A 12.6-V battery is in series with a resistance of 0.330 Ω and an inductor.?
A 12.6-V battery is in series with a resistance of 0.330 Ω and an inductor.
I got part a right, but I can't seem to figure out the rest..... help please.
a) After a long time, what is the current in the circuit?
I(t=infinity) = 38.1 A
(b) What is the current after one time constant?
I(t=τ) = A
How is the time constant defined for an RL series circuit? How does the current change as a function of time in such a circuit
(c) What's the voltage drop across the inductor at this time?
ΔVL = V
Use Kirchoff's Loop rule to describe the relative sizes of the potential differences across the battery, inductor, and resistor.
(d) Find the inductance if the time constant is 0.110 s.
L = H
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Verified answer
b) Current is given by
i = (E/R)*(1 - e^(-t/tL))
where tL is the inductive time constant. If t = tL then
i = (E/R)*(1 - e^(-1))
You can crank that out.
In an RL series circuit, tL = L/R
c) If the current is i, the voltage across it is i*R. The voltage across the circuit is E. So the voltage across the inductor is E - i*R.
d) tL = L/R = 0.110 s
Solve for L. The L = H threw me at first -- I think it's telling you that the units are henries.