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25 Mar 2008 21:17:05 IST
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Easy one::::::electricity::::::AC circuit::::::
None
An AC source at 100V(rms) and 50Hz having only internal resistance and can supply a maximum current of 10A(rms) is connecyed to a capacitor of capacitance 1/pie mF.The current through the capacitor is
a)10A at its peak value
b)5root(2) at peak value
c)greater than 10A(rms)
d)none of these.
Comments (5)
26 Mar 2008 01:17:59 IST
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2 people liked this
Here's a different (and easier) approach. Just thought of sharing it with you people.
Let us try to solve the problem in complex number form, as in AC circuits, most of the things are represented by phasors, which are themselves rotating vectors, and as we know, complex numbers have a great similarity with vectors.
Notes:
In complex number form, we have:
Xc = -j/wC
XL = jwL
R = R ohms.
(j = iota. We use j so as to not to confuse it with current i)
Hence, here, Xc = -j10. (put values)
R = 10 ohms
Hence, net impedance = Zeq = 10-j10 ohms
hence, the phasor current I = V/Zeq = 100/(10-j10) = 5+j5 A
The magnitude of the phasor current gives us the magnitude of rms current.
Hence, Irms=
50 = 5 2 A. Hence, Ipeak = 10A.
The angle tan-1((5/5) = 45 gives us the angle that the phasor current makes with the x axis in the phasor diagram.
The beauty of using complex numbers here is that you can treat various impedances as normal resistances, and simply add if they are in series, and simply take reciprocal and add if they are in parallel. Just remember that the magnitude of the final ans gives you rms value.
This is not needed at school level, but just posted it in case some ppl are curious about easier methods.
Let us try to solve the problem in complex number form, as in AC circuits, most of the things are represented by phasors, which are themselves rotating vectors, and as we know, complex numbers have a great similarity with vectors.
Notes:
In complex number form, we have:
Xc = -j/wC
XL = jwL
R = R ohms.
(j = iota. We use j so as to not to confuse it with current i)
Hence, here, Xc = -j10. (put values)
R = 10 ohms
Hence, net impedance = Zeq = 10-j10 ohms
hence, the phasor current I = V/Zeq = 100/(10-j10) = 5+j5 A
The magnitude of the phasor current gives us the magnitude of rms current.
Hence, Irms=
50 = 5 2 A. Hence, Ipeak = 10A.The angle tan-1((5/5) = 45 gives us the angle that the phasor current makes with the x axis in the phasor diagram.
The beauty of using complex numbers here is that you can treat various impedances as normal resistances, and simply add if they are in series, and simply take reciprocal and add if they are in parallel. Just remember that the magnitude of the final ans gives you rms value.
This is not needed at school level, but just posted it in case some ppl are curious about easier methods.
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we get R = 10 ohms..
C = 1/ pi * 10^(-3)
=> wC = 2* pi * 50 * 1/ pi * 10 ^(-3)
= 0.1
so, X = reactance = 10 ohms
=> Z = root(10^2 + 10^2) = 10 rt2
so, Irms = 100/10rt2 = 5 rt 2
=> Ipeak = rt 2 * 5 rt 2
= 10 A