Inductance, volt drop and time constant? Help

AI Thread Summary
The discussion revolves around calculating various parameters of an RL circuit consisting of a 0.2H inductor and a 330-ohm resistor connected to a 12V DC supply. The time constant of the circuit is determined to be 606 microseconds. After two time constants, the voltage drop across the inductor is approximately 1.66V, and after three time constants, it drops to about 0.62V. A correction was made regarding the time value for calculating resistance when the time constant is set to 0.55ms. Overall, the calculations provided align with the expected results, barring minor rounding errors.
ilovescience85
Messages
25
Reaction score
0
Good evening all, I am looking for some clarity on a question I am answering at the moment.

An inductor of negligible resistance and an inductance of 0.2H is connected in series with a 330ohm resistor to a 12v d.c. Supply determine:

A - the time constant of the circuit
B - the voltage drop across the inductor after two time constants
C - the voltage drop across the inductor after three time constants
D - the resistance of a 0.2H coil used to replace the inductor if the circuits time constant falls to 0.55s

Attempt so far
I = V/R = 12/330 = 36.36mA
A - t=L/R = 0.2/330 = 606microseconds
B - I'm trying to use Ve exp -Rt/L but it does seem to be giving me any answers that make sense?!
C - same as B
D - R= L/t = 0.2/0.00055 = 363.64 ohms

Any help would be greatly appreciated.

Thanks
 
Physics news on Phys.org
I moved your thread to our homework section.

B - I'm trying to use Ve exp -Rt/L but it does seem to be giving me any answers that make sense?!
What do you get?

At (D), your value for t in the formula does not match the value given in the problem statement. I guess that should read 0.55ms?
 
Last edited:
Hi sorry for the delay i had a busy weekend.

When using the above equation i get as follows;

B - 12 exp -(330 *0.0012)/0.2 = 12 exp -1.98 = 1.66v
C - 12 exp -(330 *0.0018)/0.2 = 12 exp -2.97 = 0.62v

Regarding D yes sorry that should have read 0.55ms
 
ilovescience85 said:
When using the above equation i get as follows;

B - 12 exp -(330 *0.0012)/0.2 = 12 exp -1.98 = 1.66v
C - 12 exp -(330 *0.0018)/0.2 = 12 exp -2.97 = 0.62v
Apart from rounding errors, it is correct. Note that you can directly plug in -2 and -3 in the exponents, without calculating the time.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

Similar threads

Back
Top