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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 :

cjmejiab

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,

[tex]e(t) = e* sin(wt)[/tex]

[tex]e(t) = 6.04 * sin(Φ + π)[/tex]

[tex]e(t) = 6.04 * sin(32.5 + 180)[/tex]

[tex]e(t) = -3.245 V[/tex]

Now, Using Kirchoff Voltage Law,

[tex]e(t) - VR- VL - VC = 0[/tex]

[tex]-3.24 - 0 - VL - 5.32 = 0[/tex]

Finally we have the potential difference across the inductor.

[tex]VL = - 8.56 v[/tex]

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