how about Vout=((1/10k + 1/500k)^(-1))*500k*V2

I make it a different answer.

1. Once Vout settles, after the application of V1, what value must V3 become?

2. Then derive an expression for V3 based on the voltage divider formed by the 10K and 500K resistors

3. Equate 1. and 2. above and you're there

PS: try to simplify your answers as well - tidy up the maths by not leaving (1/a + 1/b)^-1 type expressions

 

I expect your answer to contain a term of the form \(\Large{e^{\frac{-t}{RC}}}\).

hgmjr

 

I expect your answer to contain a term of the form \(\Large{e^{\frac{-t}{RC}}}\).

hgmjr

jstrike

At the instant the step is applied, what will the capacitor look like as a circuit element?

What will it look like as time moves on and what will it look like after a few time constants?

 

I expect your answer to contain a term of the form \(\Large{e^{\frac{-t}{RC}}}\).

hgmjr

Iplugged in values for the resistors and capacitor and got -100.1u(t) where do i go from here?

I make it a different answer.

1. Once Vout settles, after the application of V1, what value must V3 become?

2. Then derive an expression for V3 based on the voltage divider formed by the 10K and 500K resistors

3. Equate 1. and 2. above and you're there

PS: try to simplify your answers as well - tidy up the maths by not leaving (1/a + 1/b)^-1 type expressions

so your saying
V3=V1
and
V3=(500/510)V2

 

Iplugged in values for the resistors and capacitor and got -100.1u(t) where do i go from here?



so your saying
V3=V1
and
V3=(500/510)V2

your voltage divider expression is incorrect

the voltage across the divider is what? (hint: it's NOT V2)

 

your voltage divider expression is incorrect

the voltage across the divider is what? (hint: it's NOT V2)

Is it Vout? but that doesnt make sense to me Im treating V2 like the input going across the 10k then being divided into v3 and across the 500k

 

The voltage across the divider is Vout - V2. So the voltage at V3 can be calculated by the normal voltage divider calculation THEN we need to add V2 to this result to get V3 relative to ground (since Vout, V1 and V2 are presumed relative to some 0v point)

 

The voltage across the divider is Vout - V2. So the voltage at V3 can be calculated by the normal voltage divider calculation THEN we need to add V2 to this result to get V3 relative to ground (since Vout, V1 and V2 are presumed relative to some 0v point)

so i set V1=50/51(Vo-V2)
=Vo=51V1/50+V2

 

You keep making the same error - the voltage in question is across the 10K resistor not the 500K resistor

V3 = (Vout - V2)10/510 + V2

The first part (Vout - V2)10/510 is the voltage across the 10K resistor BUT we need that voltage PLUS V2 since the bottom of the 10K resistor is at V2 volts.

Imagine a battery underneath the 10K resistor. It is easy to see then that the voltage at V3 is the battery voltage (V2) plus the actual voltage across the 10K resistor.

 

You keep making the same error - the voltage in question is across the 10K resistor not the 500K resistor

V3 = (Vout - V2)10/510 + V2

The first part (Vout - V2)10/510 is the voltage across the 10K resistor BUT we need that voltage PLUS V2 since the bottom of the 10K resistor is at V2 volts.

Imagine a battery underneath the 10K resistor. It is easy to see then that the voltage at V3 is the battery voltage (V2) plus the actual voltage across the 10K resistor.

V1=(10/510)(Vo-V2)?

sorry about all of the questions i was just kind of throwninto these problems without much guidance

 

Almost there........

V1=(10/510)(Vo-V2) is correct for the voltage ACROSS the 10K resistor.....BUT.....

the voltage at point V3 (which will equal V1) is the 10K voltage PLUS V2 volts

You effectively have 2 batteries in series: V2 + the 10K resistor volts

So V3 = V1 = (10/510)(Vo-V2) + V2

Now just solve for Vo

 

ok thank you very much

and on the other one (6)
What do I use for omega when solving?

 

You don't use omega.

At the instant the step occurs, the capacitor will be uncharged and so will look like a short circuit. So what happens to the voltage at the -ve terminal?

After that C will charge up dependent on the time constant, so the voltage at the -ve terminal will do what over time? So what will Vout become?

Finally, once C is fully charged, it will look like an open circuit and so what will the output do?

 

You don't use omega.

At the instant the step occurs, the capacitor will be uncharged and so will look like a short circuit. So what happens to the voltage at the -ve terminal?

After that C will charge up dependent on the time constant, so the voltage at the -ve terminal will do what over time? So what will Vout become?

Finally, once C is fully charged, it will look like an open circuit and so what will the output do?

so right when it occurs the -ve terminal voltage is: (Vin-Vout)/500k
Voltage at the -ve terminal will get smaller with time causing Vout to get smaller
When C is fully charged Vo=(500/10)*Vin

 

You got it - well done. Remember the signs though - this is an inverting amplifier so Vout will go negative as the input goes positive and will then decay back towards zero, to end up at Vin 500/10. If the input is too large the output will saturate of course.

 

so right when it occurs the -ve terminal voltage is: (Vin-Vout)/500k
Voltage at the -ve terminal will get smaller with time causing Vout to get smaller
When C is fully charged Vo=(500/10)*Vin

Looks like you have made a breakthrough. Be sure to heed thyristor's caution about signs.

hgmjr

 

* Vo=-(500/10)*Vin
so Vo=-100v