Solve 2.00 & 4.60µF Capacitors Voltage: Help Needed

In summary, there are two capacitors, one with a capacitance of 2.00 µF and a charge of 520 V, and the other with a capacitance of 4.60 µF and a charge of 570 V. When the plates of opposite signs are connected, the voltage for each capacitor can be found by using the equation C=Q/V. Despite trying different values and equations, the correct answer has not been achieved. However, a similar question with 2.4uF and 880V and 4uF and 560V capacitors resulted in a voltage of 20V. The correct answer is still unknown.
  • #1
ncsuwx
3
0
Here is the question

A(n) 2.00 µF capacitor is charged to 520 V and a 4.60 µF capacitor is charged to 570 V

What is the voltage for each capacitor if plates of opposite signs are connected?

Any help would be helpful. I have bene trying to get this and haven't been able to all night. I only have one submission left to get the right answer soany suggestions would be greatly appreciated.
 
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  • #2
The equation is C=Q/V. Just by pluggin in values does not work for me and i can't figure out what I am supposed to do.
 
  • #3
I still can't figure it out. I have been trying and still can't get the right answer. I have an answer similar for a question for 2.4uF and 880V and 4uF and 560V capacitors. The voltage is 20V for that one. I can't even get that answer
 

Related to Solve 2.00 & 4.60µF Capacitors Voltage: Help Needed

1. How do I solve for the voltage of two capacitors with different capacitance values?

To solve for the voltage of two capacitors with different capacitance values, you can use the formula V = Q/C, where V is the voltage, Q is the charge, and C is the capacitance. For two capacitors in series, the total capacitance (C) is equal to the reciprocal of the sum of their individual capacitances (1/C = 1/C1 + 1/C2). For two capacitors in parallel, the total capacitance is equal to the sum of their individual capacitances (C = C1 + C2). Once you have calculated the total capacitance, you can plug it into the formula to solve for the voltage.

2. What units should I use when solving for the voltage of capacitors?

When solving for the voltage of capacitors, you should use units of volts (V) for voltage, coulombs (C) for charge, and farads (F) for capacitance. Make sure to also use consistent units for all values in the formula to avoid errors in your calculation.

3. How does the voltage change when capacitors are connected in series versus parallel?

In a series circuit, the voltage across each capacitor is divided based on their individual capacitance values. This means that the total voltage across all capacitors in a series circuit will be less than the voltage supplied by the source. In a parallel circuit, the voltage across each capacitor is the same and equal to the source voltage. This means that the total voltage across all capacitors in a parallel circuit will be equal to the voltage supplied by the source.

4. Can the voltage of capacitors be negative?

No, the voltage of capacitors cannot be negative. Capacitors store energy in the form of electric charge, and the voltage is a measure of the potential difference between the plates of the capacitor. A negative voltage would indicate that one plate has a lower potential than the other, which is not possible in a capacitor.

5. What factors can affect the voltage of capacitors?

The voltage of capacitors can be affected by various factors, such as the capacitance value, the type of dielectric material used, the size and shape of the capacitor, and the voltage supplied by the source. Other external factors, such as temperature and humidity, can also impact the voltage of capacitors.

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