An alternating source drives a series RLCcircuit with emf amplitude of, 6.00Vat a phase angle of+30.0°. When the potential difference across the capacitor reaches its maximum positive value of+5.00V, what is the potential difference across the inductor (sign included)?

Short Answer

Expert verified

The potential difference across the inductor is -8.00V.

Step by step solution

01

The given data

  1. Emf amplitudeεm=6.00V
  2. Phase angle,ϕ=30°
  3. Potential difference across capacitorVC=5.00V
02

Understanding the concept of phasor diagram

Using the phasor diagram and applying the Pythagorean Theorem, we can find the potential difference across the inductor. The theorem applies to the right-angled triangle relating hypotenuse, perpendicular, and base of the triangle.

03

Calculation of the potential difference

We have the following phasor diagram:

From the phasor diagram, we can apply the Pythagorean Theorem considering equation (i) to the diagram as follows:

εm2=VR2+VL-VC2VL=Vc+εm2-VR2=Vc+εm2-εm2cos2ϕ(VR=εmcosϕ,from diagram)=Vc+εmsinϕ

For the given values, we get

VL=5.00V+6.00Vsin30°=8.00V

But VLand VCare 180°out of the phase.

Therefore, the value of the potential difference across the inductor is -8.00V.

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