Question

Four capacitors with capacitance 3.0 pF, 2.0 pF, 5.0 pF and X pF are connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, what is the value of the X capacitor? (A). 6.0 pF (B). 5.0 pF (C). 3.0 pF (D). 2.0 pF (E). 0.83 pF

Answer 1

A. 6pF

Step-by-step explanation:

If unknown capacitance C1, C2, C3 and C4 are connected in series to one another, their equivalent capacitance of the circuit will be expressed as shown

`(1)/(C_t) = (1)/(C_1) +(1)/(C_2) +(1)/(C_3) +(1)/(C_4) \\`

Given the capacitance's 3.0 pF, 2.0 pF, 5.0 pF and X pF connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, then to get X, we will apply the formula above;

`(1)/(0.83) = (1)/(3.0) +(1)/(2.0) +(1)/(5.0) +(1)/(C_4) \\\\\\1.205 = 0.333+0.5+0.2+(1)/(C_4) \\\\1.205 = 1.033 + (1)/(C_4) \\\\(1)/(C_4) = 1.205-1.033\\\\(1)/(C_4) = 0.172\\\\C_4 = (1)/(0.172)\\ \\C_4 = 5.8pF\\\\`

C₄ ≈ 6pF

Hence the value of the X capacitor is approximately 6pF