A series RLC circuit takes a maximum current of 0.3 A at 200 V, 50 Hz. If the voltage
across the capacitor is 290V at resonance. Determine R,L,C and Q of the coil.
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Explanation:
V=5sin(100 t)
Compare witer v=V
o
sinwt
We get
V
o
=5 volt,w=100
⇒X
L
=wL=100×0.5=50√2
X
c
=
wc
1
=
100×200×10
−6
1
=50√2
R=20√2
X
L
∼X
c
=0⇒ resonance
impedance z=
(X
L
∼X
c
)
2
+R
2
+
0+20
2
=20√2
⇒ max current I=
Z
V
o
=
20
S
Amp
⇒ max voltage across inductance will be
V
2
I
X
L
=
20
5
×50√2=12.5 volt
it is more than 5 volt because its inductor
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