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:

V$_{L}$  =- 8.78v

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

Answer 2

Answer: -8.81 V

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