NCERT Solutions For Class 12 Physics Chapter 7: Alternating Current

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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

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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.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:
- attract with a force \( \frac{F}{2} \)
- repel with a force \( \frac{F}{2} \)
- repel with a force \( F \)
- attract with a force \( F \)
View Solution
2.
The energy of an electron in an orbit in hydrogen atom is \( -3.4 \, \text{eV} \). Its angular momentum in the orbit will be:
- \( \dfrac{3h}{2\pi} \)
- \( \dfrac{2h}{\pi} \)
- \( \dfrac{h}{\pi} \)
- \( \dfrac{h}{2\pi} \)
View Solution
3.
A circular coil of 100 turns and radius \( \left(\frac{10}{\sqrt{\pi}}\right) \, \text{cm}\) carrying current of \( 5.0 \, \text{A} \) is suspended vertically in a uniform horizontal magnetic field of \( 2.0 \, \text{T} \). The field makes an angle \( 30^\circ \) with the normal to the coil. Calculate:
the magnetic dipole moment of the coil, and
the magnitude of the counter torque that must be applied to prevent the coil from turning.
View Solution
4.
Four long straight thin wires are held vertically at the corners A, B, C and D of a square of side \( a \), kept on a table and carry equal current \( I \). The wire at A carries current in upward direction whereas the current in the remaining wires flows in downward direction. The net magnetic field at the centre of the square will have the magnitude:
- \( \dfrac{\mu_0 I}{\pi a} \) and directed along OC
- \( \dfrac{\mu_0 I}{\pi a \sqrt{2}} \) and directed along OD
- \( \dfrac{\mu_0 I \sqrt{2}}{\pi a} \) and directed along OB
- \( \dfrac{2\mu_0 I}{\pi a} \) and directed along OA
View Solution
5.
Consider a cylindrical conductor of length \( l \) and area of cross-section \( A \). Current \( I \) is maintained in the conductor and electrons drift with velocity \( \vec{v}_d \, (|\vec{v}_d| = \frac{eE}{m} \tau) \), where symbols have their usual meanings. Show that the conductivity of the material of the conductor is given by \[ \sigma = \frac{n e^2 \tau}{m}. \]
View Solution
6.
The figure represents the variation of the electric potential \( V \) at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at:
- P
- Q
- R
- S
View Solution