Question

b. Determine the equivalent resistance of the circuit 42) 2. Given the arrangement of capacitors given to the probably not answer these questions in the exact a. What is the equivalent capacitance of this right, answer the following questions. You will 24 V order they are asked Cs 2.0 uF arrangement? b. Cs What is voltage across each capacitor? e on each capacitor? 1.2 uF Ca | 3.5 μF age als to 10% of its original value?

#2

Answer 1

In below circuit, there are two C₃ , so first I took C₃= 2.8 uF then below one would be C₆ = 2 uF. Please consider it accordingly. In case of any doubt, please comment.

In below circuit, there are two c_3, so first I took C_3 = 2.8 mu F then below one would be c_6 = 2 mu f. please consider it accordingly. In case of any doubt, please comment c_23 = c_2 + c_3 = 2.2 + 2.3 c_23 = 5 mu f c_546 = c_5 + c_3 times c_4/c_3 + c_4 = 1.2 + 2 times 3.5/2 + 3.5 = 2.473 mu f (a) net capacitance c = c_1 + c_23 c_54 = 11 times 5 times 2.473/11 times 5 + 5 times 2.473 + 2.473 times 11 c = 1.4383 mu f b) charge through circuit, Q = cr = 1.4383 times 24 = 34.52 mu c charge through c_1, Q_1 = Q = 34.52 mu c voltage across c_1, v_1 = Q_1/c_1 = 34.52/11 = 23.138 V charge through c_2, Q_2, Q(C-2/c_2 + C_30 = 34.52(2.2/2.2 + 2.8) = 15.188 mu c voltage across c_2, v_2 = Q_2/C_2 = 15.188/2.2 = 6.904 V

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