Differentiation with electricity help

[Thepiman]

A circuit consists of 230V supply, a switch, a 2mH inductor and a 12k ohm resistor in series


When the switch is closed at time t=o, a current i begins to flow in the circuit: The current is modelled by the following equation:


i= v/r (1-e^-Rt/L)


Determine the rate of change of the current at the following times:
i) t= o ns
ii) t= 1 ns
iii) t= 50 ns


I have subbed the values in and come up with answers of
i) i= 0 amps
ii) i= 1.146x10^-4 amps
iii) i= 4.968x10^-3 amps


But is this right? Do i need to insert the values in first and simplify the equation? Then differentiate it then sub in values for t?


Any help would be appreciated.

 

[HallsofIvy]

A circuit consists of 230V supply, a switch, a 2mH inductor and a 12k ohm resistor in series


When the switch is closed at time t=o, a current i begins to flow in the circuit: The current is modelled by the following equation:


i= v/r (1-e^-Rt/L)


Determine the rate of change of the current at the following times:
i) t= o ns
ii) t= 1 ns
iii) t= 50 ns


I have subbed the values in and come up with answers of
i) i= 0 amps
ii) i= 1.146x10^-4 amps
iii) i= 4.968x10^-3 amps

The problem did NOT ask for the current, it asked for the rate of change of current.

But is this right? Do i need to insert the values in first and simplify the equation? Then differentiate it then sub in values for t?

What do you mean by "insert the values in first and simplify the equation"? The only values given are values of t and once you insert those, there is no variable, t, left in the equation.

Any help would be appreciated.

 

[stapel]

The current is modelled by the following equation: i= v/r (1-e^-Rt/L)

Determine the rate of change of the current...

I have subbed the values in...

You "subbed the values in" to what? The equation is for the current, not the rate of change of the current. Please show your work. Thank you!

 

[Bob Brown MSEE]

Your stated procedure is correct, do it that way

But is this right? No

Do i need to insert the values (other than t) in first and simplify the equation? Yes
Then differentiate it Yes
then sub in values for t? Yes

Any help would be appreciated.

Good. What do you get after you differentiate the expression for i?

 

[Thepiman]

Good. What do you get after you differentiate the expression for i?

I got after differentiating for i:

i= 0.01916 (1-e^-6000000t)

got that far.

 

[Thepiman]

You "subbed the values in" to what? The equation is for the current, not the rate of change of the current. Please show your work. Thank you!

I subbed the values of time as 0 ns, 1ns and 50ns.

for t= 0 nano seconds:
i= v/r (1-e^-Rt/L)
i= 230/12x10^3 (1-e^-12x10^3x1x0/2x10^-3)
i= 0.01919 (0)
i= 0 amps

for t= 1 nano second:
i= v/r (1-e^-Rt/L)
i= 230/12x10^3 (1-e^-12x10^3x1x10^9/2x10^-3)
i= 0.01919 (5.982x10^-3)
i= 1.146^-4 amps

for t= 50 nano seconds:
i= v/r (1-e^-Rt/L)
i= 230/12x10^3 (1-e^-12x10^3x50x10^9/2x10^-3)
i= 0.01919 (0.25918)
i= 4.968x10^-3 amps

But I think that is wrong. Have I found the current and not the change in current?

My attempt to differentiate the formula is:

i= v/r (1-e^-Rt/L)
i= 230/12x10^3 (1-e^-12x10^3xt/2x10^-3)
i= 0.01916 (1-e^-6000000t)

On the right lines?

 

[HallsofIvy]

I subbed the values of time as 0 ns, 1ns and 50ns.

for t= 0 nano seconds:
i= v/r (1-e^-Rt/L)
i= 230/12x10^3 (1-e^-12x10^3x1x0/2x10^-3)
i= 0.01919 (0)
i= 0 amps

for t= 1 nano second:
i= v/r (1-e^-Rt/L)
i= 230/12x10^3 (1-e^-12x10^3x1x10^9/2x10^-3)
i= 0.01919 (5.982x10^-3)
i= 1.146^-4 amps

for t= 50 nano seconds:
i= v/r (1-e^-Rt/L)
i= 230/12x10^3 (1-e^-12x10^3x50x10^9/2x10^-3)
i= 0.01919 (0.25918)
i= 4.968x10^-3 amps

But I think that is wrong. Have I found the current and not the change in current?

My attempt to differentiate the formula is:

i= v/r (1-e^-Rt/L)
i= 230/12x10^3 (1-e^-12x10^3xt/2x10^-3)
i= 0.01916 (1-e^-6000000t)

On the right lines?

Are you saying that you do not know what differentiate" means? All you have done is put in the given values for v, r, R, and L. You have not differentiated at all.