physics problem please helpp?

Question

Consider the circuit in Figure P20.46. Taking = 6.00 V, L = 3.00 mH, and R = 9.00 .

Figure P20.46
http://www.webassign.net/pse/p32-19.gif

(a) What is the inductive time constant of the circuit?
ms
(b) Calculate the current in the circuit 250 µs after the switch is closed.

A
(c) What is the value of the final steady-state current?
A
(d) How long does it take the current to reach 80% of its maximum value?
ms

Answer 1

A series R-L circuit has
voltage across the resistor
vR(t)=V*(1-e^(t/T))
since the resistor is linear and obeys Ohm's law
IR(t)=vR(t)/R

Since the inductor and resistor are in series, the current is the same everywhere

BTW the voltage across the inductor is
vL(t)=e^(t/T)

T is the time constant, which means for time elapsed after switch closure the function will vary as e or 1-e at one time constant.

T is L/R

For this circuit L/R=3/9 ms

j

Answer 2

in reality greater beneficial physics jargon ... i'm not even specific if it extremely is genuinely anymore ... Photons and gluons are strictly massless. they're the gauge fields of unbroken symmetries (in evaluation to the susceptible vector bosons). because of the solid stress and the self-interactions of the gluons (the solid stress originates from a non-Abelian group -- so there is self interaction between the gauge fields -- the gauge fields themselves are charged), there could be gluon condensation and the condensation looks super. As they're massless, they don't precise now artwork at the same time with Higgs boson on the tree-element. At greater beneficial-loop stages, there are interactions because you will possibly have each and each form of digital debris (i.e. debris which on the instant at the instant are not 'on-shell'). an identical probably is going for the interactions with gravitons (the particle interpretation of the gravitational field). Photons curve around super our bodies. Photons can not get away black holes as devoid of delay as they're interior the shape horizon of the black holes. even regardless of the reality that, those 2 themes are fairly self holding, different than the curvature of the photon trajectory is delivered approximately by employing the black hollow.

Answer 3

(a) time constant(tau) = L/R = 0.003/9.0 = 333.3uS

(b) instantaneous current = Imax * e^(-t/tau), where Imax is simply V/R, e is natural log, t is in seconds(250us), tau is our answer for (a), so (6/9) * e^(250E-6/(0.003/9)) = 314.9mA

(c) Final steady state current is equal to V/R = 6/9 = 666.7mA

(d) The time it takes to get to 80% of maximum current is 74.4us