# Stuck on finding the natural response

Discussion in 'Homework Help' started by p75213, Nov 22, 2011.

1. ### p75213 Thread Starter Member

May 24, 2011
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Hello,
I am attempting this problem but have got stuck on finding the natural response to the circuit at t>0.
When the independent current source is turned of we get a short circuit from the top to bottom. I am struggling to come up with the appropriate KCL and KVL equations.

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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No we don't. Presumably you mean turned on....?

Through which elements does the current flow at t=0+ when the switch first closes? The source will adopt whatever voltage is required to match its terminal load conditions.

Through which elements does the current flow at t=∞?

3. ### p75213 Thread Starter Member

May 24, 2011
39
0
At t=0+ i(0+)=i(0-)=0
At t=$\infty$ i=2

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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And - at t=0+ all of the 2A flows into the left hand branch.

5. ### p75213 Thread Starter Member

May 24, 2011
39
0
For the LHS I used v(t)=v($\infty$)+(v(0)-v($\infty$))e$^{-t/RC}$
And for the RHS i(t)=i($\infty$)+(i(0)-i($\infty$))e$^{-tR/L}$

The result is the same as the answer given however in both cases I get the exponent = -2t rather than -5t.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The solution is somewhat convoluted ...

Denote the current in the left branch as i1(t) and the voltage at the top node as v1(t)

$v_1(t)=10i_1+20\int i_1 dt=4i+2\frac{di}{dt}$

also

$i=2-i_1$ & $\frac{di}{dt}=-\frac{di_1}{dt}$

$v_1(t)=10i_1+20\int i_1 dt=4(2-i_1)-2\frac{di_1}{dt}$

$10i_1+20\int i_1 dt=4(2-i_1)-2\frac{di_1}{dt}$

and

$\frac{di_1}{dt}+7i_1+10\int i_1 dt=4$

$\frac{d^2i_1}{dt^2}+7\frac{di_1}{dt}+10 i_1=0$

Initial conditions: i1(0)=2, i(0)=0, v1(0)=20

The real roots of the CE are -5 and -2

So

$i_1(t)=K_1e^{-2t}+K_2e^{-5t}$

Using i1(0)=2

$K_1+K_2=2$

Using v1(0)=20 and i1 as defined above, along with the relationship

$v_1(t)=8-4i_1-2\frac{di_1}{dt}$

The values of K1 & K2 can then be found and so on ...

As it turns out the value of K1=0 which accounts for the existence of the exponential solution being in -5t only.

This problem is far more readily solved by the use of Laplace transforms.

7. ### p75213 Thread Starter Member

May 24, 2011
39
0
Your right - it is convoluted. Haven't got to Laplace Transforms yet. That's towards the end of the book. I notice Khan Academy has included them in the differentials section. This is a good free learning resource for anybody interested.