Determine the capacitive reactance for a 20 uF capacitor that is across a 20 volt, 60 Hz source

Physics

1 answer:

aleksandr82 [10.1K]3 years ago

4 0

Answer:

Capacitive reactance is 132.6 Ω.

Explanation:

It is given that,

Capacitance, $C=20\ \mu F=20\times 10^{-6}\ F=2\times 10^{-5}\ F$

Voltage source, V = 20 volt

Frequency of source, f = 60 Hz

We need to find the capacitive reactance. It is defined as the reactance for a capacitor. It is given by :

$X_C=\dfrac{1}{2\pi fC}$

$X_C=\dfrac{1}{2\pi \times 60\ Hz\times 2\times 10^{-5}\ F}$

$X_C=132.6\ \Omega$

So, the capacitive reactance of the capacitor is 132.6 Ω. Hence, this is the required solution.

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A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$