- If E
_{1}= Asinωt and E_{2}= Asin(ωt - θ), thenAt ωt = 0, E_{1}= 0 &s; E_{2}= Asin(- θ) = - Asinθ.

From, the expression of E_{1}= Asinωt and E_{2}= A.sin(ωt - θ), it is clear that, E_{2}crosses zero t = θ ⁄ ω sec after that of E_{1}.

Therefore, it can be concluded that E_{1}leads E_{2}by θ. - Two sinusoidal quantities are said to be phase quadrature, when their phase difference is
Two sinusoidal quantities are said to be phase quadrature, when their phase difference is 90°.
- The equation for 25 cycles electric current sine wave having rms value of 30 amps, will be
General equation of sinusoidal electric current quantity is I
_{m}sinωt = I_{m}sin2πft.

Where, I_{m}is the maximum ampletude of the electric current wave, f is the frequency or cycle per second.

Here, rms value of electric current is 30 A.

∴ Maximum amplitude of the electric current wave form I_{m}= √2 × 30 = 42.4 A and frequency f is here 25 Hz. Hence, the electric current equation will be 42.4sin2π25t = 42.4sin50πt. - What will be the rms value of rectangular wave with amplitude 10 V
RMS value means, root mean square value of a wane. A rectangular voltage wave has constant amplitude in both positive and negative direction. Hence the mean value and its amplitude will be identical. Therefore square root of square of the mean value of amplitude is same as amplitude of the wave.
- The equation of an emf is given by e = I
_{m}[√(R^{2}+ 4ω^{2}L^{2})]sin2ωt. The amplitude of the wave will beThe general form of emf equation is e = V_{m}sinωt ......(1)

Where, V_{m}is the amplitude of the voltage wave.

Here, the given equation is e = I_{m}[(R^{2}+ 4ω^{2}L^{2})^{1/2}]sin2ωt ......(2)

Now, Comparing equation (1) &s; (2) we can conclude that amplitude of the given emf equation is I_{m}[(R^{2}+ 4ω^{2}L^{2})^{1/2}]. - The RMS value of sinusoidal voltage wave v = 200sinωt, is
The peak value of given voltage wave is 200 V. Therefore the RMS value will be 200 ⁄ √2 = 100√2 volts.
- If one cycle of ac waveform occurs every milli - second, the frequency will be
One cycle of ac waveform occurs every milli second means 1000 cycles of that waveform occur in one second. The numbers of cycles of waveform per second is the frequency of that waveform.
- If emf in a given circuit is given by e = 100sin628t, then maximum value of voltage and frequency will be
The emf equation is e = 100sin628t .....(1)

Again the general form of emf equation is e = V_{m}sin2πft ......(2)

Where, V_{m}is the voltage amplitude and f is the frequency.

Comparing, equations (1) &s; (2) we get,

V_{m}= 100 V and 2πf = 628 ⇒ 628/(2 × 3.24) ⇒ f = 628/6.28 = 100 Hz. - The value of supply voltage for 400 W, 4 ohm load is

Where, V is supply voltage and W is wattage rating of the lamp.

Here, W = 400 W and R = 4 Ω

⇒ V = 40 Volts. - Say A point has an absolute potential of 40 V and point B has an absolute potential of - 10 V, then what will be the value of V
_{BA}?V_{BA}is defined as V_{B}− V_{A}= -10 - 40 = - 50 V. - The rms value of the voltage U(t)= 3 + 4cos(3t)
- In the figure, the potential difference between points P and Q is

. - A coil of negligible resistance has an induction of 100 mH. The electric current passing through the coil changes from 2 A to 4 A at a uniform rate in 0.1 sec the voltage across the coil during this time would be ___ V.
- What is represented by the hypotenuse of impedance triangle?
Impedance triangle means the right angle triangle formed by the vectors representing the resistance drop, reactance drop & the impedance drop of the circuit carrying an alternating current.
- The phase angle difference between electric current and voltage is 90°, the power will be
The expression of active power P = VIcosθ.

Where, V is voltage, I is electric current and θ is the angle between electric current and voltage.

here, this θ = 90°

∴ Power P = VIcos90° = 0 [Since, cos90°= 0]. - Kirchhoff's laws are valid for
Linear circuits obey Ohm's law. Kirchhoff's laws are valid for those elements that obey Ohm's law.
- For the circuit shown below the value of R is adjusted so as to make the electric current in R
_{L}equal to zero. Calculate the value of R.

As per Wheatstone bridge principle: 10 / 4 = 5 / R ⇒ R = 4 / 10 × 5 ohms ⇒ R = 2 ohms. - In the circuit shown in figure if I
_{1}= 1.5 A, then I_{2}will be

- In the circuit shown in the figure the voltage across the 2 Ω resistor is

- The value of electric current I flowing in the 1 Ω resistor in the circuit shown in the given figure will be

Design with by SARU TECH