Prove that the power consumed by a pure capacitor is zero.
2 Answers
To prove that the power consumed by a pure capacitor is zero, we can analyze the behavior of a capacitor in an AC circuit.
### Key Concepts
1. **Capacitance**: A capacitor stores energy in an electric field when a voltage is applied across its plates. The capacitance \( C \) is defined as the ability of a capacitor to store charge per unit voltage:
\[
C = \frac{Q}{V}
\]
where \( Q \) is the charge and \( V \) is the voltage across the capacitor.
2. **Voltage and Current Relationship**: In an AC circuit, the voltage \( V(t) \) across a capacitor and the current \( I(t) \) through it are related by the following equation:
\[
I(t) = C \frac{dV(t)}{dt}
\]
3. **Power in AC Circuits**: The instantaneous power \( P(t) \) consumed by an element in an AC circuit is given by:
\[
P(t) = V(t) \cdot I(t)
\]
### Power Calculation for a Capacitor
1. **Voltage and Current Expressions**: Assume the voltage across the capacitor is sinusoidal, given by:
\[
V(t) = V_0 \sin(\omega t)
\]
where \( V_0 \) is the peak voltage and \( \omega \) is the angular frequency.
2. **Current Calculation**: Using the relationship between voltage and current, we find the current:
\[
I(t) = C \frac{dV(t)}{dt} = C \frac{d}{dt}(V_0 \sin(\omega t)) = C V_0 \omega \cos(\omega t)
\]
3. **Instantaneous Power**: Now, substituting \( V(t) \) and \( I(t) \) into the power equation:
\[
P(t) = V(t) \cdot I(t) = (V_0 \sin(\omega t)) \cdot (C V_0 \omega \cos(\omega t))
\]
This simplifies to:
\[
P(t) = C V_0^2 \omega \sin(\omega t) \cos(\omega t)
\]
4. **Using a Trigonometric Identity**: We can simplify further using the trigonometric identity \( \sin(a) \cos(a) = \frac{1}{2} \sin(2a) \):
\[
P(t) = \frac{1}{2} C V_0^2 \omega \sin(2\omega t)
\]
5. **Average Power Over a Cycle**: To find the average power consumed by the capacitor over one complete cycle, we can compute:
\[
P_{\text{avg}} = \frac{1}{T} \int_0^T P(t) \, dt
\]
where \( T \) is the period of the AC signal. Since \( \sin(2\omega t) \) oscillates between -1 and 1 over a full cycle, the average value of \( \sin(2\omega t) \) over one complete cycle is zero:
\[
\int_0^T \sin(2\omega t) \, dt = 0
\]
Therefore, we conclude that:
\[
P_{\text{avg}} = 0
\]
### Conclusion
From our analysis, we see that the average power consumed by a pure capacitor in an AC circuit is zero. This implies that while the capacitor alternately stores energy in its electric field and releases it back into the circuit, it does not consume any net power over time. Thus, we conclude that the power consumed by a pure capacitor is indeed zero.
### Key Concepts
1. **Capacitance**: A capacitor stores energy in an electric field when a voltage is applied across its plates. The capacitance \( C \) is defined as the ability of a capacitor to store charge per unit voltage:
\[
C = \frac{Q}{V}
\]
where \( Q \) is the charge and \( V \) is the voltage across the capacitor.
2. **Voltage and Current Relationship**: In an AC circuit, the voltage \( V(t) \) across a capacitor and the current \( I(t) \) through it are related by the following equation:
\[
I(t) = C \frac{dV(t)}{dt}
\]
3. **Power in AC Circuits**: The instantaneous power \( P(t) \) consumed by an element in an AC circuit is given by:
\[
P(t) = V(t) \cdot I(t)
\]
### Power Calculation for a Capacitor
1. **Voltage and Current Expressions**: Assume the voltage across the capacitor is sinusoidal, given by:
\[
V(t) = V_0 \sin(\omega t)
\]
where \( V_0 \) is the peak voltage and \( \omega \) is the angular frequency.
2. **Current Calculation**: Using the relationship between voltage and current, we find the current:
\[
I(t) = C \frac{dV(t)}{dt} = C \frac{d}{dt}(V_0 \sin(\omega t)) = C V_0 \omega \cos(\omega t)
\]
3. **Instantaneous Power**: Now, substituting \( V(t) \) and \( I(t) \) into the power equation:
\[
P(t) = V(t) \cdot I(t) = (V_0 \sin(\omega t)) \cdot (C V_0 \omega \cos(\omega t))
\]
This simplifies to:
\[
P(t) = C V_0^2 \omega \sin(\omega t) \cos(\omega t)
\]
4. **Using a Trigonometric Identity**: We can simplify further using the trigonometric identity \( \sin(a) \cos(a) = \frac{1}{2} \sin(2a) \):
\[
P(t) = \frac{1}{2} C V_0^2 \omega \sin(2\omega t)
\]
5. **Average Power Over a Cycle**: To find the average power consumed by the capacitor over one complete cycle, we can compute:
\[
P_{\text{avg}} = \frac{1}{T} \int_0^T P(t) \, dt
\]
where \( T \) is the period of the AC signal. Since \( \sin(2\omega t) \) oscillates between -1 and 1 over a full cycle, the average value of \( \sin(2\omega t) \) over one complete cycle is zero:
\[
\int_0^T \sin(2\omega t) \, dt = 0
\]
Therefore, we conclude that:
\[
P_{\text{avg}} = 0
\]
### Conclusion
From our analysis, we see that the average power consumed by a pure capacitor in an AC circuit is zero. This implies that while the capacitor alternately stores energy in its electric field and releases it back into the circuit, it does not consume any net power over time. Thus, we conclude that the power consumed by a pure capacitor is indeed zero.
To prove that the power consumed by a pure capacitor is zero, we need to analyze the relationship between voltage, current, and power in a purely capacitive circuit.
### 1. **Voltage and Current Relationship in a Capacitor**
For a capacitor, the current \( I(t) \) is related to the rate of change of voltage \( V(t) \) across it. The relationship is given by:
\[
I(t) = C \frac{dV(t)}{dt}
\]
where:
- \( I(t) \) is the current through the capacitor at time \( t \).
- \( C \) is the capacitance of the capacitor.
- \( V(t) \) is the voltage across the capacitor at time \( t \).
### 2. **Instantaneous Power in a Capacitor**
The instantaneous power \( P(t) \) consumed by the capacitor is given by the product of the voltage \( V(t) \) across it and the current \( I(t) \) through it:
\[
P(t) = V(t) \times I(t)
\]
Substituting the expression for \( I(t) \) from the voltage-current relationship:
\[
P(t) = V(t) \times C \frac{dV(t)}{dt}
\]
### 3. **Power in a Pure Capacitor**
To understand the power consumed over time, we consider the energy \( W \) stored in the capacitor. The energy stored in a capacitor is given by:
\[
W(t) = \frac{1}{2} C V(t)^2
\]
Now, the instantaneous power \( P(t) \) can also be interpreted as the rate of change of energy stored in the capacitor with respect to time:
\[
P(t) = \frac{dW(t)}{dt}
\]
Substituting \( W(t) = \frac{1}{2} C V(t)^2 \) into this equation:
\[
P(t) = \frac{d}{dt} \left( \frac{1}{2} C V(t)^2 \right)
\]
Applying the chain rule:
\[
P(t) = C V(t) \times \frac{dV(t)}{dt}
\]
This is the same expression we derived earlier. The key thing to notice here is that this power can be positive or negative depending on whether \( V(t) \) and \( \frac{dV(t)}{dt} \) have the same or opposite signs.
### 4. **Average Power Over a Complete Cycle**
In AC circuits, we are often interested in the average power over a complete cycle. For a pure capacitor, the voltage and current are 90 degrees out of phase. This means that when the voltage is at its maximum (or minimum), the current is zero, and when the current is at its maximum (or minimum), the voltage is zero.
The average power over a complete cycle \( T \) is given by:
\[
P_{\text{avg}} = \frac{1}{T} \int_0^T P(t) \, dt
\]
Substituting \( P(t) = C V(t) \frac{dV(t)}{dt} \):
\[
P_{\text{avg}} = \frac{1}{T} \int_0^T C V(t) \frac{dV(t)}{dt} \, dt
\]
This can be simplified as:
\[
P_{\text{avg}} = \frac{C}{T} \int_0^T V(t) \frac{dV(t)}{dt} \, dt
\]
Recognizing that the integral of the derivative of a function over a full period is zero:
\[
\int_0^T V(t) \frac{dV(t)}{dt} \, dt = \frac{1}{2} V(t)^2 \Big|_0^T = 0
\]
Hence:
\[
P_{\text{avg}} = 0
\]
### **Conclusion**
The average power consumed by a pure capacitor over a complete AC cycle is zero. This implies that a pure capacitor does not dissipate energy as heat or any other form; instead, it alternately stores and releases energy, resulting in no net energy consumption.
### 1. **Voltage and Current Relationship in a Capacitor**
For a capacitor, the current \( I(t) \) is related to the rate of change of voltage \( V(t) \) across it. The relationship is given by:
\[
I(t) = C \frac{dV(t)}{dt}
\]
where:
- \( I(t) \) is the current through the capacitor at time \( t \).
- \( C \) is the capacitance of the capacitor.
- \( V(t) \) is the voltage across the capacitor at time \( t \).
### 2. **Instantaneous Power in a Capacitor**
The instantaneous power \( P(t) \) consumed by the capacitor is given by the product of the voltage \( V(t) \) across it and the current \( I(t) \) through it:
\[
P(t) = V(t) \times I(t)
\]
Substituting the expression for \( I(t) \) from the voltage-current relationship:
\[
P(t) = V(t) \times C \frac{dV(t)}{dt}
\]
### 3. **Power in a Pure Capacitor**
To understand the power consumed over time, we consider the energy \( W \) stored in the capacitor. The energy stored in a capacitor is given by:
\[
W(t) = \frac{1}{2} C V(t)^2
\]
Now, the instantaneous power \( P(t) \) can also be interpreted as the rate of change of energy stored in the capacitor with respect to time:
\[
P(t) = \frac{dW(t)}{dt}
\]
Substituting \( W(t) = \frac{1}{2} C V(t)^2 \) into this equation:
\[
P(t) = \frac{d}{dt} \left( \frac{1}{2} C V(t)^2 \right)
\]
Applying the chain rule:
\[
P(t) = C V(t) \times \frac{dV(t)}{dt}
\]
This is the same expression we derived earlier. The key thing to notice here is that this power can be positive or negative depending on whether \( V(t) \) and \( \frac{dV(t)}{dt} \) have the same or opposite signs.
### 4. **Average Power Over a Complete Cycle**
In AC circuits, we are often interested in the average power over a complete cycle. For a pure capacitor, the voltage and current are 90 degrees out of phase. This means that when the voltage is at its maximum (or minimum), the current is zero, and when the current is at its maximum (or minimum), the voltage is zero.
The average power over a complete cycle \( T \) is given by:
\[
P_{\text{avg}} = \frac{1}{T} \int_0^T P(t) \, dt
\]
Substituting \( P(t) = C V(t) \frac{dV(t)}{dt} \):
\[
P_{\text{avg}} = \frac{1}{T} \int_0^T C V(t) \frac{dV(t)}{dt} \, dt
\]
This can be simplified as:
\[
P_{\text{avg}} = \frac{C}{T} \int_0^T V(t) \frac{dV(t)}{dt} \, dt
\]
Recognizing that the integral of the derivative of a function over a full period is zero:
\[
\int_0^T V(t) \frac{dV(t)}{dt} \, dt = \frac{1}{2} V(t)^2 \Big|_0^T = 0
\]
Hence:
\[
P_{\text{avg}} = 0
\]
### **Conclusion**
The average power consumed by a pure capacitor over a complete AC cycle is zero. This implies that a pure capacitor does not dissipate energy as heat or any other form; instead, it alternately stores and releases energy, resulting in no net energy consumption.