Alternating
4
Current
RLC Circuits
Jayant Nagda
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B.Tech, IIT Bombay
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AC Circuits In general,
If ε = ε0 Sinωt ; I = I0 Sin(ωt + Φ)
Where I =
Z: impedance of circuit, Unit is Ω
Type of Circuit Phase Impedance
Purely Resistive Φ = 00 Z=R
Purely Capacitive Φ = π/2 Z = XC = 1/(ωC)
Purely Inductive Φ = - π/2 Z = XL= ωL
Solving AC Circuit For Series Circuits:
● Treat inductor & capacitor as resistance equal to
XL= ⍵L and XC= 1/⍵C
R XL XC
~
● In series circuit, every element has same current
through it hence voltage across circuit elements will be:
VR = iR VL = iXL VC = iXC
Solving AC Circuit For Series Circuits:
V↑
● In series, Add Voltages like vectors.
Plot current along x-axis, VR along x-axis,
VL along +y-axis(as VL leads current by 𝜋/2),
i and VC along -ve y-axis (as VC lags current by 𝜋/2)
Find Vnet-y along y-axis.
● Combining VR & Vnet-y like vectors gives
resultant voltages and its angle with current
gives phase difference b/w voltage and current
R-L Circuit
R L
Here
VR = iR VL = iXL
i
~ V↑
E = E0sinωt
i
R-L Circuit R L V↑ V
VR VL VL
i
Φ
~ VR
i
E = E0sinωt
Here
VR = iR VL = iXL
R-L Circuit R L V↑ V
VR VL VL
i
Φ
~ VR
i
E = E0sinωt
Voltage leads current by Φ
i = i0sin(ωt - Φ)
Where i0 = E0/Z
In an A.C. circuit, a resistance of R ohm is connected
in series with an inductance L. If phase angle between voltage
and current be 450, the value of inductive reactance will be
A. R/4 B. R C. R/2 D. cannot be found with
given data
When 100 V DC is supplied across a solenoid, a current of
1 amperes flows in it. When 200 V AC is supplied across the
same coil, the current drops to 0.5 ampere. If the frequency of
ac source is 50 Hz, then the impedance and inductance of the
solenoid are -
A. 200 Ω and 0.55 henry
B. 100 Ω and 0.86 henry
C. 200 Ω and 1.0 henry
D. 100 Ω and 0.93 henry
In a series L–R circuit (L = 35 mH and R = 11 Ω), a variable emf
source (V = V0 sin ωt) of Vrms = 220 V and frequency 50Hz is applied.
Find the current amplitude in the circuit and phase of current
with respect to voltage. Draw current-time graph on given graph
(π = 22/7). Current in the circuit has Amplitude:
[IIT- 2004]
V = V0sinωt A. 20 A, leads V by π/4
T/2 3T/2 B. 10 A, leads V by π/4
O
T/4
C. 20 A, trails V by π/4
D. 10 A, trails V by π/4
R-C Circuit Here
R C VR = iR VC = iXC
VR VC V
i
~
E = V0sinωt i
R-C Circuit Here
VR = iR VC = iXC
R C
VR
VR VC i
i Φ
~
E = V0sinωt VC
Current leads voltage by Φ
i = i0sin(ωt + Φ)
Where i0 = V0 / Z
In given circuit, when a dielectric slab is introduced
between the plates of the capacitor , intensity of bulb
A. increases Bulb
B. decreases
C. may increase or decrease
~
D. can't say V = V0 sin ωt
Integer Type Question
A series R-C combination is connected to an AC voltage
of angular frequency ω = 500 radian/s. If the impedance
of the R-C circuit is R√1.25 , the time constant (in millisecond)
of the circuit is [JEE 2011]
An AC source of angular frequency ω is fed across a resistor R
and a capacitor C in series. The current registered is I. If now
the frequency of source is changed to ω/3 (but maintaining
the same voltage), the current in the circuit is found to be
halved. Calculate the ratio of reactance to resistance at the
original frequency ω -
A. √(3/5)
B. √(2/5)
C. √(1/5)
D. √(4/5)
Ans : A
L-C Circuit Here
VL = iXL VC = iXC
VL
↑V
O i
~
V = V0sinωt VC
L-C Circuit Here
VL = iXL VC = iXC
VL
↑V
O i
~
V = V0sinωt VC
V = VL - VC = i(XL- XC)
V/i = Z = XL - XC , Φ = π/2
L-C Circuit (a) If XL > XC , voltage leads the current by π/2
↑V
VL- VC
O i
~
V = V0sinωt
(b) If XC > XL , current leads the voltage by π/2
↑V
O i
VC- VL
L-C Circuit
(c) If XL = XC
↑V
VL
~ O i
V = V0sinωt
VC ⇒ω=
Condition of Resonance
Z=0
&
io = ∞
L-C Circuit
variation of impedance Z with freq
~
V = V0sinωt ω
L-C Circuit
variation of impedance Z with freq
~
V = V0sinωt ω
ω = ω0
ω = ω0 ω
L-C Circuit
If XL = XC , Φ = 0 voltage & current are in same phase
XL = XC
~ ωL = 1/ωC
V = V0sinωt
resonant angular frequency
resonant frequency
L, C and R represent the physical quantities, inductance,
capacitance and resistance respectively. The combination(s)
which have the dimensions of frequency are
[IIT 1984]
A. 1/RC B. R/L C. 1/√LC D. C/L
Ans : A,B,C
R-L-C Circuit
VR = i R ; VL = i XL ; VC = i XC
R L C
↑V
VR VL VC
VL
~
V = V0sinωt V i
R
V
C
R-L-C Circuit
V
R L C
VL-VC
VR VL VC Φ
VR i
~
V = V0sinωt
The value of alternating emf E in the given circuit will be -
A. 100 V
B. 20 V
C. 220 V
D. 140 V
R-L-C Circuit
(a) If XL > XC ; voltage leads current by Φ
V
Hence i = i0sin( ωt - Φ)
VL-VC
Φ
VR i
(b) If XC > XL ; current leads voltage by Φ
VR
Hence i = i0sin( ωt + Φ) Φ i
VC-VL
V
In an LCR series a.c. circuit, the voltage across each of the
components, L, C and R is 50V. The voltage across the LC
combination will be
[AIEEE 2004]
A. 100 V B. 50√2 V C. 50 V D. 0 V (zero)
Ans : D
R-L-C Circuit In series resonance, Z = ?
R L C
VR VL VC
~
V = V0sinωt
i0
i0(max) = V0/R
ω
ω = ω0
R-L-C Circuit In series resonance, impedance of circuit is minimum
and equal to resistance, Z = R and current is maximum.
R L C i0
VR VL VC i0(max) = V0/R
~
V = V0sinωt
ω
ω = ω0
(a) If ω < ωo ; XC > XL i.e. before resonance current leads
(b) If ω > ωo ; XL > XC i.e. after resonance voltage leads
Series Resonance Circuit is called “ Acceptor Circuit ”
Amplitude at Resonance
Amplitude at Resonance
Tacoma Narrows Bridge
In a series resonant LCR circuit, the voltage across R is
100 volts and R = 1 kΩ with C = 2μF. The resonant frequency
ω is 200 rad/s. At resonance the voltage across L is
[AIEEE 2006]
A. 2.5 × 10–2 V B. 40 V C. 250 V D. 4 × 10–3 V
Ans : C
Quality Factor
At resonance, I is maximum. Hence VL or VC is maximum.
Voltage across inductor
or capacitor at resonance
Quality factor =
applied voltage
In a series LCR circuit L = 1H, C = 6.25 μF and R = 1 ohm.
Its quality factor is -
A. 400 B. 200 C. 125 D. 25
Ans : A
Parallel AC Circuits Solution is similar to series circuits.
iC C iC XC
iL L iL XL
iR R
iR
R
~ ~
V,ω
Current in various branches is different
but Potential is same.
take potential along x-axis
& add current like vectors.
Parallel AC Circuits take potential along x-axis
& add current like vectors.
IC leads V by 𝝅/2
i = i0 sin(ωt + π/2)
V IR is in phase with V
IL lags V by 𝝅/2 i = i0 sin(ωt)
i = i0 sin(ωt - π/2)
Parallel AC Circuits take potential along x-axis
& add current like vectors.
iC C
iR = V/R ; iL = V/XL ; iC = V/XC iC
iL L
iR V
R
iR
~ iL
V,ω
Impedance of Parallel circuit iR = V/R ; iL = V/XL ; iC = V/XC
iC - iL i
Φ
V
iR
Solve the circuit for i delivered by the AC source
0.8A L A. 0.2 A, π/2 ahead of voltage
B. 0.2 A, π/2 behind of voltage
C
0.6A
C. 1.4 A, π/2 ahead of voltage
D. 1.4 A, π/2 behind of voltage
~
i=?
Solve the circuit for i delivered by the AC source
Solution: iC = 0.6A
0.8 L
C
0.6
V
~
iL = 0.8A i=?
Current from source is 0.2 A
i = 0.2 A
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