Question

In a series LCR circuit with an AC source, R = 300 Ω, C = 20 μF, L = 1.0 henry, ε_rms = 50 V and ν = 50/π Hz. Find (a) the rms current in the circuit and (b) the rms potential difference across the capacitor, the resistor and the inductor. Note that the sum of the rms potential differences across the three elements is greater than the rms voltage of the source.

Answer 1

Given:
Resistance in series LCR circuit, R = 300 Ω
Capacitance in series LCR circuit, C = 20 μF= 20 × 10⁻⁶ F
Inductance in series LCR circuit, L = 1 Henry
RMS value of voltage, ε_rms   = 50 V
Frequency of source, f = 50/ ` pi`Hz
Reactance of the inductor (X_L) is given by,
`X_C = omegaL = 2pifL`
`rArr X_C = 500 Ω`
(a) Impedance of  an LCR circuit (Z) is given by,
`Z = sqrt(R^2 + (X_C - X_L )^2 `
`rArr Z = sqrt((300)^2 + (500 - 100)^2`
`rArr Z = sqrt((300)^2 + (400)^2`
⇒  Z = 500
RMS value of current I_rms is given
`I_{rms} = epsilon_{rms}/Z`
`rArr I_{rms} = 50/500`
`rArr I_{rms} = 0.1A`
(b) Potential across the capacitor `(V_C)` is given by,
`V_C = I_{rms} xx X_C`
`rArr V_C = 0.1xx500 = 50V`
Potential difference across the resistor `(V_R)` is given by
`V_R =I_{rms}xxR`
`V_R = 0.1xx300xx = 30V`
Potential difference across the inductor `(V_L)` is given by,
`V_L = I_{rms]xxXL`
` rArr V_L =0.1xx100 = 10V `
R.M.s potential = 50V
Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
Sum of the potential drops > RMS potential applied