Two capacitors A and B having capcitance 2micro farad and 6 micro farad are charged separately to same potential of 120 volt.Now pisitive plate of A is connected to negative plate of B and negative plate of A is connected to positve plate of B. Find final charge on each capacitor.
=> First, we calculate the net charge stored in the initial circuit.
=> Q stored in C1 = CV = 2 x 120 = 240 micro Coul.
=> Q stored in C2 = CV = 6 x 120 = 720 micro Coul.
That gives us a total charge of 960 micro Coul.
Try using the following hints now to get at the answer yourself:
- In the new circuit, both capacitors will be considered to be in parallel
- The net charge will remain 960 micro Coul. due to conservation of charge
I'd be happy to help further if you can't get to the answer, but do attempt it yourself first!
Just to be clear, the point I was making by 'will be in parallel' is that they will have a common potential difference (V). So you have the values of C, a common value for V, and the total value of Q. Can you recall a certain equation which includes C, V, and Q?
To find the final charge on each capacitor, we can use the principle of conservation of charge. According to this principle, the total charge before and after connecting the capacitors remains the same.
Let's break down the solution step by step:
Step 1: Find the initial charge on each capacitor.
Since both capacitors are charged separately to the same potential of 120 volts, we can use the formula Q = CV to find the charge on each capacitor. Here, Q represents the charge, C represents the capacitance, and V represents the potential.
For capacitor A:
Q_A = C_A * V = 2 µF * 120 V = 240 µC
For capacitor B:
Q_B = C_B * V = 6 µF * 120 V = 720 µC
Step 2: Connect the capacitors in series.
When we connect the capacitors in series, the positive plate of capacitor A is connected to the negative plate of capacitor B, and the negative plate of capacitor A is connected to the positive plate of capacitor B.
+------------+
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+-----| Capacitor A |------+
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+++ +++
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| +------------+ |
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+----------------------------------+
Step 3: Calculate the equivalent capacitance for the series combination.
In a series combination of capacitors, the reciprocal of the equivalent capacitance (1/C_eq) is equal to the sum of the reciprocals of the individual capacitances (1/C_A + 1/C_B).
1/C_eq = 1/C_A + 1/C_B
Substituting the values, we get:
1/C_eq = 1/2 µF + 1/6 µF
1/C_eq = (1/2 + 1/6) µF
1/C_eq = 4/6 µF
1/C_eq = 2/3 µF
Therefore, C_eq = 3/2 µF = 1.5 µF
Step 4: Calculate the final charge on each capacitor after the connection.
Since the capacitors are connected in series, they share the same charge Q.
Q = Q_A = Q_B
So, the final charge on both capacitors is 240 µC.
Final charge on capacitor A = Q_A = 240 µC
Final charge on capacitor B = Q_B = 240 µC
Thus, the final charge on each capacitor is 240 µC.