Three capacitors 5uF, 3uF and 2uF are connected in series to a 100V d.c. Draw the circuit diagram.
Calculate the energy stored by the system
for each
C = q/V so V = q/C
They each have the same charge q
V1 = q / (5 *10^-6)
V2 = q / (3 *10^-6)
V3 = q / (2 *10^-6)
so
10 = 10^6 q [ 1 / 5 +1/3 + 1/2 ] =10^6 q * 31/30
q = 10^-6 *(300/31)
V1 = 300/(5*31) = 1.935
V2 = 300 / (3*31) = 3.225
V3 = 300 /(2*31) = 4.839 (whew, the sum is 10 volts :)
You now have the charge and the voltage on each capacitor
energy stored in each = (1/2) q V in Joules
so just add them up
Whoops, I used ten volts. They said 100
multiply the charges by 10 and the voltages by 10 so the energy by100.
To draw the circuit diagram, we can represent the capacitors as individual components connected one after the other in a series configuration.
Here is the circuit diagram for the given scenario:
________
| |
-------------| 5uF |--------| 3uF |--------| 2uF |--------------
The positive terminal of the 100V DC source is connected to one end of the 5uF capacitor, and the negative terminal is connected to one end of the 2uF capacitor. The other ends of the capacitors are connected in series.
To calculate the energy stored by the system, we can use the formula:
Energy stored in a capacitor = 1/2 * C * V^2,
where C is the capacitance and V is the voltage applied across the capacitor.
For each capacitor, we can calculate the energy stored using the given formula:
Energy stored in 5uF capacitor = 1/2 * 5 * 10^-6 * (100)^2
Energy stored in 5uF capacitor = 0.25 Joules
Energy stored in 3uF capacitor = 1/2 * 3 * 10^-6 * (100)^2
Energy stored in 3uF capacitor = 0.15 Joules
Energy stored in 2uF capacitor = 1/2 * 2 * 10^-6 * (100)^2
Energy stored in 2uF capacitor = 0.1 Joules
Finally, the total energy stored by the system is the sum of the energies stored in each capacitor:
Total Energy = Energy stored in 5uF capacitor + Energy stored in 3uF capacitor + Energy stored in 2uF capacitor
Total Energy = 0.25 Joules + 0.15 Joules + 0.1 Joules
Total Energy = 0.5 Joules
To draw the circuit diagram, you would need to arrange the capacitors in series, which means connecting them one after the other. The positive (+) terminal of the first capacitor should be connected to the negative (-) terminal of the second capacitor, and so on. The remaining terminals of the first capacitor and the last capacitor in the series should be connected to the power source (100V d.c.). Here is a simple schematic representation:
100V d.c. ----[5uF]----[3uF]----[2uF]----
| | |
---- ---- ----
To calculate the energy stored by the system, you can use the formula:
E = 1/2 * C * V^2
Where:
E = energy stored in joules
C = capacitance in farads
V = voltage across the capacitor in volts
Since the capacitors are connected in series, the total capacitance of the system will be the reciprocal of the sum of the reciprocals of the individual capacitances.
1/C_total = 1/C1 + 1/C2 + 1/C3
Substituting the respective values:
1/C_total = 1/5uF + 1/3uF + 1/2uF
Calculating the sum and taking the reciprocal will give you the total capacitance:
C_total = 1 / (1/5uF + 1/3uF + 1/2uF)
Once you have the total capacitance, you can calculate the energy stored using the formula mentioned earlier:
E = 1/2 * C_total * V^2
Substitute the values of C_total (in farads) and V (in volts) to find the energy stored.