A. What is the effective capacitance of the circuit? B. What is the total energy

Question

A. What is the effective capacitance of the circuit?

B. What is the total energy stored by the 38 F capacitor on the right hand side of the circuit?

Explanation / Answer

in the given figure above let the capacitors are in the order and give them names as c1 =36uf, c2=36uf, c3=34uf, c4=38uf as moving from left to right from the figure we can see that c1 and c2 capacitors are parallel => effective capacitance between c1 and c2 = c12 = c1+c2 =36+36 = 72uf similarly we can see that c3 and c4 are parallel => effective capacitance between c3 and c4 = c34 =c3+c4 = 34+38 = 72uf we can see that the whole circuit is reduced to a simple circuit which consists of 98v voltage and two capacitors c12 and c34 as c12 and c34 are seen in series effective capacitance = c12*c34/c12+c34 = 36uf effective capacitance of above circuit = 36uf b)in the effective capacitance calculation stage where the circuit consists of effective capacitors c12 and c34 capacitors w we can easily divide the potential difference across each capacitor to be 49v as both the capacitors are of same value while going backwards i.e. dividing the c34 circuit into a parallel combination of 2 capacitors there potential difference across them doesnt change as they are divided parallel => 72uf divided into 34uf and 38uf parallelly voltage across 38uf = 49v energy stored in 38uf capacitor = .5cv^2 = .5*38*49*49 = 40817 J = 40.817 kJ

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