EE101: RLC Circuits (with DC sources) - IIT Bombay
[Pages:47]EE101: RLC Circuits (with DC sources)
M. B. Patil mbpatil@ee.iitb.ac.in ee.iitb.ac.in/~sequel
Department of Electrical Engineering Indian Institute of Technology Bombay
M. B. Patil, IIT Bombay
Series RLC circuit
iR VR
V0
L
VL
C
VC
M. B. Patil, IIT Bombay
Series RLC circuit
iR VR
V0
L
VL
C
VC
KVL: VR + VL + VC
di = V0 i R + L dt
+1 C
Z
i dt = V0
M. B. Patil, IIT Bombay
Series RLC circuit
iR VR
V0
L
VL
C
VC
KVL: VR + VL + VC
di = V0 i R + L dt
+1 C
Z
i dt = V0
Differentiating w. r. t. t, we get,
R
di dt
+L
d2i dt 2
+
1 C
i
=
0.
M. B. Patil, IIT Bombay
Series RLC circuit
iR VR
V0
L
VL
C
VC
KVL: VR + VL + VC
di = V0 i R + L dt
+1 C
Z
i dt = V0
Differentiating w. r. t. t, we get,
R
di dt
+L
d2i dt 2
+
1 C
i
=
0.
i.e.,
d2i dt 2
+
R L
di dt
+
1 LC
i
= 0,
a second-order ODE with constant coefficients.
M. B. Patil, IIT Bombay
Parallel RLC circuit
iR
iL
iC
I0
R
LC V
M. B. Patil, IIT Bombay
Parallel RLC circuit
iR
iL
iC
I0
R
LC V
1
1Z
dV
KCL: iR + iL + iC = I0 R V + L V dt + C dt = I0
M. B. Patil, IIT Bombay
Parallel RLC circuit
iR
iL
iC
I0
R
LC V
1
1Z
dV
KCL: iR + iL + iC = I0 R V + L V dt + C dt = I0
Differentiating w. r. t. t, we get,
1 R
dV dt
+ 1 V +C L
d2V dt 2
= 0.
M. B. Patil, IIT Bombay
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