Circuit I L Circuit II 2L C/2 ? sin(cot) ? sin(cot) Consider two RLC circuits wi

Question

Circuit I L Circuit II 2L C/2 ? sin(cot) ? sin(cot) Consider two RLC circuits with identical generators and resistors. Both circuits are driven at the resonant frequency. Circuit II has twice the inductance and 1/2 the capacitance of circuit I as shown above. 1) Compare the peak voltage across the resistor in the two circuits Submit You currently have O submissions for this question. Only 10submission are allowed You can make 10 more submissions for this question. (Survey Question) 2) Briefly explain your reasoning Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question.

Explanation / Answer

for the given circuit

identical genrators, emf = Esin(wt)

L2 = 2*L1

C1 = 2*C2

a. peak voltage rein resistor = Vm

now, V = IZ

Z = sqrt(R^2 + (Xl - Xc)^2)

Xl = wL

Xc = 1/wC

at resonant frequency

Z = R

hence

V = IR

Vm = IR = V = E

hence peak voltage is same for both the circuits

Vi = Vb

b. because the circuit is osscilating at resonant frequency, the peak voltage of both the resistors is the same as the effect of the inductor and the capacitor cancel each other at resonant frequency

c. peak voltage across inductor = Vl

Vl = -L*di/dt = -L*(E/R)*w*cos(wt)

pak volateg

Vl = L*Ew/R

w = 1/sqrt(LC)

w1 = 1/sqrt(L1C1)

w2 = 1/sqrt(2L1*C1/2) = 1/sqrt(L1C1)

hence

Vl = L(E/R)1/sqrt(LC) = (E/R)*sqrt(L/C)

L1/C1 = L2/4C2

hence

Vl > Vr

p[eak voltage is not same same for both the circuits across inductor]

d. the peak voltage across inductor is the same in the both circuits as the ratio of inductance and the capacitance is not the same for both the circuits, leading to same resonant frequency of the circuit,but different peak voltage

e. voltage across capacitor Vc = q/C

q = integral(Idt) = (E/Rw)

Vc = (E/RwC)

as cL > cR

HENCE

vR > vL ( for capacitor peak voltage)

f. the reason is the mathematical reason provided above

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