Question

An alternating source drives a series RLC circuit with an emf amplitude of 6.04 V, at a phase angle of +30.3°. When the potential difference across the capacitor reaches its maximum positive value of +5.32 V, what is the potential difference across the inductor (sign included)?

Answer 1

-8.56V

Step-by-step explanation:

Our values are given by,

e = 6.04 V

Φ = 30.3

VC = 5.32

We can calculate the voltage across the circuit with the emf formula, that is,

`e(t) = e* sin(wt)`

`e(t) = 6.04 * sin(Φ + π)`

`e(t) = 6.04 * sin(32.5 + 180)`

`e(t) = -3.245 V`

Now, Using Kirchoff Voltage Law,

`e(t) - VR- VL - VC = 0`

`-3.24 - 0 - VL - 5.32 = 0`

Finally we have the potential difference across the inductor.

`VL = - 8.56 v`