Inductive & Capacitive Circuits. Subhasish Chandra Assistant Professor Department of Physics Institute of Forensic Science, Nagpur

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1 Inductive & Capacitive Circuits Subhasish Chandra Assistant Professor Department of Physics Institute of Forensic Science, Nagpur

2 LR Circuit LR Circuit (Charging) Let us consider a circuit having an inductance L and a resistance R connected in a series with a battery of e.m.f E. The current increases gradually from zero to a maximum value of I 0. The time in which the current is increasing, a back e.m.f is introduced in the inductance. Let I be the value of the current at any instant. The rate at which the current increases is di/dt. The potential drop across the resistance is RI Back e.m.f introduced in the inductance is L di dt Back e.m.f. opposes the e.m.f. E due to the battery. The net e.m.f is E L di dt Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 2

3 LR Circuit Applying Kirchhoff s Voltage law, we get E L di dt = RI E L di dt = RI E R I = L di R dt di E R I = R L dt Integrating both sides, we get ln E R I = R L t + K where, K is the constant of integration When t = 0, I = 0 and we get K = ln E R dx x = ln x Hence, we get ln E R I = ln E R R L t ln E R I = ln E R R L t E ln R I E R = R L t Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 3 E R I E R = e R L t E R I = E R e R L t I = E R R 1 e L t I = I 0 1 e R L t I0 = E/R is the Final Steady Value of Current

4 LR Circuit LR Circuit (I vs t) (Charging) We get At t = 0, I = 0 The current increases exponentially with time The fraction L/R is called the time constant of the circuit. If t = L/R = τ, then I = I 0 1 e R L t = I 0 1 e I = I 0 ( 1 e 1 ) I = I 0 R L L R Time constant of LR circuit is defined as the time during which the current rises by 63.21% of the maximum steady value. As t, the current tends asymptotically to the steady state value I 0 = E/R Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 4

LR Circuit LR Circuit (Discharging) After the steady current has been established, the steady e.m.f is removed. The current begins to decay exponentially and during this time a back e.m.f -L di/dt is introduced in the inductance.

5 LR Circuit LR Circuit (Discharging) After the steady current has been established, the steady e.m.f is removed. The current begins to decay exponentially and during this time a back e.m.f -L di/dt is introduced in the inductance. Applying Kirchhoff s Voltage law, we get L di dt = RI di I = R L dt Integrating both sides, we get ln I = R L t + K where, K is the constant of integration When t = 0, I = I 0 and we get K = ln I 0 Hence, we get ln ln I = ln I 0 R L t Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 5 I I 0 = R L t

6 LR Circuit I = e R I 0 L t I = I 0 e R L t For t = L/R = τ, R L I = I 0 e t = I 0 e R L L R I = I 0 ( e 1 ) I = I 0 Time constant of LR circuit can also be defined as the time during which the current decreases by 36.79% of the maximum steady value. LR Circuit (I vs t)(discharging) The current decays exponentially from its maximum value I0 to zero. Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 6

7 LR Circuit The reactance for inductance is X L = ωl = 2πfL The current is given as, I = E Z = E 0 Z sin(ωt φ) LR Circuit (AC) Let us consider a circuit having an inductance L and a resistance R connected in a series with a power supply giving alternating e.m.f E. Let the angular frequency of the alternating e.m.f. be ω Let the maximum magnitude of e.m.f be E 0 The alternating e.m.f be E 0 sin ωt where, Z is the impedance of the circuit and is given by Z = R 2 + ω 2 L 2 φ is the phase difference between the current and voltage and is given by φ = tan 1 X L R = tan 1 ω L R The current lags with respect to the e.m.f. Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 7 I = E 0 sin(ωt φ) R 2 + ω 2 2 L

8 Circuit Circuit (Charging) Let us consider a circuit having a capacitance C and a resistance R connected in a series with a battery of e.m.f E. The charge increases gradually from zero to a maximum value of q0. Let q be the value of the charge at any instant. The rate at which the current increases is dq/dt. The potential drop across the resistance is RI = R dq/dt The potential drop across the capacitance is q/c The capacitor plate charges to a constant maximum of q 0, when the potential difference across the capacitor is E. Applying Kirchhoff s Voltage law, we get R dq dt + q C = E R dq dt + q C = q 0 C E = q 0 C Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 8

9 dq q 0 q = 1 dt Integrating both sides, we get ln( q 0 q) = t + K where, K is the constant of integration When t = 0, q = 0 and we get K = ln( q 0 ) Hence, we get ln( q 0 q) = t ln ( q 0 ) ln q 0 q = t q 0 q q 0 q 0 = e t Circuit q = q 0 1 e t Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 9 Circuit (I vs t) (Charging)

10 Circuit The current is given as, I = dq dt = d q 0 1 e t dt I = q 0 e t The steady current is given as I 0 = E R = q 0 I = I 0 e t Hence, we observe that, The charge in a circuit exponentially increases with time to a steady maximum value q 0 The current decreases from a steady maximum value I 0 to zero exponentially. The fraction is called the capacitative time constant of the circuit. If t = = τ, then q = q 0 1 e t = q 0 1 e q = q 0 ( 1 e 1 ) q = q 0 Time constant of circuit is defined as the time during which the charge rises to 63.21% of the maximum steady value. Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 10

11 Circuit Circuit (Discharging) After the steady current has been established, the steady e.m.f is removed. The charge in the capacitor begins to decay exponentially through the resistance and a current flows through the resistance. Applying Kirchhoff s Voltage law, we get R dq dt + q C = 0 dq q = dt Integrating both sides, we get lnq = 1 t + K where, K is the constant of integration When t = 0, q = q 0 and we get K = lnq 0 Hence, we get lnq = lnq 0 1 t ln q q 0 = 1 t q = q 0 e t Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 11

12 The current is given as, I = dq dt = d q 0 e t dt I = q t 0 = I 0 e t e Hence, we observe that, Circuit The charge in a circuit exponentially discharges with time from a steady maximum value q 0 to zero The current changes from a steady maximum value -I 0 to zero exponentially. For t = = τ, q = q 0 e t = q 0 e = q 0 e 1 = q 0 Time constant of circuit can also be defined as the time during which the charge falls to 36.79% of the maximum steady value. Circuit (I vs t) (Discharging) Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 12

13 Circuit The reactance for capacitance is X C = 1/ωC = 1/2πfC The current is given as, I = E Z = E 0 Z sin(ωt +φ) Circuit (AC) Let us consider a circuit having an capacitance C and a resistance R connected in a series with a power supply giving alternating e.m.f E. Let the angular frequency of the alternating e.m.f. be ω Let the maximum magnitude of e.m.f be E 0 The alternating e.m.f be E 0 sin ωt I = sin(ωt +φ) 1 where, R 2 + ω 2 C 2 Z is the impedance of the circuit and is given by Z = R 2 + ( 1/ω 2 C 2 ) φ is the phase difference between the current and voltage and is given by φ = tan 1 X C R = tan 1 1 ω The current leads with respect to the e.m.f. Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 13 E 0

14 LCR Circuit (Series - AC) LCR Circuit Let us consider a circuit having an capacitance C, inductance L and a resistance R connected in a series with a power supply giving alternating e.m.f E. Let the angular frequency of the alternating e.m.f. be ω Let the maximum magnitude of e.m.f be E 0 The alternating e.m.f be E 0 sin ωt The reactance for capacitance is XC = 1/ωC = 1/2πfC The reactance for inductance is XL = ωl = 2πfL The current is given as, where, I = E Z = E 0 Z sin(ωt φ) Z is the impedance of the circuit and is given by Z = R 2 + ω L 1 2 ωc Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 14 I = E 0 R 2 + ω L 1 2 ωc sin(ωt φ)

15 LCR Circuit LCR Circuit (Series - AC) φ is the phase difference between the current and voltage and is given by X φ = tan 1 L X ω L 1 C R = tan 1 ωc R The phase difference depends on the relative values of X L and X C. Hence, we get three scenarios, X L > X C i.e. ω L > 1 ωc The phase angle φ will be positive and the current lags behind the applied e.m.f. The potential difference across the inductance is greater than that of the capacitor and the circuit will behave as an inductive circuit. X L < X C i.e. ω L < 1 ωc The phase angle φ will be negative and the current leads the applied e.m.f. The potential difference across the inductance is lesser than that of the capacitor and the circuit will behave as an capacitive circuit. Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 15

16 LCR Circuit X L = X C i.e. ω L = 1 ωc The phase angle φ will be zero and the current is in phase with the applied e.m.f. The potential difference across the inductance and capacitance is equal in magnitude but opposite in phase. Hence, the entire potential drops across the resistance and it behaves as a purely resistive circuit. In this case the current is maximum i.e. The resonant frequency f0 is given as, X L = X C ω 0 L = 1 ω 0 C ω 0 = f 0 = 2π 1 LC 1 LC I 0 = E 0 R The frequency at which this occurs is called as the resonant frequency. The circuit is said to be a series resonant circuit. X L < X C Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 16 X L > X C

17 LCR Circuit The reactance for capacitance is X C = 1/ωC = 1/2πfC The reactance for inductance is X L = ωl = 2πfL The current is given as, Z is the impedance of the circuit and is LCR Circuit (Parallel - AC) given by 1 Z = jωc + 1 R + jω L φ is the phase difference between the current and voltage and is given by Let us consider a circuit having an capacitance C connected in parallel with an inductance L and a resistance R. A power supply giving alternating e.m.f E is connected to this combination. Let the angular frequency of the alternating e.m.f. be ω Let the maximum magnitude of e.m.f be E 0 The alternating e.m.f be E 0 sin ωt I = E Z = E 0 sin(ωt φ) Z where, φ = tan 1 X I L 1 I Z 2 LR R I 1 Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 17 Z LR ω L I 1 = tan 1 R 2 + ω 2 L I 2 2 R I 1 R 2 + ω 2 L 2

18 LCR Circuit For resonance, the current through the capacitance is equal to quadrature component of the current through the series combination of inductance and resistance. At this condition supply current I is in-phase with the supply voltage E. This condition can be represented as, 2 Z LR E = E i X L X C Z LR Z LR = X C X L R 2 + ω 0 2 L 2 = 1 ω 0 C iω 0 L The resonant frequency f 0 is given as, f 0 = 1 2π The current is minimum and impedance is maximum at resonance. Such a circuit rejects the currents at resonance and are known as rejector circuits, filter circuits or anti-resonant circuits. X L < X C 1 LC R2 L 2 X L > X C R 2 + ω 0 2 L 2 = L C R 2 + 2π f 0 2 L 2 = L C Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne 18

Chapter 33. Alternating Current Circuits

Chapter 33. Alternating Current Circuits Chapter 33 Alternating Current Circuits 1 Capacitor Resistor + Q = C V = I R R I + + Inductance d I Vab = L dt AC power source The AC power source provides an alternative voltage, Notation - Lower case