Q: Q.96. An oscillating circuit consists of a capacitor with capacitance C, a coil of inductance L with negligible resistance, and a switch. With the switch disconnected, the capacitor was charged to a voltage Vm, and then at the moment t = 0 the switch was closed. Find: (a) the current I (t) in the circuit as a function of time; (b) the emf of self-inductance in the coil at the moments when the electric energy of the capacitor is equal to that of the current in the coil.

After the switch was closed, the circuit satisfies−LdtdI=Cqdt2d2q+ω02q=0⇒q=CVmcosω0twhere we have used the fact that when the switch is dosed we must haveV=Cq=Vm,I=dtdq=0 at t=0.Thus (a)I=dtaq=−CVmω0sinω0t=−VmLCsinω0t(b) The electrical energy of the capacitor is 2Cq2αcos2ω0t and of the inductor is 21LI2αsin2ω0tThe two are equal whenω0t=4πAt that instant the emf of the self-inductance is−Ldtdi=Vmcosω0t=Vm/2

Q: Q.97. In an oscillating circuit consisting of a parallel-plate capacitor and an inductance coil with negligible active resistance the oscillations with energy W are sustained. The capacitor plates were slowly drawn apart to increase the oscillation frequency η-fold. What work was done in the process?

In the oscillating circuit, letq=qmcosωtbe the change on the condenser wherechanges η2 fold and so does y i.e. changes from y0 initially to η2y0 finally. Now the P.D.across the condenser is V=Cqmcosωt=ε0Syqmcosωtand hence the electric field between the plates isE=e0SqmcosωtThus, the chaige on the plate being qmcosωt, the force on the plate isF=ε0sqm2cos2ωtSince this force is always positive and the plate is pulled slowly we can use the average forceFˉ=2ε0Sqm2and work done is A−Fˉ(η2y0−y0)−(η2−1)2e0sqE2y02ε0Sqm2y0=2C0qm2=Wthe initial stored energy. Thus.A=(η2−1)W

Q: Q.98. In an oscillating circuit shown in Fig. 4.27 the coil inductance is equal to L = 2.5 mH and the capacitor have capacitances C1= 2.0 μF and C2 = 3.0 μF. The capacitors were charged to a voltage V = 180 V, and then the switch Sw was closed. Find: (a) the natural oscillation frequency; (b) the peak value of the current flowing through the coil.

The equations of the L−C circuit are Ldtd(I1+I)=C1C1V−∫I1dt=C2C2V−∫I2dt Differentiating again L(I1+I2)=−C11I1=−C21I2 , ThenI1=C1−C2C1I,I2=C1+C2C2I, I=I1+I2 so L(C1+C2)I+I=0 or I=I0sin(ω0t+α) where ω02=L(C1+C2)1 (Part a) (Hence T=ω02π=0.7 ms )At t=0,I=0 so α=0 I=I0sinω0t The peak value of the current is 70 and it is related to the voltage V by the first equationor LI=V−∫Idt/(C1+C2) +Lω0I0cosω0t=V−C1+C21∫0tI0sinω0tdt (The P.D. across the inductance is V at t=0 )Hence=V+C1+C21⋅ω0I0(cosω0t−1) I0=(C1+C2)ω0V=VLC1+C2=8.05 A

Q: Q.99. An electric circuit shown in Fig. 4.28 has a negligibly small active resistance. The left-hand capacitor was charged to a voltage V0 and then at the moment t = 0 the switch Sw was closed. Find the time dependence of the voltages in left and right capacitors.

q1+q2=CV0Cq1+LdtdI−Cq2=0Also I=q˙1=−q˙2. ThusLI¨+C2I=0Hence I¨+ω02I=0ω02=LC2,The solution of this equation subject toI=0 at t=0is I=I0sinω0t.Integratingq1=A−ω0I0cosω0tq2=B+ω0I0cosω0tFinally substituting in (1)CA−B−ω0C2I0cosω0t+LI0ω0cosω0t=0ThusA=B=2CV0 and 2CV0+ω0I0=0Soq1=2CV0(1+cosω0t)q2=2CV0(1−cosω0t)

Q: Q.100. An oscillating circuit consists of an inductance coil L and a capacitor with capacitance C. The resistance of the coil and the lead wires is negligible. The coil is placed in a permanent magnetic field so that the total flux passing through all the turns of the coil is equal to φ. At the moment t = 0 the magnetic field was switched off. Assuming the switching off time to be negligible compared to the natural oscillation period of the circuit, find the circuit current as a function of time t.

The flux in the coil isΦ(t)={Φ0t 0 The equation of the current is −LdtdI=C∫0IIdtLCdt2d2I+I=0This mean thator with ω02=LC1I=I0sin(ω0t+α)Putting in (1) −LI0ω0cos(ω0t+α)=−ω0CI0[cos(ω0t+α)−cosα]This implies cosα=0∴I=±I0cosω0t. From Faraday's lawε=−dtdΦ=−LdtdIor integrating from t=−ε to −ε where ε→0Φ=LI0 with + sign in Iso, I=LΦcosω0t

Q: Q.102. A certain oscillating circuit consists of a capacitor with capacitance C, a coil with inductance L and active resistance R, and a switch. When the switch was disconnected, the capacitor was charged; then the switch was closed and oscillations set in. Find the ratio of the voltage across the capacitor to its peak value at the moment immediately after closing the switch.

The equation of the circuit isLdt2d2Q+RdtdQ+CQ=0 This has the solution, Q=Qme−βtsin(ωt+α) where β=2LR,ω=ω02−β2,ω02=LC1 Now, I=dtdQ=0 at t=0 so, Qme−βt(−βsin(ωt+α)+ωcos(ωt+α))=0 at t=0 Thus, ωcosα=βsinα or, α=tan−1βω Now, Vm=CQm and V0=P.D. at t=0 =CQmsinα ∴ VmV0=sinα=ω2+β2ω=ω0ω=1−β2/ω02=1−4L2R2C

Q: Q.103. A circuit with capacitance C and inductance L generates free damped oscillations with current varying with time as I=Ime−βtsinωt . Find the voltage across the capacitor as a function of time, and in particular, at the moment t = 0.

We write −dtdQ=I=Ime−βtsinωt =gmIme−βt+iωt (gm means imaginary part)ThenQ=gmIm−β+iωe−βt+iωt Q=gmImβ−iωe−βt+iωt =gmImβ2+ω2(β+iω)e−βt+iωt =Ime−βtβ2+ω2βsinωt+ωcosωt =Ime−βtβ2+ω2sin(ωt+δ),tanδ=βω (An arbitrary constant of integration has been put equal to zero.) ThusV=CQ=ImCLe−βtsin(ωt+δ) V(0)=ImCLsinδ=ImCLω2+β2ω =ImC(1+β2/ω2)L

Q: Q.104. An oscillating circuit consists of a capacitor with capacitance C = 4.0 μF and a coil with inductance L = 2.0 mH and active resistance R = 10Ω. Find the ratio of the energy of the coil's magnetic field to that of the capacitor's electric field at the moment when the current has the maximum value.

I=Ime−βtsinωt β=2LR,ω0=LC1,ω=ω02−β2 I=−q˙,q= charge on the capacitorThen, q=Ime−βtω2+β2sin(ωt+δ),tanδ=βω Thus, WM=21LIm2e−2βtsin2ωt WE=2CIm2ω2+β2e−2βtsin2(ωt+δ)=2LIm2e−2βtsin2(ωt+δ) Current is maximum when dtde−βtsinωt=0 Thus, −βsinωt+ωcosωt=0 or, tanωt=βω=tanδ i.e., and henceWEWM=sin2(ωt+δ)sin2(ωt)=sin22δsin2δ=4cos2δ1 =4β2/ω021=4β2ω02=LC1×R2L2=CR2L=5 (WM is the magnetic energy of the inductance coil and WE is the electric energy of t capacitor.)

Question 1

Q.66. A coil of inductance L connects the upper ends of two vertical copper bars separated by a distance l. A horizontal conducting connector of mass m starts falling with zero initial velocity along the bars without losing contact with them. The whole system is located in a uniform magnetic field with induction B perpendicular to the plane of the bars. Find the law of motion x (t) of the connector.

Accepted Answer

As the connector moves, an emf is set up in the circuit, and a current flows, since the emf isξ=−Blx˙ , we must have :-Blx˙+LdtdI​=0 so, I=Blx/L provided x is measured from the initial position.We then havemx¨=−LBlx​,B⋅l+mg for by Lenz’s law, the induced current will oppose downward sliding. Finallyx¨+mL(Bl)2​x=g on puttingω0​=mL​Bl​ x¨+ω02​x=g A solution of this equation is x=ω02​g​+Acos(ω0​t+α) But x=0 and x˙=0 at t=0 . This givesx=ω02​g​(1−cosω0​t)

Question 2

Q.95. An oscillating circuit consisting of a capacitor with capacitance C and a coil of inductance L maintains free undamped oscillations with voltage amplitude across the capacitor equal to Vm. For an arbitrary moment of time find the relation between the current I in the circuit and the voltage V across the capacitor. Solve this problem using Ohm's law and then the energy conservation law.

Accepted Answer

Since there are no sources of emf in the circuit, Ohm's law readscq​=−LdtdI​where q= change on the capacitor, I=dtdq​= current through the coil. Thendt2d2q​+ω02​q=0,ω02​=LC1​The solution fo this equation isq=qm​cos(ω0​t+α)From the problem Vm​=Cqm​​ ThenandI=−ω0​CVm​sin(ω0​t+α)V=Vm​cos(ω0​t−α)V2+ω02​C2I2​=Vm2​or, V2+CLI2​=Vm2​By energy conservation 21​LI2+2Cq2​= constantWhen the P.D. across the capacitor takes its maximum value Vmt​​ the current I must be zero. Thus "constant" =21​CVm2​ Hence CLI2​+V2=Vm2​ once again.

Question 3

Q.96. An oscillating circuit consists of a capacitor with capacitance C, a coil of inductance L with negligible resistance, and a switch. With the switch disconnected, the capacitor was charged to a voltage Vm, and then at the moment t = 0 the switch was closed. Find: (a) the current I (t) in the circuit as a function of time; (b) the emf of self-inductance in the coil at the moments when the electric energy of the capacitor is equal to that of the current in the coil.

Accepted Answer

After the switch was closed, the circuit satisfies−LdtdI​=Cq​dt2d2q​+ω02​q=0⇒q=CVm​cosω0​twhere we have used the fact that when the switch is dosed we must haveV=Cq​=Vm​,I=dtdq​=0 at t=0.Thus (a)I=dtaq​=−CVm​ω0​sinω0​t=−Vm​LC​​sinω0​t(b) The electrical energy of the capacitor is 2Cq2​αcos2ω0​t and of the inductor is 21​LI2αsin2ω0​tThe two are equal whenω0​t=4π​At that instant the emf of the self-inductance is−Ldtdi​=Vm​cosω0​t=Vm​/2​

Question 4

Q.97. In an oscillating circuit consisting of a parallel-plate capacitor and an inductance coil with negligible active resistance the oscillations with energy W are sustained. The capacitor plates were slowly drawn apart to increase the oscillation frequency η-fold. What work was done in the process?

Accepted Answer

In the oscillating circuit, letq=qm​cosωtbe the change on the condenser wherechanges η2 fold and so does y i.e. changes from y0​ initially to η2y0​ finally. Now the P.D.across the condenser is V=Cqm​​cosωt=ε0​Syqm​​cosωtand hence the electric field between the plates isE=e0​Sqm​​cosωtThus, the chaige on the plate being qm​cosωt, the force on the plate isF=ε0​sqm2​​cos2ωtSince this force is always positive and the plate is pulled slowly we can use the average forceFˉ=2ε0​Sqm2​​and work done is A−Fˉ(η2y0​−y0​)−(η2−1)2e0​sqE2​y0​​2ε0​Sqm2​y0​​=2C0​qm2​​=Wthe initial stored energy. Thus.A=(η2−1)W

Question 5

Q.98. In an oscillating circuit shown in Fig. 4.27 the coil inductance is equal to L = 2.5 mH and the capacitor have capacitances C1= 2.0 μF and C2  = 3.0 μF. The capacitors were charged to a voltage V = 180 V, and then the switch Sw was closed. Find: (a) the natural oscillation frequency; (b) the peak value of the current flowing through the coil.

Accepted Answer

The equations of the L−C circuit are Ldtd​(I1​+I)​=C1​C1​V−∫I1​dt​=C2​C2​V−∫I2​dt​ Differentiating again L(I1​+I2​)=−C1​1​I1​=−C2​1​I2​ , ThenI1​=C1​−C2​C1​​I,I2​=C1​+C2​C2​​I, I=I1​+I2​ so L(C1​+C2​)I+I=0 or I=I0​sin(ω0​t+α) where ω02​=L(C1​+C2​)1​ (Part a) (Hence T=ω0​2π​=0.7 ms )At t=0,I=0 so α=0 I=I0​sinω0​t  The peak value of the current is 70 and it is related to the voltage V by the first equationor LI=V−∫Idt/(C1​+C2​) +Lω0​I0​cosω0​t=V−C1​+C2​1​∫0t​I0​sinω0​tdt (The P.D. across the inductance is V at t=0 )Hence=V+C1​+C2​1​⋅ω0​I0​​(cosω0​t−1) I0​=(C1​+C2​)ω0​V=VLC1​+C2​​​=8.05 A

Question 6

Q.99. An electric circuit shown in Fig. 4.28 has a negligibly small active resistance. The left-hand capacitor was charged to a voltage V0 and then at the moment t = 0 the switch Sw was closed. Find the time dependence of the voltages in left and right capacitors.

Accepted Answer

q1​+q2​=CV0​Cq1​​+LdtdI​−Cq2​​=0Also I=q˙​1​=−q˙​2​. ThusLI¨+C2I​=0Hence I¨+ω02​I=0ω02​=LC2​,The solution of this equation subject toI=0 at t=0is I=I0​sinω0​t.Integratingq1​=A−ω0​I0​​cosω0​tq2​=B+ω0​I0​​cosω0​tFinally substituting in (1)CA−B​−ω0​C2I0​​cosω0​t+LI0​ω0​cosω0​t=0ThusA=B=2CV0​​ and 2CV0​​+ω0​I0​​=0Soq1​=2CV0​​(1+cosω0​t)q2​=2CV0​​(1−cosω0​t)

Question 7

Q.100. An oscillating circuit consists of an inductance coil L and a capacitor with capacitance C. The resistance of the coil and the lead wires is negligible. The coil is placed in a permanent magnetic field so that the total flux passing through all the turns of the coil is equal to φ. At the moment t = 0 the magnetic field was switched off. Assuming the switching off time to be negligible compared to the natural oscillation period of the circuit, find the circuit current as a function of time t.

Accepted Answer

The flux in the coil isΦ(t)={Φ0​t<0t>0​ The equation of the current is −LdtdI​=C∫0I​Idt​LCdt2d2I​+I=0This mean thator with ω02​=LC1​I=I0​sin(ω0​t+α)Putting in (1) −LI0​ω0​cos(ω0​t+α)=−ω0​CI0​​[cos(ω0​t+α)−cosα]This implies cosα=0∴I=±I0​cosω0​t. From Faraday's lawε=−dtdΦ​=−LdtdI​or integrating from t=−ε to −ε where ε→0Φ=LI0​ with + sign in Iso, I=LΦ​cosω0​t

Question 8

Q.102. A certain oscillating circuit consists of a capacitor with capacitance C, a coil with inductance L and active resistance R, and a switch. When the switch was disconnected, the capacitor was charged; then the switch was closed and oscillations set in. Find the ratio of the voltage across the capacitor to its peak value at the moment immediately after closing the switch.

Accepted Answer

The equation of the circuit isLdt2d2Q​+RdtdQ​+CQ​=0 This has the solution, Q=Qm​e−βtsin(ωt+α) where β=2LR​,ω=ω02​−β2​,ω02​=LC1​ Now, I=dtdQ​=0  at t=0 so, Qm​e−βt(−βsin(ωt+α)+ωcos(ωt+α))=0 at t=0 Thus, ωcosα=βsinα or, α=tan−1βω​ Now, Vm​=CQm​​ and V0​=P.D. at t=0 =CQm​​sinα ﻿∴ Vm​V0​​=sinα=ω2+β2​ω​=ω0​ω​=1−β2/ω02​​=1−4L2R2C​​﻿

Question 9

Q.103. A circuit with capacitance C and inductance L generates free damped oscillations with current varying with time as I=Im​e−βtsinωt . Find the voltage across the capacitor as a function of time, and in particular, at the moment t = 0.

Accepted Answer

We write −dtdQ​=I=Im​e−βtsinωt =gmIm​e−βt+iωt (gm means imaginary part)ThenQ=gmIm​−β+iωe−βt+iωt​ Q=gmIm​β−iωe−βt+iωt​ =gmIm​β2+ω2(β+iω)e−βt+iωt​ =Im​e−βtβ2+ω2βsinωt+ωcosωt​ =Im​e−βtβ2+ω2​sin(ωt+δ)​,tanδ=βω​ (An arbitrary constant of integration has been put equal to zero.) ThusV=CQ​=Im​CL​​e−βtsin(ωt+δ) V(0)=Im​CL​​sinδ=Im​CL​​ω2+β2​ω​ =Im​C(1+β2/ω2)L​​

Question 10

Q.104. An oscillating circuit consists of a capacitor with capacitance C = 4.0 μF and a coil with inductance L = 2.0 mH and active resistance R = 10Ω. Find the ratio of the energy of the coil's magnetic field to that of the capacitor's electric field at the moment when the current has the maximum value.

Accepted Answer

I=Im​e−βtsinωt β=2LR​,ω0​=LC1​​,ω=ω02​−β2​ I=−q˙​,q= charge on the capacitorThen, q=Im​e−βtω2+β2​sin(ωt+δ)​,tanδ=βω​  Thus, WM​=21​LIm2​e−2βtsin2ωt WE​=2CIm2​​ω2+β2e−2βtsin2(ωt+δ)​=2LIm2​​e−2βtsin2(ωt+δ) Current is maximum when dtd​e−βtsinωt=0 Thus, −βsinωt+ωcosωt=0 or, tanωt=βω​=tanδ i.e., and henceWE​WM​​=sin2(ωt+δ)sin2(ωt)​=sin22δsin2δ​=4cos2δ1​ =4β2/ω02​1​=4β2ω02​​=LC1​×R2L2​=CR2L​=5 (WM is the magnetic energy of the inductance coil and WE is the electric energy of t capacitor.)

Question 11

Q.105. An oscillating circuit consists of two coils connected in series whose inductances are L1  and L2, active resistances are R1 and R2, and mutual inductance is negligible. These coils are to be replaced by one, keeping the frequency and the quality factor of the circuit constant. Find the inductance and the active resistance of such a coil.

Accepted Answer

Clearly L=L1​+L2​,R=R1​+R2​

Question 12

Q.106. How soon does the current amplitude in an oscillating circuit with quality factor Q = 5000 decrease η = 2.0 times if the oscillation frequency is v = 2.2 MHz?

Accepted Answer

Q=βTπ​ or β=QTπ​ Nowβt=lnη so, πvQlnη​=0.5 ms

Question 13

Q.107. An oscillating circuit consists of capacitance C = 10μF, inductance L = 25 mH, and active resistance R = 1.0 Ω. How many oscillation periods does it take for the current amplitude to decrease e-fold?

Accepted Answer

Current decreases e-fold in timet=β1​=R2L​ sec=RT2L​ oscillations=R2L​2πω​ =πRL​LC1​−4L2R2​​=2π1​R2C4L​−1​=15.9 oscillations

Question 14

Q.109. In a circuit shown in Fig. 4.29 the battery emf is equal to δ = 2.0 V, its internal resistance is r = 9.0Ω, the capacitance of the capacitor is C = 10μF, the coil inductance is L = 100 mH, and the resistance is R = 1.0 Ω. At a certain moment, the switch Sw was disconnected. Find the energy of oscillations in the circuit (a) immediately after the switch was disconnected; (b) t = 0.30 s after the switch was disconnected.

Accepted Answer

At t = 0 current through the coil =R+rε​ P.D. across the condenser =R+rε​ (a) At t = 0, energy stored = W0=21​L(R+rε​)2+21​C(R+rεR​)2=21​ε2(R+r)2(L+CR˙2)​=20 mJ (b) The current and the change stored decreases as e−tR/2L so energy decreases as e−tR/L∴W=W0​e−tR/L=0.10 mJ

Question 15

Q.131. A series circuit consisting of a capacitor and a coil with active resistance is connected to a source of harmonic voltage whose frequency can be varied, keeping the voltage amplitude constant. At frequencies ω1 and ω2  the current amplitudes are n times less than the resonance amplitude. Find: (a) the resonance frequency; (b) the quality factor of the circuit.

Accepted Answer

At resonanceω0​L=(ω0​C)−1 or ω0​=LC​1​, and (Im​)nes​=RVm​​ Now, nRVm​​=R2+(ω1​L−ω1​C1​)2​Vm​​=R2+(ω2​L−ω2​C1​)2​Vm​​ Then, ω1​L−ω1​C1​=n2−1​R ω2​L−ω2​C1​=+n2−1​R (assuming ω2​>ω1​ )or ω1​−ω1​ω02​​=−ω2​+ω2​ω02​​=−n2−1​LR​ or, ω1​+ω2​=ω1​ω2​ω02​​(ω1​+ω2​)⇒ω0​=ω1​ω2​​ and ω2​−ω1​=n2−1​LR​ β=2LR​=2n2−1​ω2​−ω1​​ andQ=4β2ω02​​−41​​=(ω2​−ω1​)2(n2−1)ω1​ω2​​−41​​

Question 16

Q.133. A circuit consisting of a capacitor and a coil connected in series is fed two alternating voltages of equal amplitudes but different frequencies. The frequency of one voltage is equal to the natural oscillation frequency (ωo) of the circuit, the frequency of the other voltage is η times higher. Find the ratio of the current amplitudes (l0/l) generated by the two voltages if the quality factor of the system is equal to Q. Calculate this ratio for Q = 10 and 100, if η = 1.10.

Accepted Answer

At resonanceω=ω0​ ﻿Im​(ω0​)=RVm​​﻿ Then Im​(ηω0​)=R2+(ηω0​L−ηω0​C1​)2​Vm​​ =R2+(η−ηl​)2CL​​Vm​​=1+(Q2+4l​)(η−η1​)2CL​​Vm​​

Question 17

Q.135. Find the effective value of current if its mean value is l0 and its time dependence is (a) shown in Fig. 4.34; (b) I∼∣sinωt∣ .

Accepted Answer

(a) I(t)=I1​Tt​0≤t<T I(t±T)=I(t) Now mean current⟨I⟩=T1​∫0T​I1​Tt​dt=I1​T2T2/2​=I1​/2 Then I1​=2I0​ Since ⟨I⟩=I0​  Now mean square current ⟨I2⟩=4I02​T1​∫0T​T2t2​dt=34I02​​ so effective current =3​2I0​​ (b) In this case I=I1​∣sinωt∣ and I0​=T1​∫0​I1​∣sinωt∣dt =2π1​I1​∫02π​∣sinθ∣dθ=πI1​​∫0π​sinθdθ=π2I1​​ so, I1​=2πI0​​ Then, the mean square current  =⟨I2⟩=4Tπ2I02​​∫0T​sin2ωtdt  =4π2I02​​×2π1​∫02π​sin2θdθ=8π2I02​​ so effective current =8​πI0​​

Question 18

Q.136. A solenoid with inductance L = 7 mH and active resistance R = 44Ω is first connected to a source of direct voltage Vo  and then to a source of sinusoidal voltage with effective value V = V0. At what frequency of the oscillator will the power consumed by the solenoid be η = 5.0 times less than in the former case?

Accepted Answer

Pd.c.​=RV02​​ Pa.c.​=R2+ω2L2​V02​​⋅R2+ω2L2​R​=1+(RωL​)2v02​/R​=ηPd,c.​​ Thus RωL​=η−1​ or ω=LR​η−1​ v=2πLR​η−1​=2 kH of on putting the values.

Question 19

Q.137. A coil with inductive resistance XL = 30Ω and impedance Z = 50Ω is connected to the mains with effective voltage value V = 100 V. Find the phase difference between the current and the voltage, as well as the heat power generated in the coil.

Accepted Answer

Z=R2+XL2​​ or R0​=Z2−XL2​​ The, tanθ=Z2−XL2​​XL​​ So, cosφ=zz2−XL2​​​=1−(ZXL​​)2​  The current lags by φ behind the voltage.φ=cos−11−(ZXL​​)2​=37∘ also P=VIcosφ=z2V2​z2−XL2​​=.160 kW

Question 20

Q.138. A coil with inductance L =- 0.70 H and active resistance r = 20Ω is connected in series with an inductance-free resistance R. An alternating voltage with effective value V = 220 V and frequency w=314s−1 is applied across the terminals of this circuit. At what value of the resistance R will the maximum heat power be generated in the circuit? What is it equal to?

Accepted Answer

P=(R+r)2+ω2L2V2(R+r)​ This is maximum when R+r=ωL forP=R+r+R+r(ωL)2​V2​=[R+r​−R+r​ωL​]2+2ωLV2​ Thus R=ωL−r for maximum power and Pmax​=2ωLV2​ Substituting the values, we get R=200Ω and Pmax​=.114 kW

Chapter 3 : PART THREE. ELECTRODYNAMICS