This question concerns your ability to manipulate combinations of capacitors. Devise a network
of capacitors whose total equivalent capacitance is 2μF while the total voltage difference from
one end to the other end is 400 V.
Your challenge is to do this with only 1 type of capacitor, namely those with capacitances of 2μF
each. In addition, the voltage drop over each and every one of them should be 100 V.
To devise a network of capacitors that satisfies the given conditions, we need to use a combination of series and parallel connections.
First, let's determine how many capacitors we need. The total equivalent capacitance should be 2μF, and each capacitor has a capacitance of 2μF. Therefore, we need a total of 1μF of capacitance.
To achieve this capacitance, we can use a combination of series and parallel connections. Let's start by connecting capacitors in series. When capacitors are connected in series, the total capacitance is given by the reciprocal of the sum of the reciprocals of the individual capacitances.
1/C_total = 1/C_1 + 1/C_2 + 1/C_3 + ...
Since each capacitor has a capacitance of 2μF, we can rewrite the equation as:
1/(1μF) = 1/(2μF) + 1/(2μF) + 1/(2μF) + ...
Simplifying the equation, we get:
1/(1μF) = n/(2μF)
Where n is the number of capacitors connected in series.
Solving for n, we find:
n = 2
So, we need 2 capacitors connected in series to achieve 1μF of capacitance.
Now, let's determine the voltage drop over each capacitor. The total voltage difference from one end to the other end is given as 400V, and the voltage drop over each capacitor should be 100V.
Since the capacitors are connected in series, the total voltage difference is equal to the sum of the voltage drops over each capacitor.
Total voltage difference = Voltage drop over capacitor 1 + Voltage drop over capacitor 2 + ...
400V = 100V + 100V + ...
To achieve a total voltage of 400V, we need to connect 4 capacitors in series.
Therefore, the network configuration that satisfies the given conditions is as follows:
- Connect 2 capacitors in series to achieve a total capacitance of 1μF.
- Connect 4 of these series combinations in parallel to achieve a total capacitance of 2μF.
- Each capacitor will have a voltage drop of 100V, resulting in a total voltage difference of 400V across the network.
Note: It is important to consider the voltage ratings of the capacitors being used. Ensure that the capacitors used have a voltage rating higher than 100V to avoid damaging them.