NCERT Exemplar Solutions for Class 12 Physics Chapter 7 – Alternating Current are available here. In this article, you will get solutions from question number 7.8 to question number 7.13. These are basically multiple choice questions with multiple correct answers (or MCQ II).
These questions can be asked in competitive exams like NEET, WBJEE, JEE Main, UPSEE etc & CBSE Class 12 Physics board exams.
NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 7 (from question number 7.8 to 7.13) are given below:
As the frequency of an ac circuit increases, the current first increases and then decreases. What combination of circuit elements is most likely to comprise the circuit?
(a) Inductor and capacitor.
(b) Resistor and inductor.
(c) Resistor and capacitor.
(d) Resistor, inductor and capacitor.
Solution 7.8: (a, d)
In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?
(a) Only resistor.
(b) Resistor and an inductor.
(c) Resistor and a capacitor.
(d) Only a capacitor.
Solution 7.9: (c, d)
As, the current increases on increasing the frequency of supply, this is possible in case of a capacitive circuit.
For a capacitive circuit, XC = 1/(ωC) = 1/(2πfC)
As, frequency f increases, Xc decreases.
Therefore, circuit must have a capacitor; however it may or may not have a resistor.
Electrical energy is transmitted over large distances at high alternating voltages. Which of the following statements is (are) correct?
(a) For a given power level, there is a lower current.
(b) Lower current implies less power loss.
(c) Transmission lines can be made thinner.
(d) It is easy to reduce the voltage at the receiving end using step-down transformers.
Solution 7.10: (a, b, d)
We have to transmit energy (power) over large distances at high alternating voltages, so current flowing through the wires will be low because for a given power (P).
P = Erms Irms
Irms is low when Erms is high.
Power loss = I2rms R = low (So, Irms is low)
For an LCR circuit, the power transferred from the driving source to the driven oscillator is P = I2Z cos φ.
(a) Here, the power factor cos φ ≥ 0, P ≥ 0.
(b) The driving force can give no energy to the oscillator (P = 0) in some cases.
(c) The driving force cannot syphon out (P < 0) the energy out of oscillator.
(d) The driving force can take away energy out of the oscillator.
Solution 7.11: (a, b, c)
Power transferred, P = I2Z cos φ
As power factor, cos ϕ = R/Z
Here, R > 0 and Z > 0
⇒ cos ϕ > 0
⇒ P > 0
Question 7.12: When an AC voltage of 220 V is applied to the capacitor C
(a) the maximum voltage between plates is 220 V.
(b) the current is in phase with the applied voltage.
(c) the charge on the plates is in phase with the applied voltage.
(d) power delivered to the capacitor is zero.
On application of AC voltage to the capacitor, the charge on plates is in phase with the applied voltage because the charge on plates is in phase with applied voltage.
The current developed leads the applied voltage by a phase angle of 90o.
Therefore, average power delivered to the capacitor per cycle is, Pav = Vrms Irms cos 90o = 0.
Question 7.13: The line that draws power supply to your house from street has
(a) zero average current.
(b) 220 V average voltage.
(c) voltage and current out of phase by 90°.
(d) voltage and current possibly differing in phase φ such that |φ| < π/2
AC current is used in power supply to our house from street.
AC currents are having zero average value over a cycle.
As the power line is having some resistance, so power factor, cos ϕ = R/Z ≠0.
In also means, φ ≠ π/2 and | φ | < π/2.