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
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Answered by
19
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.
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Answered by
0
Answer:
USING KIRCHHOF'S LAW . Q=CV AND V=Q/C
CORRECT ANSWER IS 12V
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