For the circuit shown bellow, determine the current i1,i2,i3 using nodal analysis
Redraw the above circuit, we have
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Fig. 5(a)
Considering two nodes V1 and V2, we have
(V1-24)/2+V1/6+V2/8=3 Or 16V1+3V2=360 …(1)
The difference between two nodes is
V1-V2=4 … (2)
Solving equation (1) and (2), we have
16V1+3(V1-4)=360 V1=372/19=19.58 V …(3)
Putting the value of V1 from equation (3) in to equation (2), we have
V2=19.58-4=15.58 V
The current i1=(24-19.58)/2=2.2 Amp.
i2=(-19.58)/6=-3.26 Amp. And
i3=15.58/8=1.95 Amp
Derive the expression of quality factor of series R-L-C circuit at resonance.
Draw a mathematical expression for RMS value of a sinusoidal voltage \(v=V_{m}\sin{{\omega t}}\).
Explain
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Fig.22(a).png)
Fig. 20(a)