Question
Three capacitors, each of capacitance 120 pF, are each charged to 500 V and then connected in series. Determine (a) the potential difference between the end plates, (b) the charge on each capacitor, and (c) the energy stored in the system.
how?
8=D
2. Three capacitors each of 100nF are charged to 500V and then connected in series,
determine (a) The potential difference between the end plate (b) The charge on each capacitor �
Is too see the work
To answer this question, we need to apply the principles of series combination of capacitors. Let's break it down step by step:
Step 1: Determine the total capacitance (Ceq) of the series combination.
When capacitors are connected in series, their effective capacitance (Ceq) is given by the reciprocal of the sum of the reciprocals of individual capacitances. In this case, since all three capacitors have the same capacitance, we can use the formula:
1/Ceq = 1/C1 + 1/C2 + 1/C3
= 1/120pF + 1/120pF + 1/120pF
Calculating this expression will give us the value of Ceq.
Step 2: Calculate the potential difference (Veq) across the series combination.
When capacitors are connected in series, the total voltage across the combination is equal to the sum of the voltages across each individual capacitor.
In this case, the potential difference (Veq) across the series combination is 500V since each capacitor is charged to 500V.
Step 3: Calculate the potential difference (V) across each capacitor.
Since the capacitors are in series, the potential difference across each capacitor is the same and is equal to Veq. Therefore, the potential difference across each capacitor is also 500V.
Step 4: Calculate the charge (Q) on each capacitor.
The charge on each capacitor in a series combination is the same. To calculate it, we can use the formula:
Q = C × V
where C is the capacitance and V is the potential difference across each capacitor.
Step 5: Calculate the energy stored in the system.
The energy stored in a capacitor is given by the formula:
E = (1/2) × C × V^2
Now, using the values obtained from the previous steps, we can calculate the solutions to the given questions:
(a) The potential difference between the end plates is 500V.
(b) The charge on each capacitor is calculated by using Q = C × V.
(c) The energy stored in the system is calculated by using E = (1/2) × C × V^2.
So, by following these steps and using the given values, we can find the answers to the questions.
a. 1500v
b. q=C/V=120E-12/500 C
c. add the energies of each 1/2 q^2/V for each of the three, V=500, q=above