Advertisements
Advertisements
Question
Calculate the resultant capacitances for each of the following combinations of capacitors.
Advertisements
Solution
Parallel combination of capacitor 1 and 2
Cp = C0 + C0 = 2C0
Series combination of capacitor Cp and 3
`1/"C"_"s" = 1/"C"_"p" + 1/"C"_3 = 1/(2 "C"_0) + 1/"C"_0 = ("or") 1/"C"_"s" = 3/2 "C"_0 ("or") "C"_"s" = 2/3 "C"_0`
`1/"C"_("s"_1) = 1/"C"_1 + 1/"C"_2 = 1/"C"_0 + 1/"C"_0 = 1/"C"_0` (or)
`1/"C"_("s"_1) = 2/"C"_0 ("or") "C"_("S"_1) = "C"_0/2`
Similarly 3 and 4 are series combination
`1/"C"_("s"_1) = 1/"C"_3 + 1/"C"_4 = 1/"C"_0 + 1/"C"_0 = 2/"C"_0 ("or") "C"_("S"_2) = "C"_0/2`
`"C"_("S"_1) and "C"_("S"_2)` are in parallel combination
`"C"_"p" = "C"_("S"_1) + "C"_("S"_2) = "C"_0/2 + "C"_0/2 ("or") "C"_"p" = (2"C"_0)/2 "C"_"p" = "C"_0`- Capacitor 1, 2 and 3 are in parallel combination
Cp = C0 + C0 + C0 = 3C0
Cp = 3C0
- Capacitar C1 and C2 are in combination
`1/"C"_("s"_1) = ("C"_1 + "C"_2)/("C"_1"C"_2)`
`"C"_("s"_1) = ("C"_1"C"_2)/("C"_1 + "C"_2)`
Similarly C3 and C4 are in series combination
`1/"C"_("s"_2) = 1/"C"_3 + 1/"C"_4 = ("C"_3 + "C"_4)/("C"_3"C"_4)`
`"C"_("s"_2) = ("C"_3"C"_4)/("C"_3 + "C"_4)`
`"C"_("s"_1) and "C"_("s"_2)` are in parallel combination across RS:
CP = `"C"_("S"_1) + "C"_("S"_2)`
`= ("C"_1"C"_2)/("C"_1 + "C"_2) + ("C"_3"C"_4)/("C"_3 + "C"_4)`
`= ("C"_1"C"_2 ("C"_3 + "C"_4) + "C"_3"C"_4("C"_1 + "C"_2))/(("C"_1 + "C"_2)("C"_3 + "C"_4))`
`"C"_"P" = ("C"_1"C"_2"C"_3 + "C"_1"C"_2"C"_4 + "C"_3"C"_4"C"_1 + "C"_3"C"_4"C"_2)/(("C"_1 + "C"_2)("C"_3 + "C"_4))` - Capacitor 1 and 2 are series combination
`1/"C"_("s"_1) = 1/"C"_1 + 1/"C"_2 = 1/"C"_0 + 1/"C"_0 = 1/"C"_0`
`1/"C"_("s"_1) = 2/"C"_0` (or) `"C"_"s"_1 = "C"_0/2`
Similarly 3 and 4 are series combination
`1/"C"_("s"_1) = 2/"C"_0 ("or") "C"_("s"_2) = "C"_0/2`
Three capacitors are in parallel combination
`"C"_"P" = "C"_0/2 + "C"_0 + "C"_0/2 + (2"C"_0)/2 + "C"_0/2 = (4"C"_0)/2`
`"C"_"P" = 2 "C"_0`
APPEARS IN
RELATED QUESTIONS
A capacitor of capacitance C is charged fully by connecting it to a battery of emf E. It is then disconnected from the battery. If the separation between the plates of the capacitor is now doubled, how will the following change?
(i) charge stored by the capacitor.
(ii) Field strength between the plates.
(iii) Energy stored by the capacitor.
Justify your answer in each case.
A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected? (a) The electric field between the plates of the capacitor Justify your answer by writing the necessary expressions.
Two metal spheres of capacitance C1 and C2 carry some charges. They are put in contact and then separated. The final charges Q1 and Q2 on them will satisfy
The outer cylinders of two cylindrical capacitors of capacitance 2⋅2 µF each, are kept in contact and the inner cylinders are connected through a wire. A battery of emf 10 V is connected as shown in figure . Find the total charge supplied by the battery to the inner cylinders.
A capacitor of capacitance 10 μF is connected to a battery of emf 2 V. It is found that it takes 50 ms for the charge of the capacitor to become 12.6 μC. Find the resistance of the circuit.
Suppose the space between the two inner shells is filled with a dielectric of dielectric constant K. Find the capacitance of the system between A and B.
Three circuits, each consisting of a switch 'S' and two capacitors, are initially charged, as shown in the figure. After the switch has been closed, in which circuit will the charge on the left-hand capacitor
(i) increase,
(ii) decrease, and
(iii) remains the same? Give reasons.
If the voltage applied on a capacitor is increased from V to 2V, choose the correct conclusion.
- 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.
Two plates A and B of a parallel plate capacitor are arranged in such a way, that the area of each plate is S = 5 × 10-3 m 2 and distance between them is d = 8.85 mm. Plate A has a positive charge q1 = 10-10 C and Plate B has charge q2 = + 2 × 10-10 C. Then the charge induced on the plate B due to the plate A be - (....... × 10-11 )C
