Question

consider the combination of two capacitor c1 and c2 with c2>c1,when connected in parallel,the equivalent capacitance is 8 times the equivalent capacitance of the same connected in series. calculate the ratio of capacitor c2/c1?​

Answer 1

Step-by-step explanation:

Current flowing through the circuit, I = 2A Applying Kirchoff’s Ist law at junction P, I = I1 + I2 I2 = I – I1 …(1) Applying Kirchoff’s Ind law at junction PQSP 5 I1 + 10 Ig – 15 I2 = 0 5 I1 + 10 Ig -15(I – I1) = 0 20 I1 + 10 Ig = 15 I 20 I1 + 10 Ig = 15 x 2 ÷ by 10 21 I1 + Ig = 3 … (2) Applying Kirchoff’s II law at junction QRSQ 10(I1 – Ig ) – 20(I – I1 – Ig ) – 10 Ig = 0 10 I1 – 10g – 20(I – I1 – Ig ) – 10 Ig = 0 10 I1 – 10 – 20 I + 20 I1 -20Ig – 10 Ig = 0 30 I1 – 40 Ig = 20 I ÷ by 10 ⇒ 3 I1 – 4 Ig = 20 I 3 I1 – 4 Ig = 2 I 3 I1 – 4 Ig = 2 x 2

hope this helps :)