Advertisements
Advertisements
Question
Derive the expression for resultant capacitance, when the capacitor is connected in parallel.
Advertisements
Solution
Consider three capacitors of capacitance C1,C2 and C3 connected in parallel with a battery of voltage V as shown in figure (a).
(a) capacitors in parallel-
(b) equivalent capacitance with the same total charge
Since corresponding sides of the capacitors are connected to the same positive and negative terminals of the battery, the voltage across each capacitor is equal to the battery’s voltage. Since capacitance of the capacitors is different, the charge stored in each capacitor is not the same. Let the charge stored in the three capacitors be Q1,Q2, and Q2 respectively. According to the law of conservation of total charge, the sum of these three charges is equal to the charge Q transferred by the battery,
Q = Q1 + Q2 + Q3 ….. (1)
Now, since Q = CV, we have
Q = C1V + C2 V + C3 V ….. (2)
If these three capacitors are considered to form a single capacitance CP which stores the total charge Q as shown in figure (b), then we can write Q = CPV. Substituting this in equation (2), we get
Cp V = C1 V + C2 V + C3 V
Cp = C1 + C2 + C3
Thus, the equivalent capacitance of capacitors connected in parallel is equal to the sum of the individual capacitance. The equivalent capacitance Cp in a parallel connection is always greater than the largest individual capacitance. In a parallel connection, it is equivalent as area of each capacitance adds to give more effective area such that total capacitance increases.
APPEARS IN
RELATED QUESTIONS
As `C = (1/V) Q` , can you say that the capacitance C is proportional to the charge Q?
Following operations can be performed on a capacitor:
X − connect the capacitor to a battery of emf ε.
Y − disconnect the battery.
Z − reconnect the battery with polarity reversed.
W − insert a dielectric slab in the capacitor.
(a) In XYZ (perform X, then Y, then Z) the stored electric energy remains unchanged and no thermal energy is developed.
(b) The charge appearing on the capacitor is greater after the action XWY than after the action XYZ.
(c) The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.
(d) The electric field in the capacitor after the action XW is the same as that after WX.
Find the charge supplied by the battery in the arrangement shown in figure.
Find the equivalent capacitances of the combinations shown in figure between the indicated points.
The figure show a network of five capacitors connected to a 10V battery. Calculate the charge acquired by the 5μF capacitor.
Obtain an expression for equivalent capacitance when three capacitors C1, C2 and C3 are connected in series.
- Charge on each capacitor remains same and equals to the main charge supplied by the battery.
- Potential difference and energy distribute in the reverse ratio of capacitance.
- Effective capacitance is even les than the least of teh individual capacitances.
The displacement current of 4.425 µA is developed in the space between the plates of the parallel plate capacitor when voltage is changing at a rate of 106 Vs-1. The area of each plate of the capacitor is 40 cm2. The distance between each plate of the capacitor is x × 10-3 m. The value of x is ______.
(Permittivity of free space, ε0 = 8.85 × 10-12C2N-1m-2).
Read the following paragraph and answer the questions.
| A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference between them. The capacitance of a capacitor depends on the geometrical configuration (shape, size and separation) of the system and also on the nature of the insulator separating the two conductors. They are used to store charges. Like resistors, capacitors can be arranged in series or parallel or a combination of both to obtain the desired value of capacitance. |
- Find the equivalent capacitance between points A and B in the given diagram.
- A dielectric slab is inserted between the plates of the parallel plate capacitor. The electric field between the plates decreases. Explain.
- A capacitor A of capacitance C, having charge Q is connected across another uncharged capacitor B of capacitance 2C. Find an expression for (a) the potential difference across the combination and (b) the charge lost by capacitor A.
OR
Two slabs of dielectric constants 2K and K fill the space between the plates of a parallel plate capacitor of plate area A and plate separation d as shown in the figure. Find an expression for the capacitance of the system.
The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K1 and K2 with thickness `3/8 d and d/2`, respectively, are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K1 = 1.25 K, the value of K1 is:
