V = i*R + L * di/dt, I have 175A/ns, L=9.5pH, and 5VDC. How do I solve this?? The answer is suppose to be 3.338V

Question

V = i*R + L * di/dt, I have 175A/ns, L=9.5pH, and 5VDC.  How do I solve this??  The answer is suppose to be 3.338V

myininaya · Answer

I don't know some of these units...
what is 175A/ns? <-this sounds like a rate because of the /ns part
so is that suppose to be di/dt?

myininaya · Answer

And 5VDC not sure what variable that is

anonymous · Answer

175 Amps per nanosecond.

anonymous · Answer

5VDC is volts direct current

perl · Answer

there shouldnt be a ph there, that measures acidity

perl · Answer

oh pH is picoHenries?

perl · Answer

pH = picoHenrys

anonymous · Answer

Lol.  yes it does.  It also stands for picoHenrys

perl · Answer

i see

anonymous · Answer

This is for a logic gate circuit

anonymous · Answer

I am suppose to find the voltage at the terminal for the instantaneous rate of change from switching from off to on...

kropot72 · Answer

It appears that the voltage i*R is needed, with 5V being the applied voltage to a
series L - C circuit.
$$\large 175\ A/ns=175\times10^{9}\ A/s$$
Therefore the voltage across the 9.5 pico-Henry inductor is given by:
$$\large V _{L}=175\times10^{9}\times9.5\times10^{-12}=1.6625\ V$$
The voltage across the resistance is therefore 5.0000 - 1.6625 = 3.3375 V

anonymous · Answer

Oh wow.  Thank you.   How did you figure that out without that equation?  Lol.  Yours looks a lot easier than what I was looking at.

kropot72 · Answer

......series L - R circuit*
I used your equation, putting V = 5 V and then calculated L * di/dt from the given values. the voltage given by i*R can then be found.

anonymous · Answer

Yes.  How did you calculate L*di/dt?

kropot72 · Answer

di/dt is given as 175 A/ns. As shown in my post I converted that to A/s by multiplying numerator and denominator by 10^9, getting 175 * 10^9 A/s.

anonymous · Answer

Oh my.  Thank you.  I was way over thinking that.  It was right in from of me.....Lol.  When I get stuck somewhere I always think it can not be that easy.

anonymous · Answer

Thanks for all your help and your clarification, I greatly appreciate it.

kropot72 · Answer

You're welcome.
As you can see it is important to get the units correct.

anonymous · Answer

Yes for sure.  For some reason I was not looking at the ns as a relation to time....  I was just focusing on the Amps.