Question

Five capacitors are connected as shown in the figure. Initially S is opened and all capacitors are uncharged. WhenSis closed, steady state is obtained. Then find out potential difference between the points Mand N.MIHA4uF2uF 4uF31V6uF1.2uFÑECTION(C): COMBINATION OF CAPACITORS​

Answer 1

Answer:

Explanation:

=> Five capacitors 4μF, 2μF, 4μF, 6μF and 1.2μF are connected as shown in the figure.

Here, Equivalent capacitance = ½,

EMF = 24

Thus, charge passes through each capacitor = 12 μC

Volatge across 4μF = 3V

Volatge across 6μF = 2V

=> Potential difference (V) :

V = 3 + 7+ 2 = 12V

Thus, potential difference (V) between the two given points M and N is 12v.

Answer 2

Answer:

USING KIRCHHOF'S LAW . Q=CV AND V=Q/C

CORRECT ANSWER IS 12V