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Screwdriver

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## Homework Statement

The picture of the circuit is attached; I want to find [itex]|V_{A}/V_{J}|[/itex]. This seems really easy but I haven't done circuit analysis in forever.

## Homework Equations

Complex impedances, [itex]Z_{C} = 1/i\omega C, Z_{R} = R[/itex].

## The Attempt at a Solution

First [itex]R_{A}[/itex] and [itex]C_{A}[/itex] are in parallel, so the equivalent impedance is [itex]Z_{A} = Z_{C_{A}} Z_{R_{A}} /(Z_{C_{A}} + Z_{R_{A}})[/itex]. Then the circuit is a voltage divider, so the voltage across [itex]Z_{A}[/itex] is [itex]V_{A} = V_{J} Z_{A}/(Z_{A} + Z_{R_{J}})[/itex]. Therefore,

$$

\frac{V_{A}}{V_{J}} = \frac{Z_{A}}{Z_{A} + Z_{R_{J}}} = \frac{i R_{A}}{i(R_{A} + R_{J}) - C_{A} R_{A} R_{J} \omega} \implies \left| \frac{V_{A}}{V_{J}} \right| = \frac{R_{A}}{\sqrt{(R_{A} + R_{J})^2 + (C_{A} R_{A} R_{J} \omega)^2}}

$$

#### Attachments

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