Two capacitor of a capacitance 4uf and 6uf are connected in series to a 100volt DC
A.draw the circuit diagram
B.calculate the energy on either plates of each capacitor
C.potential difference across eàch capacitor
D.energy in the combined capacitor
A. Sorry, I'm not an artist, but imagine two sad-looking capacitors connected in a row like a chain. One is labeled 4uF and the other 6uF, and there's a 100V DC source connected to one end of the chain.
B. Well, I hate to break it to you, but capacitors don't have plates for eating energy. They mostly just sit there and store it. So, the energy stored on each capacitor is given by the formula: E = 0.5 * C * V^2, where C is the capacitance and V is the voltage. Plug in the values: For the 4uF capacitor, E = 0.5 * 4uF * (100V)^2. For the 6uF capacitor, E = 0.5 * 6uF * (100V)^2. Calculate those values, and you'll find the energy on each plate.
C. Since the capacitors are connected in series, they share the same voltage. So, the potential difference across each capacitor would be the same as the total voltage of the circuit, which is 100V.
D. To find the energy in the combined capacitor, you'll need to calculate the total capacitance. In series, the total capacitance (C_total) is given by the formula: 1/C_total = 1/C1 + 1/C2. Plug in the values: 1/C_total = 1/4uF + 1/6uF. Calculate that, and take the reciprocal to find the value of C_total. Then, you can use the formula E = 0.5 * C_total * V^2 to find the energy in the combined capacitor.
A. The circuit diagram for two capacitors connected in series to a 100V DC source would look like this:
------------------------ ------------------------
| | | |
--| 4μF Capacitor |--|----| 6μF Capacitor |--
| | | | | | |
| |______________________| | |______________________| |
| | |
| | |
| |-----| |
| | |----------------------------|
| |
| |
|--------------------------------|
| |
| 100V DC Source |
| |
| |
|________________________________|
B. The energy on each plate of the capacitors can be calculated using the formula: E = (1/2) * C * V^2, where E is the energy, C is the capacitance, and V is the voltage.
For the 4μF capacitor:
E1 = (1/2) * 4μF * (100V)^2
E1 = (1/2) * 4 * 10^-6 F * (100)^2 V^2
E1 = 2 * 10^-6 * 10,000 V^2
E1 = 20 mJ
For the 6μF capacitor:
E2 = (1/2) * 6μF * (100V)^2
E2 = (1/2) * 6 * 10^-6 F * (100)^2 V^2
E2 = 3 * 10^-6 * 10,000 V^2
E2 = 30 mJ
So, the energy on the plates of the 4μF capacitor is 20 mJ, and the energy on the plates of the 6μF capacitor is 30 mJ.
C. The potential difference across each capacitor in a series circuit is the same as the applied voltage, which is 100V in this case.
D. The total energy in the combined capacitor can be found by adding the energies of the individual capacitors:
E_total = E1 + E2
E_total = 20 mJ + 30 mJ
E_total = 50 mJ
A. To draw the circuit diagram, you will need to arrange the capacitors in series. The positive terminal of the first capacitor should be connected to the negative terminal of the second capacitor. The positive terminal of the second capacitor should be connected to the positive terminal of the battery, and the negative terminal of the first capacitor should be connected to the negative terminal of the battery.
---------- -------------
| | | |
| 4µF | | 6µF |
| |----| |----| |
| | | | |
---------- | 100V DC
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|
⏚
B. To calculate the energy on either plate of each capacitor, we can use the formula:
E = (1/2) * C * V^2
Where E is the energy, C is the capacitance, and V is the voltage.
For the 4µF capacitor:
E1 = (1/2) * 4µF * (100V)^2
For the 6µF capacitor:
E2 = (1/2) * 6µF * (100V)^2
Calculate the values to get the specific energy on either plate of each capacitor.
C. To calculate the potential difference (voltage) across each capacitor, we can use the formula:
V = Q / C
Where V is the potential difference, Q is the charge stored on the capacitor, and C is the capacitance.
The total charge stored on each capacitor in a series connection is the same. So, we can calculate the charge Q using the formula:
Q = C_eq * V
Where C_eq is the equivalent capacitance of the series combination, which can be calculated as:
(1 / C_eq) = (1/C1) + (1/C2)
Once you have calculated C_eq, you can substitute it into the formula for Q and then find the potential difference across each capacitor (V1 and V2) using the formula V = Q / C.
D. To calculate the energy in the combined capacitor, we can use the formula:
E_combined = (1/2) * C_combined * V^2
Where E_combined is the energy in the combined capacitor, C_combined is the equivalent capacitance of the series combination, and V is the voltage.
Using the formula for C_eq mentioned earlier, you can substitute the value of C_eq into the formula for E_combined.
When the capacitor is connected to dc source, it gets charged. The battery is then disconnected, so no more charge can flow in. On removing dielectric, capacitance decreases.
Energy stored (u= 2Cq2) will increase
Potential (V= Cq) will also increase
Electric field (E= DV) will increase