow clockwise i.e., 𝑞𝑟𝑝𝑞 And (b)  i The given figure shows the South Pole of a bar magnet moving towards the first closed loop (qrpq) and North Pole of the bar magnet moving away from the second loop (yzxy). The direction of the induced current in a closed loop is given by Lenz's law. Using Lenz's rule, the direction of the induced current in the ...