In the circuit given below, the equivalent capacitance is ______________(a) \(\frac{C}{4}\)(b) \(\frac{5C}{13}\)(c) \(\frac{5C}{2}\)(d) 3CThis question was addressed to me in unit test.This is a very interesting question from Series-Series Connection of Two Port Network in chapter Two-Port Networks of Network Theory

In the circuit given below, the equivalent capacitance is ____________

1.	In the circuit given below, the equivalent capacitance is ______________(a) \(\frac{C}{4}\)(b) \(\frac{5C}{13}\)(c) \(\frac{5C}{2}\)(d) 3CThis question was addressed to me in unit test.This is a very interesting question from Series-Series Connection of Two Port Network in chapter Two-Port Networks of Network Theory
Answer» Correct choice is (b) \(\FRAC{5C}{13}\) Easy explanation: The EQUIVALENT capacitance by applying the concept of series-parallel combination of the capacitance is, \(\frac{1}{C_{EQ}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{5C/3}\) = \(\frac{1}{C} + \frac{1}{C} + \frac{3}{5C}= \frac{1}{C}(\frac{5+5+3}{5})\) Or, CEQ = \(\frac{5C}{13}\) 123 c Energy delivered during talk time E = ∫ V(t)I(t) dt Given, I (t) = 2 A = constant = 2 ∫ V(t)dt = 2 × Shaded area = 2 × \(\frac{1}{2}\) × (10 + 12) × 60 × 10 = 13.2 kJ.

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