An inductor with an inductance of 2.10 H and a resistance of 7.50 ohm is connect
ID: 2078005 • Letter: A
Question
An inductor with an inductance of 2.10 H and a resistance of 7.50 ohm is connected to the terminals of a battery with an emf of 6.50 V and negligible internal resistance. You may want to review For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Analyzing an r-l circuit. elementof - iR - L di/dt = 0 so i = elementof/R (1 - e^(-R/L)t). Initially (t = 0), i = 0 so elementof - L di/dt = 0 di/dt = elementof/L = 6.50 V/2.10 H = 3.10 A/s Find the rate of increase of current at the instant when the current is 0.500 A.Explanation / Answer
(a) battery voltage = initialrate * inductance .
6.50 = initial rate * 2.10 .
initial rate = 3.10 A/s.
(b) rate of current increase = initial rate *e-t/T where T = L/R = 2.1/7.5 = 0.28.
also... current = (V/R) (1 - e-t/T ) so....
e-t/T = 1 - (current* R / V) = 1 - 0.500 * 7.5 / 6.5 = 0.423.
rate of current inc = 3.10* 0.423 = 1.312 A/s