circuit analysisoperational-amplifier
I am trying to find \$v_0\$ in this circuit (voltage across \$R4\$):
simulate this circuit – Schematic created using CircuitLab
The issue I am having is that I would have known that the current going into the 16 ohm resistor would have been the same as the one going in the negative terminal, but that's only if they are both going into the negative terminal. I know that its 0 volts to ground on the right side. And 0 volts in between the + and -.
Can someone help me find \$v_0\$?
No, you will get the same result in both cases, which is 0. You can see this is true by definition since the output is grounded.
Even with the output not grounded, flipping the capacitor won't make any difference. The polarity of a capacitor does not decide the polarity of the voltage applied to it. It only decides which polarity of applied voltage will cause the capacitor to blow up. This is assuming a type of cap to which polarity matters, such as electrolytic or tantalum.
In your circuit, the right side of the capacitor will always be at or lower than the left side due to the fact that only positive voltage is applied into the integrator. This particular integrator inverts as a side effect of its topology. Note that this circuit won't work unless the opamp has a negative supply. Without a negative supply, the opamp can't swing below ground and therefore can't make the expected output voltage.
After step 6, your circuit will be as shown below.
simulate this circuit – Schematic created using CircuitLab
Which can be further reduced as a 16V battery in series with 3k and 5k resistors.
Now, you will have 10V across this 5k resistance, ie, 10V across each parallel 10k. Splitting 10k into 6k+4k, you will have 4V across 4k resistance. From the polarity of v1 marked, V1= -4V.
Best Answer
Golden Rules. With negative feedback in place...
Because no current flows into the positive terminal, current must flow from the voltage source, through the series resistors 16 ohm and 24 ohm. So you should be able to figure out that the voltage at the input terminal is 7.5 * 24 / 40 = 4.5V. It follows that the voltage at the negative terminal = 4.5V. Now you've got another voltage divider from the negative terminal through 8 ohms in series with 12 ohms. The voltage across the 12 ohm resistor is therefore 4.5 * 12 / 20 = 2.7V.