A 2μF capacitor is charged to 50V and then connected in parallel with a 4μF capacitor charged to 100 V.?
Determine (a) the charge and potential difference of the system, (b) the charge on
each capacitor of the system, (c) the energy stored in the system, (d) the total energy stored in
the original two capacitors before the parallel connection is made.
To solve this problem, we need to consider the properties of capacitors and how they behave when connected in parallel.
(a) To determine the charge and potential difference of the system, we need to calculate the total capacitance and the total charge.
The total capacitance (C_total) of capacitors connected in parallel is the sum of the individual capacitances (C_1 and C_2) of the capacitors:
C_total = C_1 + C_2
Given that C_1 = 2μF and C_2 = 4μF, we can calculate C_total:
C_total = 2μF + 4μF
= 6μF
The total charge (Q_total) of the capacitors is the sum of the individual charges (Q_1 and Q_2) of the capacitors:
Q_total = Q_1 + Q_2
To find the charges, we can use the formula for the charge on a capacitor:
Q = C * V
where Q is the charge, C is the capacitance, and V is the voltage.
For the first capacitor (C_1), Q_1 = C_1 * V_1:
Q_1 = 2μF * 50V
= 100μC
For the second capacitor (C_2), Q_2 = C_2 * V_2:
Q_2 = 4μF * 100V
= 400μC
Now we can calculate the total charge:
Q_total = Q_1 + Q_2
= 100μC + 400μC
= 500μC
The potential difference (V_total) of the system is the same as the potential difference across each individual capacitor when connected in parallel. Thus, V_total is equal to the higher voltage value in the system, which is 100V.
So, the charge of the system is 500μC, and the potential difference across the system is 100V.
(b) To determine the charge on each capacitor, we already calculated the charges earlier:
The charge on the first capacitor (C_1) is Q_1 = 100μC.
The charge on the second capacitor (C_2) is Q_2 = 400μC.
(c) The energy stored in the system is given by the formula:
E = (1/2) * C_total * (V_total)^2
Plugging in the values, we get:
E = (1/2) * 6μF * (100V)^2
= 300μJ
So, the energy stored in the system is 300μJ.
(d) Before the parallel connection was made, each individual capacitor had energy stored in them. The energy stored in a capacitor is given by the formula:
E = (1/2) * C * (V^2)
Using this formula for each capacitor:
For the first capacitor (C_1), E_1 = (1/2) * 2μF * (50V)^2:
E_1 = 0.05mJ
For the second capacitor (C_2), E_2 = (1/2) * 4μF * (100V)^2:
E_2 = 2mJ
Adding the energy of each individual capacitor together:
Total energy = E_1 + E_2
= 0.05mJ + 2mJ
= 2.05mJ
So, the total energy stored in the original two capacitors before the parallel connection is made is 2.05mJ.