A sinusoidal voltage of peak value 230 2 V with variable frequency is applied to a series LCR circuit in which R = 5 Ω , L = 250 π mH and C = 400 π μF . a Find the frequency of the source at which resonance occurs.

The resonance frequency is given by ω = 1 LC = 1 250 π × 400 π = 1 100000 π 2 = 32 . 04 rad s - 1

A sinusoidal voltage of peak value 230 2 V with variable frequency is applied to a series LCR circuit in which R = 5 Ω , L = 250 π mH and C = 400 π μF . a Find the frequency of the source at which resonance occurs.

The resonance frequency is given by ω = 1 LC = 1 250 π × 400 π = 1 100000 π 2 = 32 . 04 rad s - 1

Give any two reasons for small energy losses in transformer and write how these can be minimised.

) Hys1teresis Loss: The repeated magnetisation and demagnetisation of the iron core caused by the alternating input current, produces loss in energy called hysteresis loss. This loss can be minimised by using a core with a material having the least hysteresis loss. Alloys like Mu-metal and silicon steel are used to reduce hysteresis loss. 2) Copper loss: The current flowing through the primary and secondary windings lead to Joule heating effect. Hence some energy is lost in the form of heat. Thick wires with considerably low resistance are used to minimise this loss.

A series LCR circuit with R = 1 kΩ , L = 1 . 0 mH , C = 0 . 001 μF is connected to a sinusoidal voltage of peak value 200 2 V . When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one cycle.

frequency of supply = natural frequency Resonance impedance (z) = R . 1 kΩ = 1000 Ω ∴ v 0 = 200 2 = v rim = v 0 2 = 200 V I rims = V rims z = 200 1000 = 0 . 2 A ∴ pavg = v rims . I rims . cosϕ for resonance , ϕ = 0 Pavg = 200 × 0 . 2 × 1 = 40 W

Find the reactance of a capacitor having a capacitance 0 . 5 μF at 100 Hz .

In alternating current AC circuit the opposition to the flow of current is known as impedance. In DC it is only resistance. The impedance offered by capacitor or inductor is known as reactance. Capacitive reactance is given by X C = 1 2 πfC C = 0 . 5 × 10 - 6 F f = 100 Hz X C = ? ∴ X C = 2 × 3 . 142 × 100 × 0 . 5 × 10 - 6 - 1 10 4 3 . 142 = 3182 . 686 Ohms

How does the L - C circuit produce oscillation? Explain.

Whenever we connect a charged capacitor to an inductor the electric current and charge on the capacitor in the circuit undergoes LC Oscillations. The process continues at a definite frequency and if no resistance is present in the LC circuit, then the LC Oscillations will continue indefinitely.

An applied e . m . f . signal consists of superposition of a d.c. source and a.c. source of high frequency. The circuit consists of an inductor L and a capacitor C in series. Show that d.c. signal appears across C and a.c. signal appears across L .

Given, Capacitance of the capacitor is C, and inductance of the inductor is L Assume ω is the angular frequency. Expression for Capacitive reactance, X C = 1 ω C Expression for inductive reactance, X L = ω L For a high frequency, the inductive reactance for a.c. X L = ω L = ∞ and, capacitive reactance for a.c X C = 1 ω C = 0 Hence, we can see that capacitor is not offering any resistance for a.c and thus a.c. signal will only appear across L. For a d.c. voltage, the angular frequency can be taken zero as the voltage is not changing with time. Hence the inductive reactance for d.c. X L = ω L = 0 and, capacitive reactance for d.c X C = 1 ω C = ∞ Hence, d.c signal appears across capacitor only because both inductor and capacitor are connected in series and inductance reactance is zero so voltage drop will occur only in capacitor.

What is r . m . s . value of a.c.?

Root-mean-square or R.M.S. value of Alternating current is defined as that value of steady current, which would generate the same amount of heat in a given resistance is given time, as is done by A.C. current, when maintained across the same resistance for the same time.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 7 - Alternating Current>Manipur Board-2020>Q 7

EASY

12th Manipur Board

IMPORTANT

Earn 100

A sinusoidal voltage of peak value $230 \sqrt{2} V$ with variable frequency is applied to a series LCR circuit in which $R = 5 Ω, L = \frac{250}{π} mH$ and $C = \frac{400}{π} μF$ .

$(a)$ Find the frequency of the source at which resonance occurs.

Important Questions on Alternating Current

EASY

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 7 - Alternating Current>Manipur Board-2020>Q 8

A sinusoidal voltage of peak value $283 V$ with variable frequency is applied to a series LCR circuit in which $R = 3 Ω, L = 25.48 mH$ and $C = 786 μF$