Question

1. Consider the circuit below. The resistor is 150 8. The capacitor is 60 microfarads. The inductor is 2 H. The maximum AC voltage is 50 V and it oscillates at 10 Hertz. Determine the impedance of the circuit. Determine the current in the circuit. Determine what the frequency would need to be for the current to be maximized. 2. Consider a rectangular wire coil that is 50 cm wide by 15 cm tall. There are ten loops in the coil. Notice the 20 a resistor in the bottom section of the coil. The rectangular loop is pulled apart by the sides so that it stretches into a very thin loop in 3 seconds as shown below. There is a 4 T magnetic field pointed out of the page. 50cm 15cm Determine the current magnitude induced in the resistor. Determine the direction of the current through the resistor shown.

Answer 1

0 с L 70 m R=1502 C = 604F =60x1067 L = 2H Vm=sov f =10HZ Sabedance of circuit - 2= V RP+(@L-be)? J R² + (a1 fl-- I 21. fcl 2 2 Viso? +) 2) 2X3014 x 10.X2 2x3.14x10 x60x10 2 150++ (-139.79)2 z=205.042 50 205.04 Z Maximum value of current in the circuit - Imax = Vm Imax = 0.244 A Average Value (ring value) of current- Irms = Imax => Ilims = 0.173A Ja current in the circuit becomes maximum at resonant frequency which is given by- fr + F fx = 1454H2 DIT JLC = = 0.209 J2 2x3-14 2x60x106

l- - socm = 0.50mm be = 15cm 0.15m R=zor z no.og loobs n = 10 B=4T mint o Area of loop A= lb Net J le bassing through the loop CSwitally) n.BA nBlb After 3 seconds flux through the loop- 0 Indeced einf in Coil € - 10-nBlb) = mBlb де At at Induced Current Linduced E mBlb R Root the - linduced = lox 4x0.50X0.15 20 X 3 Iinduced = 0.05A => when look shrinks then no. of magnetic field lines (which is directed out thee Bagel decreases then by lenz' lew direction induced moegnet- ic Jield will be also "out of the Boge so that it Can oppose this change. Then by "eight hand ouile for loops direction of coreut will be "counterclockcorse" of

Direction of current in the resistor is from "left to right".