Question

A series RLC circuit is connected to a 4.70 kHz oscillator with a peak voltage of 3.90 V. It consists of a 4.40 mH inductor, a 360. nF capacitor, and a 37.0 Ω resistor. If ε = ε₀, what is the instantaneous current i?

Answer 1

Solution Instantaneous current in a RLC circuit in series is given by sinut I(t) = I max where Vmax Vmax 11 Imax = 2 R2+(XL - XL) 2 TR2+(21-1-2750) given us Vm= 3.90 Volt f = 4.7x 100 Hz - 3 L = 4.4 x 10 H C = 360 X10 R: 37 2 So 3.9 Imax = 1 (37)2 + (2x3-14x4.7x102x4.4816 19x47810*44450 2x3.14x4.7X10 X10 09 3. 9 3.9 3.9 Imer = = 47.55 1369+ 1369 + 892:2 (129-87- -) Inex = 0.082 A Now instantaneous instantaneous current Ilt) - Imor sin wt - T sin (277 ft) max I (4) =(0.082) sin ( 217 4.78109+) Amp I (!!=(0.082) sin (9.4 Xio'r t) Amp