and Capacitance E XAMPLE . The proof clearly goes through for any number of capacitors arranged in a similar way. Equation ( . ), for n capacitors arranged in series, generalises to n n ...
... Q Q Q ( . ) Following the same steps as for the case of two capacitors, we get the general formula for effective capacitance of a series combination of n capacitors: n ... ( .
) . . Capacitors in parallel Figure . (a) shows two capacitors arranged in parallel.
In this case, the same potential difference is applied across both the capacitors. But the plate charges (± Q ) on capacitor and the plate charges (± Q ) on the capacitor are not necessarily the same: Q = C V , Q = C V ( . ) The equivalent capacitor is one with charge Q = Q + Q ( . ) and potential difference V .
Q = CV = C V + C V ( . ) The effective capacitance C is, from Eq. ( . ), C = C + C ( .
) The general formula for effective capacitance C for parallel combination of n capacitors [Fig. . (b)] follows similarly, Q = Q + Q + ... + Q n ( .
) i.e., CV = C V + C V + ... C n V ( . ) which gives C = C + C + ... C n ( .
) Example . A network of four µ F capacitors is connected to a V supply, as shown in Fig. . .
Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.) FIGURE . Parallel combination of (a) two capacitors, (b) n capacitors. FIGURE .
E XAMPLE . Solution (a) In the given network, C , C and C are connected in series. The effective capacitance C ′ of these three capacitors is given by ′ For C = C = C = µ F, C ′ = ( / ) µ F. The network has C ′ and C connected in parallel.
Thus, the equivalent capacitance C of the network is C = C ′ + C = µ F = . µ F (b) Clearly, from the figure, the charge on each of the capacitors, C , C and C is the same, say Q . Let the charge on C be Q ′ . Now, since the potential difference across AB is Q / C , across BC is Q / C , across CD is Q / C , we have V Q Q Q .
Also, Q ′ / C = V. This gives for the given value of the capacitances, F . Q µ and F . Q µ ′