NCERT Exemplar Solutions for Class 12 Physics - Chapter 7: Alternating Current (SA)
NCERT Exemplar Solutions for Class 12 Physics Chapter 7 Alternating Current are available here. Here, you will get solutions of short answer type questions (i.e. question number 7.21 to 7.26).
We have already provided solutions of multiple choice questions with single correct answer (MCQ I), multiple choice questions with multiple correct answers (MCQ II) and very short answer type questions (VSA). You can access them from links given below
NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 7 (from question number 7.21 to 7.26) are given below:
Question 7.21: A device ‘X’ is connected to an AC source. The variation of voltage, current and power in one complete cycle is shown in Fig 7.5.
(a) Which curve shows power consumption over a full cycle?
(b) What is the average power consumption over a cycle?
(c) Identify the device ‘X’.
Because the curve of power will be having maximum amplitude, equals to multiplication of amplitudes of voltage (V) and current (I) curve.
Because the full cycle of the graph consists of one positive and one negative symmetrical area.
(c) The average power is zero; the device may be inductor (L) or capacitor (C) or the series combination of L and C.
Question 7.22: Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current?
An AC current changes direction with the source frequency and the attractive force would average to zero. Thus, the AC ampere must be defined in terms of some property that is independent of the direction of current. Joule’s heating effect is such property and hence it is used to define rms value of AC.
Question 7.23: A coil of 0.01 Henry inductance and 1 ohm resistance is connected to 200 volt, 50 Hz ac supply. Find the impedance of the circuit and time lag between max. alternating voltage and current.
Question 7.24: A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the type of transformer being used.
PL = 60 W, IL = 0.54 A
VL = 60/0.54 = 110 V
The transformer is step-down and have ½ input voltage.
Hence, ip = (½) × I2 = 0.27 A.
Question 7.25: Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency.
A capacitor does not allow flow of direct current through it as the resistance across the gap is infinite. When an alternating voltage is applied across the capacitor plates, the plates are alternately charged and discharged. The current through the capacitor is a result of this changing voltage (or charge). Thus, a capacitor will pass more current through it if the voltage is changing at a faster rate, i.e. if the frequency of supply is higher. This implies that the reactance offered by a capacitor is less with increasing frequency; it is given by 1/ωC.
Question 7.26: Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.
An inductor opposes flow of current through it by developing a back emf according to Lenz’s law. The induced voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice versa. Since the induced emf is proportional to the rate of change of current, it will provide greater reactance to the flow of current if the rate of change is faster, i.e. if the frequency is higher. The reactance of an inductor, therefore, is proportional to the frequency, being given by ωL.