Revisions to Calculating Thevenin's resistance in circuits with a Dependent Source

deleted 27 characters in body

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edited Oct 27, 2024 at 2:36

periblepsis

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Let me draw the circuit and show the test current:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The solution is as follows:

idep = vx/r2 cccs1 = 30*idep kclx = Eq( vx/r1 + vx/r2 + vx/r3, vs/r1 + vy/r3 ) kcly = Eq( vy/r3 + vy/r4 + vy/r5 + cccs1, va/r5 + vx/r3 ) kcla = Eq( va/r5 + va/r6, vy/r5 + cccs1 + itest ) vars = { vs:40, r1:2e3, r2:20e3, r3:5e3, r4:50e3, r5:10e3, r6:40e3 } ans = solve( [ kclx, kcly, kcla ], [ vx, vy, va ] ) va = ans[ va ].subs( vars ) va.subs( itest, 0 ) 280.000000000000 ans[ va ].subs(vars).subs( itest, 1 ) 20280.0000000000

So you know that \$V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$V^{^{1\:\text{A}}}_{_\text{A}}=20280\:\text{V}\$.

It follows that \$V_{_\text{TH}}=V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$R_{_\text{TH}}=\frac{V^{^{1\:\text{A}}}_{_\text{A}}\,-\,V^{^{0\:\text{A}}}_{_\text{A}}}{1\:\text{A}\,-\,0\:\text{A}}=20\:\text{k}\Omega\$.

And here's an LTspice run just to verify the above results:

As you can see, there's no difficulty using an external current source.

Let me draw the circuit and show the test current:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The solution is as follows:

idep = vx/r2 cccs1 = 30*idep kclx = Eq( vx/r1 + vx/r2 + vx/r3, vs/r1 + vy/r3 ) kcly = Eq( vy/r3 + vy/r4 + vy/r5 + cccs1, va/r5 + vx/r3 ) kcla = Eq( va/r5 + va/r6, vy/r5 + cccs1 + itest ) vars = { vs:40, r1:2e3, r2:20e3, r3:5e3, r4:50e3, r5:10e3, r6:40e3 } ans = solve( [ kclx, kcly, kcla ], [ vx, vy, va ] ) ans[ va ].subs(vars).subs( itest, 0 ) 280.000000000000 ans[ va ].subs(vars).subs( itest, 1 ) 20280.0000000000

So you know that \$V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$V^{^{1\:\text{A}}}_{_\text{A}}=20280\:\text{V}\$.

It follows that \$V_{_\text{TH}}=V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$R_{_\text{TH}}=\frac{V^{^{1\:\text{A}}}_{_\text{A}}\,-\,V^{^{0\:\text{A}}}_{_\text{A}}}{1\:\text{A}\,-\,0\:\text{A}}=20\:\text{k}\Omega\$.

And here's an LTspice run just to verify the above results:

As you can see, there's no difficulty using an external current source.

Let me draw the circuit and show the test current:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The solution is as follows:

idep = vx/r2 cccs1 = 30*idep kclx = Eq( vx/r1 + vx/r2 + vx/r3, vs/r1 + vy/r3 ) kcly = Eq( vy/r3 + vy/r4 + vy/r5 + cccs1, va/r5 + vx/r3 ) kcla = Eq( va/r5 + va/r6, vy/r5 + cccs1 + itest ) vars = { vs:40, r1:2e3, r2:20e3, r3:5e3, r4:50e3, r5:10e3, r6:40e3 } ans = solve( [ kclx, kcly, kcla ], [ vx, vy, va ] ) va = ans[ va ].subs( vars ) va.subs( itest, 0 ) 280.000000000000 va.subs( itest, 1 ) 20280.0000000000

So you know that \$V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$V^{^{1\:\text{A}}}_{_\text{A}}=20280\:\text{V}\$.

It follows that \$V_{_\text{TH}}=V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$R_{_\text{TH}}=\frac{V^{^{1\:\text{A}}}_{_\text{A}}\,-\,V^{^{0\:\text{A}}}_{_\text{A}}}{1\:\text{A}\,-\,0\:\text{A}}=20\:\text{k}\Omega\$.

And here's an LTspice run just to verify the above results:

As you can see, there's no difficulty using an external current source.

deleted 27 characters in body

Source Link

edited Oct 27, 2024 at 2:31

periblepsis

21.3k
1
13
41

Let me draw the circuit and show the test current:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The solution is as follows:

idep = vx/r2 cccs1 = 30*idep kclx = Eq( vx/r1 + vx/r2 + vx/r3, vs/r1 + vy/r3 ) kcly = Eq( vy/r3 + vy/r4 + vy/r5 + cccs1, va/r5 + vx/r3 ) kcla = Eq( va/r5 + va/r6, vy/r5 + cccs1 + itest ) vars = { vs:40, r1:2e3, r2:20e3, r3:5e3, r4:50e3, r5:10e3, r6:40e3 } ans = solve( [ kclx, kcly, kcla ], [ vx, vy, va ] )[ ans[ va ].subs(vars).subs( itest, 0 ) 280.000000000000 solve( [ kclx, kcly, kcla ], [ vx, vy, va ] )[ans[ va ].subs(vars).subs( itest, 1 ) 20280.0000000000

So you know that \$V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$V^{^{1\:\text{A}}}_{_\text{A}}=20280\:\text{V}\$.

It follows that \$V_{_\text{TH}}=V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$R_{_\text{TH}}=\frac{V^{^{1\:\text{A}}}_{_\text{A}}\,-\,V^{^{0\:\text{A}}}_{_\text{A}}}{1\:\text{A}\,-\,0\:\text{A}}=20\:\text{k}\Omega\$.

And here's an LTspice run just to verify the above results:

As you can see, there's no difficulty using an external current source.

Let me draw the circuit and show the test current:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The solution is as follows:

idep = vx/r2 cccs1 = 30*idep kclx = Eq( vx/r1 + vx/r2 + vx/r3, vs/r1 + vy/r3 ) kcly = Eq( vy/r3 + vy/r4 + vy/r5 + cccs1, va/r5 + vx/r3 ) kcla = Eq( va/r5 + va/r6, vy/r5 + cccs1 + itest ) vars = { vs:40, r1:2e3, r2:20e3, r3:5e3, r4:50e3, r5:10e3, r6:40e3 } solve( [ kclx, kcly, kcla ], [ vx, vy, va ] )[ va ].subs(vars).subs( itest, 0 ) 280.000000000000 solve( [ kclx, kcly, kcla ], [ vx, vy, va ] )[ va ].subs(vars).subs( itest, 1 ) 20280.0000000000

So you know that \$V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$V^{^{1\:\text{A}}}_{_\text{A}}=20280\:\text{V}\$.

It follows that \$V_{_\text{TH}}=V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$R_{_\text{TH}}=\frac{V^{^{1\:\text{A}}}_{_\text{A}}\,-\,V^{^{0\:\text{A}}}_{_\text{A}}}{1\:\text{A}\,-\,0\:\text{A}}=20\:\text{k}\Omega\$.

And here's an LTspice run just to verify the above results:

As you can see, there's no difficulty using an external current source.

Let me draw the circuit and show the test current:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The solution is as follows:

idep = vx/r2 cccs1 = 30*idep kclx = Eq( vx/r1 + vx/r2 + vx/r3, vs/r1 + vy/r3 ) kcly = Eq( vy/r3 + vy/r4 + vy/r5 + cccs1, va/r5 + vx/r3 ) kcla = Eq( va/r5 + va/r6, vy/r5 + cccs1 + itest ) vars = { vs:40, r1:2e3, r2:20e3, r3:5e3, r4:50e3, r5:10e3, r6:40e3 } ans = solve( [ kclx, kcly, kcla ], [ vx, vy, va ] ) ans[ va ].subs(vars).subs( itest, 0 ) 280.000000000000 ans[ va ].subs(vars).subs( itest, 1 ) 20280.0000000000

So you know that \$V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$V^{^{1\:\text{A}}}_{_\text{A}}=20280\:\text{V}\$.

It follows that \$V_{_\text{TH}}=V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$R_{_\text{TH}}=\frac{V^{^{1\:\text{A}}}_{_\text{A}}\,-\,V^{^{0\:\text{A}}}_{_\text{A}}}{1\:\text{A}\,-\,0\:\text{A}}=20\:\text{k}\Omega\$.

And here's an LTspice run just to verify the above results:

As you can see, there's no difficulty using an external current source.

added 75 characters in body

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edited Oct 26, 2024 at 7:31

periblepsis

21.3k
1
13
41

Let me draw the circuit and show the test current:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The solution is as follows:

idep = vx/r2 cccs1 = 30*idep kclx = Eq( vx/r1 + vx/r2 + vx/r3, vs/r1 + vy/r3 ) kcly = Eq( vy/r3 + vy/r4 + vy/r5 + cccs1, va/r5 + vx/r3 ) kcla = Eq( va/r5 + va/r6, vy/r5 + cccs1 + itest ) vars = { vs:40, r1:2e3, r2:20e3, r3:5e3, r4:50e3, r5:10e3, r6:40e3 } solve( [ kclx, kcly, kcla ], [ vx, vy, va ] )[ va ].subs(vars).subs( itest, 0 ) 280.000000000000 solve( [ kclx, kcly, kcla ], [ vx, vy, va ] )[ va ].subs(vars).subs( itest, 1 ) 20280.0000000000

So you know that \$V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$V^{^{1\:\text{A}}}_{_\text{A}}=20280\:\text{V}\$.

It follows that \$V_{_\text{TH}}=V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$R_{_\text{TH}}=\frac{V^{^{1\:\text{A}}}_{_\text{A}}\,-\,V^{^{0\:\text{A}}}_{_\text{A}}}{1\:\text{A}\,-\,0\:\text{A}}=20\:\text{k}\Omega\$.

And here's an LTspice run just to verify the above results:

As you can see, there's no difficulty using an external current source.

Let me draw the circuit and show the test current:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The solution is as follows:

idep = vx/r2 cccs1 = 30*idep kclx = Eq( vx/r1 + vx/r2 + vx/r3, vs/r1 + vy/r3 ) kcly = Eq( vy/r3 + vy/r4 + vy/r5 + cccs1, va/r5 + vx/r3 ) kcla = Eq( va/r5 + va/r6, vy/r5 + cccs1 + itest ) vars = { vs:40, r1:2e3, r2:20e3, r3:5e3, r4:50e3, r5:10e3, r6:40e3 } solve( [ kclx, kcly, kcla ], [ vx, vy, va ] )[ va ].subs(vars).subs( itest, 0 ) 280.000000000000 solve( [ kclx, kcly, kcla ], [ vx, vy, va ] )[ va ].subs(vars).subs( itest, 1 ) 20280.0000000000

So you know that \$V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$V^{^{1\:\text{A}}}_{_\text{A}}=20280\:\text{V}\$.

It follows that \$V_{_\text{TH}}=V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$R_{_\text{TH}}=\frac{V^{^{1\:\text{A}}}_{_\text{A}}\,-\,V^{^{0\:\text{A}}}_{_\text{A}}}{1\:\text{A}\,-\,0\:\text{A}}=20\:\text{k}\Omega\$.

And here's an LTspice run just to verify the above results:

Let me draw the circuit and show the test current:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The solution is as follows:

idep = vx/r2 cccs1 = 30*idep kclx = Eq( vx/r1 + vx/r2 + vx/r3, vs/r1 + vy/r3 ) kcly = Eq( vy/r3 + vy/r4 + vy/r5 + cccs1, va/r5 + vx/r3 ) kcla = Eq( va/r5 + va/r6, vy/r5 + cccs1 + itest ) vars = { vs:40, r1:2e3, r2:20e3, r3:5e3, r4:50e3, r5:10e3, r6:40e3 } solve( [ kclx, kcly, kcla ], [ vx, vy, va ] )[ va ].subs(vars).subs( itest, 0 ) 280.000000000000 solve( [ kclx, kcly, kcla ], [ vx, vy, va ] )[ va ].subs(vars).subs( itest, 1 ) 20280.0000000000

So you know that \$V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$V^{^{1\:\text{A}}}_{_\text{A}}=20280\:\text{V}\$.

It follows that \$V_{_\text{TH}}=V^{^{0\:\text{A}}}_{_\text{A}}=280\:\text{V}\$ and that \$R_{_\text{TH}}=\frac{V^{^{1\:\text{A}}}_{_\text{A}}\,-\,V^{^{0\:\text{A}}}_{_\text{A}}}{1\:\text{A}\,-\,0\:\text{A}}=20\:\text{k}\Omega\$.

And here's an LTspice run just to verify the above results:

As you can see, there's no difficulty using an external current source.

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answered Oct 26, 2024 at 7:25

periblepsis

21.3k
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