The electrical supply lo demesne properties in the UK is 230V, 50Hz single phase

Question

The electrical supply lo demesne properties in the UK is 230V, 50Hz single phase a.c. with a sinusoidal Maw shape. a. Why is the wave shape sinusoidal? b. Stale the r.m.s. peak and peak voltage values. (230V, 325V, 650.5V) c. If the supply is connected to a resistive load and the power in the load is measured as 3kW, stale the phase relationship of the current through the load to the silage applied across the load and the r.m.s.. peak and pack-to-peak values of the current. (0 degree, 13, 04A, 18.45A, 36.89A) The load of question 1 is modified by placing a 25mH inductor in series with the resistor. a. State the total load impedance in terms of R + jX. ((17.63 + j7.85) b. Calculate the current through the load. (11.92A) c. Calculate the voltage across the resistor and the voltage across the inductor. Draw a phasor diagram to show the phase relationship between them (210V, 93.6V, 90 degree) d. Calculate the VA. power rating and power factor of the load. (2742VA. 2504W. 0.913 lagging) The load is modified by placing a 220 mu F capacitor in series with the network of question 2. a. State the total load impedance in terms of R + jX. ((17.63 - j6.62) Ohm) b. Calculate the current through the load (12.21 A) c. Calculate the voltage across the resistor, the voltage across the inductor and the voltage across the capacitor. Draw a phasor diagram to show the phase relationship between them (215.3V, 95.8V, 176.7V) d. Calculate the VA, power rating and power factor of the load (2808VA, 2629W, 0.936 leading)

Explanation / Answer

Answer 1

(a)

The sinusoidal waveform is used because it has certain advantages.

1. it produces the minimum disturbance in the electrical circuits during operation.

2.  shape of the waveform remains same even after undergoing signal processing such as integration or differentiation.

(b)

The RMS value is the effective value of a varying voltage or current.It is the equivalent steady DC value which gives the same effect. let us take an example. if we take  a lamp connected to a 6V RMS AC supply then it will shine with same brightness althoth connected to a 6V DC supply.

maximum value VP = VRMS × 2,

= 230 × 2, = 325 V

peak to peak value of a sinusodial wave is a range of the voltage form it most negative vlaue to most positive value so it is

V p - p = 2 * Vp = 2 * 325 = 650 V

(C)

supply connected across the resistive load

so for the resistive load the phase diffrence between voltage and current is 0 degree

current through load

P = VI

3000 = 230 * I

I = 13.04 A rms

peak value of current

Ip × 2 = 13.04 × 2 = 18.45 A

peak - peak value of current

Ip-p = 2 * 18.45 = 36.90 A

please post another questions as a saperate question..

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