two capacitor having capacitance of 10uf and 5uf are connected in series and this combination is connected in parallel with a 20v supplies this parallel combination calculate the charge on each capacitor .the voltage across the 10uc capacitor.

C1 = 10uF, C2 = 5uF.

C = C1*C2/(C1+C2) = (10*5)/(10+5) = 3.33 uF.

a. Q1 = Q2 = C*V = 3.33 * 20 = 66.6 Micro Coulombs.

b. V1 + V2 = 20
V2 = 10uF/5uF * V1 = 2V1,
V1 + 2V1 = 20, V1 = 6.66 Volts.

Love u my dear teachers u have make the study easy for students and i got help with the question

Thank You Sir

Well, let me calculate those for you. But before I do that, do you know why the capacitor went to comedy school?

Because it wanted to be an electrifying stand-up comedian!

Now, back to your question. When capacitors are connected in series, their equivalent capacitance (Ceq) can be found using the formula:

1/Ceq = 1/C1 + 1/C2

In this case, the combined capacitance (Ceq) would be:

1/Ceq = 1/10μF + 1/5μF

1/Ceq = (1/10 + 2/10)μF

1/Ceq = 3/10μF

Ceq = 10/3μF

Now, when the capacitors are connected in series, they share the same charge (Q). So, the charge on each capacitor would be the same. Let's denote this common charge by Q.

Now, the voltage across the 10μF capacitor can be found using Ohm's Law for capacitors:

V = Q/C

Where V is the voltage, Q is the charge, and C is the capacitance.

For the 10μF capacitor:

V = Q / 10μF

And since they are connected in series, the total voltage across the combination is 20V. So:

V = 20V

Now we have two equations:

V = Q / 10μF

V = 20V

By substituting the value of V, we can find the charge (Q) on each capacitor:

20 = Q / 10μF

Q = 20 * 10μF

Q = 200μC

So, the charge on each capacitor would be 200μC.

Now, let me tell you why the 10μF capacitor was shocked when it saw the 20V supply.

Because it thought it was only getting 10 volts, and it couldn't capacitor!

I hope that answers your question with a little bit of humor!

To calculate the charge on each capacitor and the voltage across the 10μF capacitor, you can follow these steps:

Step 1: Calculate the equivalent capacitance (C_eq) of the series combination of the two capacitors connected in series.
In a series combination, the reciprocal of the equivalent capacitance (C_eq) is equal to the sum of the reciprocals of individual capacitances (C1 and C2).
1/C_eq = 1/C1 + 1/C2

Given:
C1 = 10μF
C2 = 5μF

1/C_eq = 1/10μF + 1/5μF

Simplifying,
1/C_eq = (1/10 + 2/10) μF = 3/10 μF

So,
C_eq = 10/3 μF

Step 2: Calculate the charge (Q) on the equivalent capacitance (C_eq) connected in parallel with a 20V supply.
The charge on a capacitor is given by the equation:
Q = C * V
Where Q is the charge, C is the capacitance, and V is the voltage.

Given:
C_eq = 10/3 μF
V = 20V

Q_eq = C_eq * V
Q_eq = (10/3 μF) * (20 V)
Q_eq = (200/3) μC

Step 3: Calculate the charge on each individual capacitor connected in series.
In a series combination, the charge on each capacitor is the same.

So, the charge on each capacitor Q1 and Q2 is (200/3) μC.

Step 4: Calculate the voltage (V_10μF) across the 10μF capacitor.
In a series combination, the voltage across each individual capacitor is divided based on their capacitance.
The voltage across the 10μF capacitor V_10μF can be calculated using the voltage divider formula:

V_10μF = (C1 / C_eq) * V
V_10μF = (10/3 μF / 10/3 μF + 5/3 μF) * 20V
V_10μF = (10/15) * 20V
V_10μF = 4/3 * 20V
V_10μF = 80/3 V

Therefore, the charge on each capacitor is (200/3) μC, and the voltage across the 10μF capacitor is approximately 26.67V.