Question

An alternating source drives a series RLC circuit with an emf amplitude of 6.34 V, at a phase angle of +33.3°. When the potential difference across the capacitor reaches its maximum positive value of +5.33 V, what is the potential difference across the inductor (sign included)?

Answer 1

Answer: -8.81 V

Step-by-step explanation:

Depending on the arrangement in a RLC circuit, we can have

From the phasor diagram

E(m)² = V(R)² + [V(L) - V(C)]², making V(L) subject of formula, we have,

V(L) = V(C) + √[E(m)² - V(R)²]

And V(R) = E(m) cosΦ, so that

V(L) = V(C) + √[E(m)² - E(m)² cos²Φ]

V(L) = V(C) + E(m)√[1 - cos²Φ]

V(L) = V(C) + E(m)sinΦ

V(L) = 5.33 + 6.34sin 33.3°

V(L) = 5.33 + 3.48

V(L) = 8.81 V

both V(L) and V(C) are 180° out of phase, so, V(C) = -8.81 V

Answer 2

V
`_(L)` =- 8.78v

Step-by-step explanation:

`E^(2) _(m)`=
`V^(2) _(R)` +
`(V_(L) -V _(C) )^(2)`

OR V
`_(L)` = V
`_(c)` +
`\sqrt{E^(2){m} - V^(2) {R} }`

where VR = E
`{m}`cosФ

V
`_(L)` = V
`_(c)` +
`\sqrt{E^(2)_(m) - E^(2)_(m) cos^(2) }`Ф

V
`_(L)` = V
`_(c)` +
`{E_(m) \sqrt{1 - cos^(2) } }`Ф

V
`_(L)` = V
`_(c)` + E
`_(m)`sinФ

substituting the given values

V
`_(L)` = 5.33 + 6.34 x sin33°

V
`_(L)` = -8.78v