00:01
For this problem on the topic of alternating current, we are shown an rlc circuit in which the ac generator is supplying 120 volts at 60 hz.
00:12
With the switch open, the current leaves the generator emf by 20 degrees, and with the switch in position 1, the current lags the generator emf by 10 degrees.
00:23
When the switch is in position 2, we are told that the current amplitude is 2 ampires and are asked to find the resistance r, the inductors, the inductors.
00:31
L and capacitance c.
00:34
Now when the switch is open, we have a series lrc circuit involving just the one capacitor near the upper right corner.
00:42
And so we have omega d times l minus 1 over omega d times c, all divided by r is equal to the tan of phi n, which is the tan of minus 20 degrees, which is minus 10 20 degrees.
01:12
Now when the switch is in position 1, the equivalent capacitance in the circuit is 2c.
01:16
And so in this case, we have omega -d times l minus 1 over 2 omega -d times c, all divided by r is equal to the tan of 5 -1, which is the 10.
01:40
Of 10 degrees.
01:43
And finally with the switch in position 2 the circuit is simply an lc circuit with the current amplitude i2 equal to the emf epsilon m over the inductance zlc not inductance but impedance at lc and this is equal to epsilon m divided by the square root of omega d times l minus 1 over omega d times c which is that's all squared so this is epsilon m divided by 1 over omega d times c minus omega d times l now we use the fact here that 1 over omega d times c is greater than omega d times l in simplifying the square root...