Consider the following circuit: Here R1 = 8.0 , R2 = 4.0 , R3 = 7.0 , L1 = 2.5 m

Question

Consider the following circuit:

Here

R1 = 8.0 , R2 = 4.0 , R3 = 7.0 ,

L1 = 2.5 mH, L2 = 3.8 mH, L3 = 5.2 mH,

C1 = 2.4 F, C2 = 3.4 F, C3 = 6.3 F.

a) What is the effective resistance of the three resistors?

b) whar is the effective inductance of the three inductors?

c) What is the effective capacitance of the three capacitors? (For F enter "mu F")

d)What is the resonant frequency of the circuit?

e) If the circuit is driven on resonance by an oscillating voltage with peak voltage V0 = 50. V, what is the peak voltage across L1?

The answer to a,b,c and d are 1.05*10^1 ohms, 4.70 mH, 1.92 muF and 1.67*10^3 HZ respectively. Pls i just need a detailed answer to question (e) only.

Explanation / Answer

a) equivalent resistance = R1 + (R2||R3)

Or R = 8 +(4*7/11) = 10.54 ohms

b) equivalent inductance = L1 + (L2||L3)

Or R = 2.5 +(3.8*5.2/9) = 4.70 mH

c) equivalent capacitance = C1|| (C2+C3)

Or R = 2.4 ||(3.4+6.3) = 1.92 muF

d) resonant freq = 1/sqrt(LC) = 1.67*10^3 HZ

e)

Z = 10.54 +j(2*3.14*1.67*10^3*4.70*10^-3) -j (1/(2*3.14*1.67*10^3*1.92*10^-6)

or Z = 10.54 + j (49.3 - 49.3) = 10.54

(Since at resonance no reactive part remains)

V =I/Z

OR I = 50sin(omega t )/10.54

so voltage across L1 = I*omega*L1 = 4.74sin (omega * t) *omega* 2.5*10^-3

here omega = 2*3.14*1670

So VL1 = 124.34 sin (omega*t)

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