# Download Problem 6.16 The parallel-plate capacitor shown in Fig. P6.16 is

Survey
Was this document useful for you?
Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Multimeter wikipedia, lookup

Integrating ADC wikipedia, lookup

Nanofluidic circuitry wikipedia, lookup

Integrated circuit wikipedia, lookup

Index of electronics articles wikipedia, lookup

Test probe wikipedia, lookup

Regenerative circuit wikipedia, lookup

Surge protector wikipedia, lookup

Power MOSFET wikipedia, lookup

Resistive opto-isolator wikipedia, lookup

Crystal radio wikipedia, lookup

Spark-gap transmitter wikipedia, lookup

Operational amplifier wikipedia, lookup

Valve RF amplifier wikipedia, lookup

Ohm's law wikipedia, lookup

Opto-isolator wikipedia, lookup

Two-port network wikipedia, lookup

Switched-mode power supply wikipedia, lookup

Current source wikipedia, lookup

Current mirror wikipedia, lookup

Oscilloscope history wikipedia, lookup

TRIAC wikipedia, lookup

Network analysis (electrical circuits) wikipedia, lookup

Rectiverter wikipedia, lookup

RLC circuit wikipedia, lookup

Transcript
```Problem 6.16 The parallel-plate capacitor shown in Fig. P6.16 is filled with a lossy
dielectric material of relative permittivity εr and conductivity σ . The separation
between the plates is d and each plate is of area A. The capacitor is connected to
a time-varying voltage source V (t).
I
A
V(t)
ε, σ
d
Figure P6.16: Parallel-plate capacitor containing a lossy dielectric material (Problem 6.16).
(a) Obtain an expression for Ic , the conduction current flowing between the plates
inside the capacitor, in terms of the given quantities.
(b) Obtain an expression for Id , the displacement current flowing inside the
capacitor.
(c) Based on your expressions for parts (a) and (b), give an equivalent-circuit
representation for the capacitor.
(d) Evaluate the values of the circuit elements for A = 4 cm2 , d = 0.5 cm, εr = 4,
σ = 2.5 (S/m), and V (t) = 10 cos(3π × 103t) (V).
Solution:
(a)
R=
(b)
d
,
σA
Ic =
V
VσA
=
.
R
d
∂D
∂ E εA ∂V
V
,
Id =
· A = εA
=
.
d
∂t
∂t
d ∂t
(c) The conduction current is directly proportional to V , as characteristic of a
resistor, whereas the displacement current varies as ∂ V/∂ t, which is characteristic
E=
of a capacitor. Hence,
R=
d
σA
C=
and
εA
.
d
I
+
ε, σ
V(t)
R
Ic
Id
+
C
V(t)
-
-
Actual Circuit
Equivalent Circuit
Figure P6.16: (a) Equivalent circuit.
(d)
0.5 × 10−2
= 5 Ω,
2.5 × 4 × 10−4
4 × 8.85 × 10−12 × 4 × 10−4
= 2.84 × 10−12 F.
C=
0.5 × 10−2
R=
```
Related documents