EE101: RLC Circuits (with DC sources)

[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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