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Circuit Analysis

electronicsLearner77

Jul 2, 2015
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I want to do circuit analysis for below circuit
upload_2022-4-23_22-49-30.png
"I" is the current source of type sine wave, The equations are

5 - 10 * i2 - 1000(i+i1+i2) = vout -> eq1
Integral(i1) dt * 10^6 = i2 -> eq2
I +i1 = i2 -> eq3
Please help me to know if the above 3 equations are correct?
 

Harald Kapp

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You should make it a habit to include units in your calculations. It is also helpful for other s who read your equations to first set up the equation with placeholders. This way we see which component is used where in the equation. Replace the placeholders by values in a second step. Also use the same labesl in the schematic and in the equations. In this case use out, not vout, or label the output vout in the schematic. Or use V(out) which would correspond to the way this is handled in LTSPICE.
Example
i is in amps, v is in volts r is in Ω. so the equation should read
step 1: V1 - R1 × i2 + R4 × (I+i1+i2) = V(out)
step 2: 5 V - 10 Ω × i2 -1000 Ω × (I+i1+i2) = V(out)

However, this is not correct.
eq1.:What about current through C2? If I is an AC source, there will be current through C2 which will influence the output voltage. If I is a DC source, you will have to consider the same integral for C2 as you do for C1 in eq.2.

eq2: Integral(i1) dt * 10^6 = i2 will be correct assuming 10^6 = 1/C1

eq3 states I + i1 = i2. Why should that be so? Consider the simple case I = 0 A. That makes i1 = i2 and it is obvious that this is not the case.

It is also helpful (though, as I recognize, not common in anglo-american schematics) to annotate the direction of voltages and currents. For v1 this is clear by the '+' and '-' in the voltage source's symbol. Equally for I by the arrow within the symbol. But for i1 and i2 this is missing. When you add the arrows like so:
upload_2022-4-25_7-57-27.png
This makes it much easier to see where values have to be added (same direction of arrow) or subtracted (opposite direction of arrow).
You don't know in advance which direction current flows, you say? No problem. You can chose an arbitrary direction. Just be consistent. If the chosen direction was "wrong" (note there is no wrong), the resulting value (e.g. current) will simply be negative which means it is opposite to the arrow's direction.
 
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