Quad Capacitor Circuit

Electricity and Magnetism Level 2

An AC power supply with a voltage ( $V_s$ ) of $15\text{ V}$ and a frequency ( $f$ ) of $30\text{ Hz}$ was connected to a circuit of four capacitors ( $C$ ).

In this circuit, $C_\text{I} = \dfrac{10}{\pi}\text{ F}$ , $C_\text{II} = \dfrac{4}{\pi}\text{ F}$ , $C_\text{III} = \dfrac{6}{\pi}F$ , and $C_\text{IV} = \dfrac{20}{\pi}\text{ F}$ .

Determine the current ( $I$ ) of this circuit in $\text{amps}$ .

Note : $X_c = \dfrac{1}{2 \pi \text{ f }C_\text{circuit}}$ and $I = \dfrac{V_s}{X_c}$ .

David's Electricity Set

The answer is 3600.

1 solution

David Hontz
Jun 5, 2016

$C_{II,III} = C_{II}+C_{III} = \frac{4}{\pi}+\frac{6}{\pi} = \frac{10}{\pi}F \\ C_{circuit} = \Big( \frac{1}{C_{I}} + \frac{1}{C_{II,III}} + \frac{1}{C_{IV}} \Big)^{-1} = \Big( \frac{\pi}{10} + \frac{\pi}{10} + \frac{\pi}{20} \Big)^{-1} = \Big( \frac{\pi}{4} \Big) ^{-1} = \frac{4}{\pi}F$ $X_c = \frac{1}{2 \pi (30Hz) (\frac{4}{\pi}F)} = \frac{1}{240}Ω$ $I = \frac{V_s}{X_c} = \frac{15V}{\frac{1}{240}Ω} = \boxed{3600 A}$

Quad Capacitor Circuit

David's Electricity Set

The answer is 3600.

1 solution

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