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