AP EAMCET201726 Apr 2017Morning ShiftPhysicsAlternating CurrentActual
When a coil is connected to AC supply of frequency 50 ~Hz , a current of 4 ~A flows in it and it consumes 240 ~W power. If the potential difference across the coil is 100 ~V , then the inductance value of the coil is
Options
- AL =(5 ) H
- BL= 5 H
- CL= 1 5 H
- DL= 1 25 H
Correct answer
C. L= 1 5 H
Step-by-step solution
I^2 R=240 ~W , I=4 ~A R= 240 16 =15 As, V=I Z=I X_L^2+R^2 V I = X_L^2+R^2 array ll & V^2 I^2 =X_L^2+R^2 or & X_L^2= 100 100 4 4 -225=400 & X_L=20=L & L= 20 2 f = 20 2 50 = 1 5 H array