NCERT Solutions For Class 12 Physics Chapter 7: Alternating Current

Jasmine Grover Content Strategy Manager

Content Strategy Manager

NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current (AC) are provided in this article. Alternating current (AC) is an electric current which periodically reverses direction and changes its magnitude continuously with time in contrast to Direct Current (DC) which flows only in one direction. AC voltage is preferable in the circuits because it can be easily and efficiently converted into another voltage using a transformer.

Unit 4 Electromagnetic Induction and Alternating Currents along with Unit 3 Magnetic Effects of Current and Magnetism has a weightage of 17 marks in the CBSE Board examinations. The NCERT Solutions for Class 12 Physics Chapter 7 deals with concepts of alternating current, LC Oscillations, AC voltage applied to an inductor, capacitor, Power in Alternating Current, etc.

Download PDF: NCERT Solutions for Class 12 Physics Chapter 7

NCERT Solutions for Class 12 Physics Chapter 7

Previous Year Questions

In the circuit shown, the symbols have their usual meanings. The cell has emf E….[BITSAT 2018]
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Class 12 Physics Chapter 7 Topics Covered

Alternating Current refers to a changing voltage electric supply. It can be efficiently transformed from one form to another.
The charge and the electric current in a charged capacitor undergo LC Oscillations when it is connected to an inductor in the circuit.
RMS refers to the Root Mean Square of instantaneous current values. RMS value of alternating current is given by the direct current that flows through a resistance.

\(I_{rms} = \frac{I_0}{\sqrt{2}} = 0.707 I_0 \)

Wattless Current is when the average power consumed in an AC circuit corresponds to Zero.

The average power dissipated in AC circuit is given by:

P = VI cos ϕ

Here, V is the voltage applied, I is the current flowing, and ϕ is the phase difference between voltage and current

Power factor is a factor through which the apparent power should be multiplied in order to get active power.

cos θ = P/\|S\| = Actual power/Apparent power

Also Read:

Chapter 7 Related Articles
AC Generator	AC Voltage applied to an Inductor	AC Voltage Applied to a Capacitor
AC Voltage applied to LCR Circuit	AC Voltage applied to Resistor	Power in AC Circuit
Capacitive reactance formula	Difference between AC and DC Motor	Phasor Representation AC

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CBSE CLASS XII Related Questions

1.
The resistance of a wire at 25°C is 10.0 \( \Omega \). When heated to 125°C, its resistance becomes 10.5 \( \Omega \). Find (i) the temperature coefficient of resistance of the wire, and (ii) the resistance of the wire at 425°C.
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2.
A charge \( -6 \mu C \) is placed at the center B of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point D at a distance of 10 cm from B. A charge \( +5 \mu C \) is moved from point ‘C’ to point ‘A’ along the circumference. Calculate the work done on the charge.
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3.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.
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4.
A small spherical shell \( S_1 \) has point charges \( q_1 = -3 \, \mu C \), \( q_2 = -2 \, \mu C \) and \( q_3 = 9 \, \mu C \) inside it. This shell is enclosed by another big spherical shell \( S_2 \). A point charge \( Q \) is placed in between the two surfaces \( S_1 \) and \( S_2 \). If the electric flux through the surface \( S_2 \) is four times the flux through surface \( S_1 \), find charge \( Q \).
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5.
A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.
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6.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).
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