Write down the equation for a sinusoidal voltage of 50 Hz and its peak value is 20 V. Draw the corresponding voltage versus time graph.
The equation for an alternating current is given by i = 77 sin 314t. Find the peak value, frequency, time period and instantaneous value at t = 2 ms.
i = 77 sin 314t ; t = 2 ms = 2×10-3 s
The general equation of an alternating current is i = Im sin ωt . On comparsion,
(i) Peak value, Im = 77 A
(ii) Frequency, f = ω/2π = 314 / 2 ×3.14 = 50 Hz
Time period, T = 1/f = 150 = 0 .02 s
(iv) At t = 2 m s,
i = 77sin(314×2×10−3 )
i = 45.24 A
A 400 mH coil of negligible resistance is connected to an AC circuit in which an effective current of 6 mA is flowing. Find out the voltage across the coil if the frequency is 1000 Hz.
L = 400 x 10-3 H; Ieff = 6 x 10-3A
f = 1000 Hz
A capacitor of capacitance 102/π µF is connected across a 220 V, 50 Hz A.C. mains. Calculate the capacitive reactance, RMS value of current and write down the equations of voltage and current.
Find the impedance of a series RLC circuit if the inductive reactance, capacitive reactance and resistance are 184 Ω, 144 Ω and 30 Ω respectively. Also calculate the phase angle between voltage and current.
XL = 184 Ω; XC = 144 Ω
R = 30 Ω
(i ) The impedance is
Impedance, Z = 50 Ω
(ii) Phase angle is
φ = 53.1
A 500 μH inductor, 80/π2 pF capacitor and a 628 Ω resistor are connected to form a series RLC circuit. Calculate the resonant frequency and Q-factor of this circuit at resonance.
L=500×10-6H; C = 80/π2 ×10−12 F; R = 628Ω
(i) Resonant frequency is
Find the instantaneous value of alternating voltage υ = 10 sin(3 π×104 t) volt at i) 0 s ii) 50 μs iii) 75 μs.
The given equation is υ = 10sin (3 π×104 t)
The current in an inductive circuit is given by 0.3 sin (200t – 40°) A. Write the equation for the voltage across it if the inductance is 40 mH.
L = 40 × 10-3 H; i = 0.1 sin (200t – 40º)
XL = ωL = 200 × 40 × 10-3 = 8 Ω
Vm = Im XL = 0.3 × 8 = 2.4 V
In an inductive circuit, the voltage leads the current by 90o Therefore,
v = Vm sin ( ωt +90º)
v = 2 . 4 sin(200t −40 + 90 º)
v = 2 . 4 sin(200t +50 º)volt