Math

Question12-47 Determine the true power and the reactive power in the previous question. True porer is ideally o W.  Meactive power =Imss2Xc=Irms22πfc=3.4nVAR\text { Meactive power }=I_{\mathrm{mss}}^{2} X_{c}=\frac{I_{\mathrm{rms}}^{2}}{2 \pi \mathrm{fc}}=3.4 \mathrm{nVAR}

Studdy Solution

STEP 1

1. The problem involves calculating true power and reactive power in an AC circuit.
2. True power is the actual power consumed by the circuit, measured in watts (W).
3. Reactive power is the power stored and released by inductors or capacitors, measured in volt-amperes reactive (VAR).

STEP 2

1. Clarify the given information and formulas.
2. Calculate the true power.
3. Calculate the reactive power using the given formula.
4. Verify the calculations.

STEP 3

Clarify the given information and formulas:
- We are given that the reactive power is calculated using the formula: $ \text{Reactive power} = I_{\text{rms}}^2 X_c = \frac{I_{\text{rms}}^2}{2 \pi f c} \] and it equals \(3.4 \, \text{nVAR}\).
- True power is stated as ideally 0W0 \, \text{W}.

STEP 4

Calculate the true power:
- True power (PP) in an ideal reactive circuit (purely capacitive or inductive) is 0W0 \, \text{W} because there is no real energy consumed, only stored and released.

STEP 5

Calculate the reactive power using the given formula:
- We are given: $ \text{Reactive power} = \frac{I_{\text{rms}}^2}{2 \pi f c} = 3.4 \, \text{nVAR} \]
- This indicates that the reactive power has already been calculated as 3.4nVAR3.4 \, \text{nVAR}.

STEP 6

Verify the calculations:
- True power is confirmed as 0W0 \, \text{W} for an ideal reactive circuit. - Reactive power is confirmed as 3.4nVAR3.4 \, \text{nVAR}.
The true power is 0W \boxed{0 \, \text{W}} and the reactive power is 3.4nVAR \boxed{3.4 \, \text{nVAR}} .

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