Wednesday

April 16, 2014

April 16, 2014

Posted by **barotseland network** on Tuesday, March 5, 2013 at 5:15pm.

a.The frequency

b.current

c.voltage drop or pd across inductance.

- MONGU COLLEGE -
**Henry**, Thursday, March 7, 2013 at 10:20pmI cannot calculate the freq. unless the reactance of the inductor or capacitor is given. I will assume a 230v.- 60 Hz supply.

a 60 Hz.

b. Xl = 2pi*F*L = 6.28*60*4 = 1508 Ohms.

Xc = 1/(2p*F*C) = 1/(6.28*60*25*10^-6=

106 Ohms.

Z = R + j(Xl-Xc)

Z = 15 + j(1508-106)

Z = 15 + j1402 = Rectangular form.

tanA = Xl/R = 1402/15 = 93.5

A = 89.4o

Z = Xl/sinA = 1402/sin89.4=1402.1 Ohms

@ 89.4o = Polar form.

I = V/Z = 230/1402.1=0.164A @ (-89.4o)

c. V = I * Xl = 0.164 * 1508 = 247.3 V.

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