JEE Mains and Advanced:

Numerical on the relationship between the emf of an AC generator and the frequency of rotation of the coil

Shortcut Method:

The emf of an AC generator is directly proportional to the frequency of rotation of the coil. This relationship can be expressed as:

$$E = \frac{\phi ZN}{60}$$

where:

• E is the emf in volts (V)
• ɸ is the magnetic flux in webers (Wb)
• Z is the number of turns in the coil
• N is the frequency of rotation in revolutions per minute (rpm)

Example:

A coil of 100 turns is rotated in a magnetic field with a flux density of 0.5 Wb. The coil rotates at a speed of 3000 rpm. Calculate the emf induced in the coil.

$$E = \frac{\phi ZN}{60} = \frac{0.5Wb \times 100 turns \times 3000 rpm}{60} = 250 V$$

Numerical on the calculation of the average and rms values of an AC current

Shortcut Method:

The average value of an AC current is zero. This is because the positive and negative half-cycles of the AC current cancel each other out.

The rms value of an AC current is given by:

$$I_{rms} = \sqrt{\frac{1}{T}\int_0^T i^2(t)~dt}$$

where:

• I(^2)_{rms} is the rms value of the current in amperes (A)
• i(t) is the instantaneous value of the current in amperes (A)
• T is the time period of the AC current in seconds (s)

Example:

An AC current has a maximum value of 10 A and a frequency of 50 Hz. Calculate the average and rms values of the current.

Average value $$= I_{avg}=0$$

$$I_{rms} = \sqrt{\frac{1}{T}\int_0^T i^2(t)~dt} = \sqrt{\frac{1}{0.02s}\int_0^{0.02s} i^2(t)~dt} =\sqrt{\frac{1}{0.02s}\times\frac{100}{2}}=5$$

Numerical on the power factor of an AC circuit

Shortcut Method:

The power factor of an AC circuit is defined as the ratio of the real power to the apparent power. The real power is the power that is actually consumed by the circuit, while the apparent power is the product of the voltage and current.

The power factor can be expressed as:

$$pf = \cos \phi$$

where:

• pf is the power factor
• ɸ is the phase angle between the voltage and current

Example:

An AC circuit has a voltage of 100 V and a current of 5 A. The power factor of the circuit is 0.8. Calculate the real power consumed by the circuit.

$$P = VI\cos \phi = 100V \times 5A \times 0.8 = 400W$$

Numerical on the impedance of an AC circuit

Shortcut Method:

The impedance of an AC circuit is defined as the ratio of the voltage to the current. The impedance is a complex quantity that has both magnitude and phase.

The magnitude of the impedance can be expressed as:

$$Z = \sqrt{R^2 + X^2}$$

where:

• Z is the impedance in ohms (Ω)
• R is the resistance of the circuit in ohms (Ω)
• X is the reactance of the circuit in ohms (Ω)

The phase angle of the impedance can be expressed as:

$$\phi = \tan^{-1}\left(\frac{X}{R}\right)$$

where:

• ɸ is the phase angle in degrees (°)
• X is the reactance of the circuit in ohms (Ω)
• R is the resistance of the circuit in ohms (Ω)

Example:

An AC circuit