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3 years ago

Calculate the equivalent capacitance of the three series capacitors in Figure 12-1

Engineering

1 answer:

GrogVix [38]3 years ago

5 0

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Calculate the equivalent capacitance of the three series capacitors in Figure 12-1

a) 0.01 μF

b) 0.58 μF

c) 0.060 μF

d) 0.8 μF

Answer:

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

Explanation:

Please refer to the attached Figure 12-1 where three capacitors are connected in series.

We are asked to find out the equivalent capacitance of this circuit.

Recall that the equivalent capacitance in series is given by

$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

Where C₁, C₂, and C₃ are the individual capacitance connected in series.

C₁ = 0.1 μF

C₂ = 0.22 μF

C₃ = 0.47 μF

So the equivalent capacitance is

$\frac{1}{C_{eq}} = \frac{1}{0.1} + \frac{1}{0.22} + \frac{1}{0.47}$

$\frac{1}{C_{eq}} = \frac{8620}{517}$

$C_{eq} = \frac{517}{8620}$

$C_{eq} = 0.0599$

Rounding off yields

$C_{eq} = 0.060 \: \mu F$

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

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3 years ago

Please answer fast. With full step by step solution.

lina2011 [118]

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8z - 2 = a (z - 1) + b (2z - 1)

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