Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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In a Series Rc Circuit with an Ac Source, R = 300 ω, C = 25 μF, ε0 = 50 V and ν = 50/π Hz. Find the Peak Current and the Average Power Dissipated in the Circuit. - Physics

Sum

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

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Solution

Given:
Resistance of the series RC circuit, R = 300 Ω
Capacitance of the series RC circuit, C = 25 μF
Peak value of voltage, ε0 = 50 V
Frequency of the AC source, ν = 50/ `pi` Hz            
Capacitive reactance (Xc) is given by,
`X_C = 1/(omegaC)`
Here, ω = angular frequency of AC source
            C = capacitive reactance of capacitance
∴ `X_C = 1/(2xx50/pixx25xx10^-6)`
⇒ `X_C = 10^4/25 Ω`
Net reactance of the series RC circuit`(Z) = sqrt(R^2 + (X_C)^2`
⇒ Z =`sqrt((300)^2 + (10^4/25)^2`
= `sqrt((300)^2 + (400)^2 = 500 Ω`
(a) Peak value of current `(I_0)` is given by,
`I_0 = (epsilon_0)/z`
⇒ `I_0 = 50/500 = 0.1 A`
(b) Average power dissipated in the circuit (P) is given by,
`P = epsilon_{rms}l_{rms} cosØ`

`epsilon_{rms] = epsilon_0/sqrt2`

`∴ P = E_0/sqrt2xx I_0/sqrt2xxR/Z`

⇒ P =`(50xx0.1xx300)/(2xx500)`

⇒ `P = 3/2 = 1.5 W`

Concept: Peak and Rms Value of Alternating Current Or Voltage
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 17 Alternating Current
Q 12 | Page 330
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