Superposition Problem- Circuits

[RoKr93]

Homework Statement

For the above circuit, find the coefficients of the linear relationship v_out = a₁v_s1 + a₂i_s2 + a₃i_s3 by superposition. Then find the power delivered to R₃ when v_s1 = 100 V, i_s2 = 2 A, and i_s3 = 4 A. Given: R₁ = 20 Ω, R₂ = 60 Ω, and R₃ = 20 Ω.

Homework Equations

Ohm's Law: V = IR

KVL: V₁ + V₂ + ... V_n = 0 for closed loops

KCL: I₁ + I₂ + ... I_n = 0 going in an out of a node

Voltage Division: V₁ = V_source * (R₁/(R₁+R₂))

The Attempt at a Solution

I think I got the first two coefficients (though I could be incorrect). I found v_out¹ = v_s1*(R₃/(R₁+R₂+R₃)) = 0.2V_s1 through voltage division. I'm pretty certain that's correct.

I also found that v_out² = I_s2*R₃ = 20I_s2, though I'm not quite as certain about that.

But I absolutely cannot wrap my head around finding v_out³ (when all independent sources are eliminated except I_s3). I understand that in that situation, the I_s2 branch essentially disappears and we are left with I_s3 in parallel with a branch with R₂ and a branch with R₁ and R₃, but we can't combine those since v_out³ must stay isolated. I'm not sure what to do.

 

[gneill]

You have a classic current divider scenario with current I_s3 being divided between two branches. You know the total resistance of each of the branches.

Alternatively, consider converting I_s3 and R₂ to their Thevenin equivalent and finding the resulting current and then the potential across R3.

 

[RoKr93]

So I want to use current division to find i₂, then multiply that by R₃ to get my voltage value?

i₂ = I_s3*(R₂/(R₁+R₂+R₃))
v_out³ = i₂R₃
v_out³ = I_s3*((R₂*R₃)/(R₁+R₂+R₃))

After substituting values in, v_out³ comes out to be 12. Unfortunately either that or one of my previous solutions is wrong- a later part of the problem states that the total v_out must be 100 V when v_s1 = 100 V, i_s2 = 2 A, and i_s3 = 4 A. And (0.2*100) + (20*2) + (12*4) is 108. Did I go about finding v_out³ the wrong way?

 

[gneill]

Check your work for the coefficient for I_s2.

 

[rezeew]

Thank you for your question. As a scientist, it is important to approach problems like these with a systematic and logical approach. It seems like you have already made some progress in finding the coefficients for the linear relationship using superposition. I will provide some additional guidance to help you find the remaining coefficient and solve the problem.

To find the third coefficient, vout3, we can use the principle of superposition again. This time, we will eliminate all independent sources except for Is3, just as you mentioned. This will leave us with two parallel branches, one with R1 and R3 and one with R2. Let's label these branches as Branch A and Branch B, respectively.

Using Ohm's Law, we can determine the voltage across each branch. For Branch A, the voltage can be calculated as vA = Is3 * R3. For Branch B, the voltage can be calculated as vB = Is3 * R2. Remember that in superposition, we treat each independent source separately, so we only consider the current flowing through each branch.

Now, we can use Kirchhoff's Voltage Law (KVL) to find the voltage at the output, vout3. Since there are no independent sources in this case, the voltage across vout3 is simply the sum of the voltages across Branch A and Branch B. This can be written as vout3 = vA + vB = Is3 * (R3 + R2).

Finally, we can use the relationship vout = a1vs1 + a2is2 + a3is3 to find the coefficient a3. Since we know the values for vout3 and Is3 when all other sources are eliminated, we can write the equation as vout3 = a3is3. This gives us a3 = (R3 + R2).

To find the power delivered to R3, we can use the formula P = IV, where I is the current through R3 and V is the voltage across R3. Using Ohm's Law, we can find the current through R3 as I = vout3 / R3. Substituting our value for vout3, we get I = Is3 * (R3 + R2) / R3. Now, we can substitute in the given values for Is3, R3, and R2 to find the power delivered to R3.

In summary, to solve