A resistor of resistance R, an inductor inductance L and a capacitor of capacitance C all are connected in series with an a.c. supply. The resistance of R is 16 ohm and for a given frequency, the inductive reactance of L is 24 ohm and capacitive reactance of C is 12 ohm. If the current in the circuit is 5 amp., find
(a) the potential difference across R, L and C
(b) the impedance of the circuit
(c) the voltage of a.c. supply
(d) phase angle

Solution

Given, an LCR circuit where all the components are connected in series with an a.c. supply.
Resistance, R = 16 $Ω$
Inductive reactance, $χ_{L}$ = 24 ohm
Capacitive reactance, $χ_{C} = 12 Ω$
Current flowing in the circuit, I = 5 A

(a) Potential difference across resistance,
V_R = iR
= 5 × 16
= 80 volt

Potential difference across inductance,
V_L = i × (ωL)
= 5 × 24
= 120 volt

Potential difference across condenser,
$V_{C} = i \times (\frac{1}{ωC}) = 5 \times 12 = 60 volt$

(b) The impedance of the circuit is given as

$Z = \sqrt{[R^{2} + {(ωL - \frac{1}{ωC})}^{2}]} = \sqrt{[(16)^{2} + (24 - 12)^{2}]} = 20 ohm$

(c) The voltage of a.c. supply is given by

V = iz
= 5 × 20
= 100 volt

(d) Phase angle

$ϕ = \tan^{- 1} [\frac{ωL - (\frac{1}{ωC})}{R}]$
$= \tan^{- 1} [\frac{24 - 12}{16}] = \tan^{- 1} (0.75) = 36 ° 46' .$

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