Two 470 µF capacitors connected in series are subjected to a total applied voltage that changes at a rate of 200 volts per second. How much current will there be ”through” these capacitors?
The correct answer and explanation is :
To calculate the current through two capacitors connected in series, we need to understand the relationship between voltage, capacitance, and current in a capacitive circuit.
Formula for current in a capacitor:
The current through a capacitor is given by the equation:
[
I = C \cdot \frac{dV}{dt}
]
Where:
- ( I ) is the current,
- ( C ) is the capacitance,
- ( \frac{dV}{dt} ) is the rate of change of voltage across the capacitor.
Step 1: Combine the capacitance in series
When capacitors are connected in series, the total capacitance ( C_{\text{total}} ) is given by the reciprocal sum:
[
\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2}
]
For two capacitors, each with capacitance ( C = 470 \, \mu F ):
[
\frac{1}{C_{\text{total}}} = \frac{1}{470} + \frac{1}{470} = \frac{2}{470}
]
Thus, the total capacitance is:
[
C_{\text{total}} = \frac{470}{2} = 235 \, \mu F
]
Step 2: Apply the given rate of change of voltage
We are told that the total applied voltage changes at a rate of ( \frac{dV}{dt} = 200 \, \text{V/s} ).
Step 3: Calculate the current
Now, using the formula for current, we substitute the values:
[
I = C_{\text{total}} \cdot \frac{dV}{dt} = 235 \, \mu F \times 200 \, \text{V/s}
]
[
I = 235 \times 10^{-6} \, \text{F} \times 200 \, \text{V/s} = 0.047 \, \text{A} = 47 \, \text{mA}
]
Final Answer:
The current through the capacitors is 47 mA.
Explanation:
In a series circuit, the same current flows through each capacitor. Since capacitors resist changes in voltage, the rate of change of voltage across them results in a current. The total capacitance of two series-connected capacitors is smaller than the individual capacitance, and the rate of voltage change directly determines the current. Thus, by knowing the rate of voltage change, the total capacitance, and using the formula ( I = C \cdot \frac{dV}{dt} ), we can calculate the current flowing through the series combination.
Here is an illustration of two 470 µF capacitors connected in series with the applied voltage changing at a rate of 200 volts per second. This shows the setup of the capacitors and the resulting current of 47 mA flowing through the circuit.