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

Answer:

A. 6pF

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

$\frac{1}{C_t} = \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3} +\frac{1}{C_4}$

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;

$\frac{1}{0.83} = \frac{1}{3.0} +\frac{1}{2.0} +\frac{1}{5.0} +\frac{1}{C_4} \\\\\\1.205 = 0.333+0.5+0.2+\frac{1}{C_4} \\\\1.205 = 1.033 + \frac{1}{C_4} \\\\\frac{1}{C_4} = 1.205-1.033\\\\\frac{1}{C_4} = 0.172\\\\C_4 = \frac{1}{0.172}\\ \\C_4 = 5.8pF$

C₄ ≈ 6pF

Hence the value of the X capacitor is approximately 6pF