In a series RC circuit with an AC source- R -300 - C -25 - F - - 0-50 V andv -50- Hz- Find the peak current and the average power dissipated in the circuit-

Given R = 300 w, C = 25 μ F = 25 × 10^ 6 F, E_0 = 50 V, f_x = 50/π Hz X_C = 1/ω C = 1/50/π× 2 π× 25 × 10^ 6 = ( 10^4/25 ) Z= √(R^2+ X_0^2) = √((300)^2+ (10^4 ...

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Byju's Answer

Standard XII

Physics

Discharging of Capacitors

In a series R...

Question

In a series RC circuit with an AC source, R = 300 $Ω, C = 25 μ F$ , $ϵ_{0} = 50 V a n d v = \frac{50}{π}$ Hz. Find the peak current and the average power dissipated in the circuit.

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Solution

Given R = 300 w, C = $25 μ F$

= $25 \times 10^{- 6} F, E_{0}$ = 50 V,

$f_{x} = \frac{50}{π}$ Hz

$X_{C} = \frac{1}{ω C}$

= $\frac{1}{\frac{50}{π} \times 2 π \times 25 \times 10^{- 6}}$

= $(\frac{10^{4}}{25})$

Z= $\sqrt{R^{2} + X_{0}^{2}}$

= $\sqrt{(300)^{2} + {(\frac{10^{4}}{25})}^{2}}$

= $\sqrt{(300)^{2} + (400)^{2}}$ = 500

(a) Peak current $\frac{E_{0}}{Z} = \frac{50}{500}$ = 0.1 A

(b) Average Power dissipated

= $E_{r m s} I_{r m s} c o s ϕ$ .

= $\frac{E_{0}}{\sqrt{2}} \times \frac{E_{0}}{\sqrt{2 Z}} \frac{R}{Z}$

= $\frac{E_{0}^{2}}{2 Z} . \frac{R}{Z}$

= $\frac{50 \times 50 \times 300}{2 \times 500 \times 500}$

= $\frac{3}{2}$ =1.5 W

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In a series RC circuit with an AC source, R = 300 Ω,C=25μF, ϵ0=50Vandv=50π Hz. Find the peak current and the average power dissipated in the circuit.

In a series RC circuit with an AC source, R = 300 $Ω, C = 25 μ F$ , $ϵ_{0} = 50 V a n d v = \frac{50}{π}$ Hz. Find the peak current and the average power dissipated in the circuit.