Inductance, volt drop and time constant? Help

[ilovescience85]

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

 

[mfb]

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?

 

[ilovescience85]

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

 

[mfb]

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.

 

[HalfLostAndSearching]

for reaching out for help with this question. Let's break down the concepts of inductance, volt drop, and time constant to better understand how to solve this problem.

Inductance is a property of an electrical circuit that describes its ability to store energy in the form of a magnetic field. It is measured in units of Henrys (H) and is denoted by the symbol L. In your circuit, the inductor has an inductance of 0.2H.

Volt drop, also known as voltage drop, is the decrease in voltage that occurs when current flows through a resistor. It is calculated using Ohm's Law: V = IR, where V is voltage, I is current, and R is resistance. In your circuit, the resistor has a resistance of 330 ohms.

The time constant of a circuit is a measure of how quickly the current or voltage in the circuit changes in response to a change in input. It is calculated using the formula t = L/R, where t is time, L is inductance, and R is resistance. In your circuit, the time constant is 606 microseconds.

Now, let's answer the given questions using this information.

A - To determine the time constant of the circuit, you correctly used the formula t = L/R. Plugging in the values of 0.2H for L and 330 ohms for R, we get a time constant of 606 microseconds.

B - To find the voltage drop across the inductor after two time constants, we can use the formula V = Ve^(-t/RC), where V is the initial voltage (12V in this case), e is the base of the natural logarithm, t is the time (2 time constants in this case), R is the resistance (330 ohms), and C is the capacitance (0.2H). Plugging in these values, we get a voltage drop of approximately 0.5V.

C - Similarly, to find the voltage drop across the inductor after three time constants, we can use the same formula with 3 time constants plugged in. This gives us a voltage drop of approximately 0.15V.

D - To determine the resistance of a 0.2H coil that would replace the inductor and result in a time constant of 0.55 seconds, we can rearrange the formula t = L/R to solve for R. This gives us R = L