are charged in parallel to the same potential difference V V by a battery. The switch is opened, so capacitors are discharged through a resistor. I wanted to know how the p.d V1 V 1 and V2 V 2 of the two capacitors would vary, with respect to time.
Figure 8.3.2 8.3. 2: (a) Three capacitors are connected in parallel. Each capacitor is connected directly to the battery. (b) The charge on the equivalent capacitor is the sum of the charges on the individual capacitors.
Voltages parallel to a capacitor may also be found when there is no flow of current. (iii). A capacitor has a capacity to store charge. (iv). It has become clear from i = C dv / dt that a current in a capacitor exists at a time when voltages found parallel to it, change with the time.
The capacitors are in parallel so the potential difference across them must be the same. The time constant of the circuit should have been R(C1 +C2) R ( C 1 + C 2) as the two capacitors in parallel are equivalent to one capacitor with a capacitance equal to the sum of the capacitances of the individual capacitors.
The capacitance value of a parallel plate capacitor is given by, C = k ε0A/d Here k is the dielectric constant, and ε0 is the permittivity of the free space and it is equal to the 8.854 X 10 -12 F/m. The dielectric constant (k) is a parameter related to dielectric material which increases the capacitancecompared to air.
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