A battery is connected in series with a 0.27-? resistor and an inductor, as show
ID: 2058462 • Letter: A
Question
A battery is connected in series with a 0.27-? resistor and an inductor, as shown in the figure below. The switch is closed at t = 0. The time constant of the circuit is 0.22 s, and the maximum current in the circuit is 8.2 A.(a) Find the emf of the battery.
=________ V
(b) Find the inductance of the circuit.
=_________ mH
(c) Find the current in the circuit after one time constant has elapsed.
=__________ A
(d) Find the voltage across the resistor after one time constant has elapsed.
=________ V
(e) Find the voltage across the inductor after one time constant has elapsed.
=__________ V
Explanation / Answer
Since you didn't include the figure, there is some key information left out (the frequency) so I can't solve the equations for you, but I will give you the equations for each and hopefully that will be enough!!
(a) emf = Voltage = IR
= (8.2) (.27) = 2.214 V
(b) inductance = L = V max / (I max X 2f) to solve this you need to know the frequency, f
L = (2.214 / 8.2) X (1/ 2f)
L = 0.27 X ( 1/ 2f)
if instead of frequency you are given angular velocity,
L = 0.27 X ( 1 / )
note: these will give you inductance in H so don't forget to convert to mH!
(c) emf = -L (I / t)
2.214 = -L (the value you calculated in b, in H) X ( (I final - 8.2) / .22 )
solve the equation for I final!
(d) V = IR Since you just found the current after an elapsed time interval, do this the same as (a)
V = I (I final from c) X 0.27
(e) Since the resistor and inductor are in series,
R final (answer from part d) - R initial (answer from part a) = Voltage in inductor
it will come out negative, however disregard the sign as resistance is not negative.
I really hope this helped!
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