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A .005-uf capacitor. a 500 ohm resistor, and a 1200-uH inductor are connected in series...

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t4trendesetter | Honors

Posted September 5, 2013 at 8:58 PM via web

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A .005-uf capacitor. a 500 ohm resistor, and a 1200-uH inductor are connected in series across a 12-V,40kHz source. Would the impedance be 704 ohm and the total current be 0.017A? and how would I find the total power?

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted September 6, 2013 at 5:00 AM (Answer #1)

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The circuit is series, it means that the current I through all components is the same. For a diagram of the voltages V(L), V(C), V(R) and I please see the figure below.

The angular frequency of the applied voltage source is

`omega =2*pi*F =2*pi*40000=251327 (rad)/sec`

The impedance of a series RLC circuit is

`Z = sqrt(R^2 + (omega*L -1/(omega*C)^2)) =`

`=sqrt(500^2 + (301.6 -795.7)^2) =sqrt(500^2 +494^2) =703 ohms`

The total current I flowing through the circuit is simply (by Ohm law)

`I = U/Z = 12/703 =0.01707 A`

Regarding power, in an alternating circuit there are 3 different types of power (see the second figure below for the power diagram).

1. The active power P is the real power i.e. the power dissipated by the resistive part of the circuit (which has the voltage in phase with the total current). It is measured in Watts (symbol W)

`P = I^2*R = 0.017^2*500 =0.146 W`

2. The reactive power Pr is the imaginar power i.e. the power dissipated on the reactive components of the circuit (inductor and capacitor). It is measured in Volt*Amps Reactive (symbol VAR)

`Pr = I^2*(1/(omega*C) -omega*L) =0.017^2*494 =0.143 VAR`

3. The apparent power Pa is the vectorial sum of the above two powers and can be regarded as the power dissipated over all components of the circuit (the power given by the total impedance). It is measured in Volt*Amps (symbol VA).

`Pa = I^2*Z = sqrt(P^2+ Pr^2) =0.017^2*703 =0.205 VA`

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