Question

A certain LR circuit with a battery has an inductor with L = 37 mH & τ = 6 ms. It is known that when t = 12 ms, I = 0.11 A. Determine the resistance of the resistor and the voltage of the battery.

The armature of a 60 Hz ac generator rotates in a 0.19 T B-field. If the area of the coils is 0.03 m², how many loops must the coil contain if the peak output is to be V = 230 V? What will the current & power output (not the power loss) be if the resistance of the coil is 0.2 Ω?

Answer 1

Given for LR 6. 17 a is 681035 for LR Circrait with a batteny has an inductore L = 37mH & t is time constant = 6ms. when time t= 12ms Connent I = 0.11 A. Time constant = = 6 ms. 375103H Roms 37 Thus the resistance R of the resiston is 6.172. Time is increases I = 0.11 A. so the current in LR cincuit is given by. ICD= V (i-e-tle) 0.11A= (1--136%mg) . (1-802) Sto (1-0.135) = V x 0.865 - OR VE 0.11 AX 6.172 0.865 0.785 V The ammature of frequency f = 60Hz, ac generator rotates in a B-field Be 0.19T. Area of the coil is A = 0.03 m² -If the peak output voltage v=230V, related with V=2TINBAT where N is the number of turns of the Goil on no. of loops of the coil.

. V. STINBAT ») a 2TBAL 230V 2x3.14 x 0.19 T X 0.03 m² x 6051 230V 2.15 v n 106.97 ~ 107 no. of tools of dools Ans) & Now current. out - Peut T = R = 0.212 J = 230v 0.2 - 1150 A & output Power P = wxI P = 1150 AX 230N p=264500 w = 2.64 x 105 w nosw. Ans).