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

Answer:

-8.56V

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.

$VL = - 8.56 v$