# Stuck on finding the natural response

#### p75213

Joined May 24, 2011
70
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.

#### t_n_k

Joined Mar 6, 2009
5,455
When the independent current source is turned of we get a short circuit from the top to bottom.
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=∞?

#### p75213

Joined May 24, 2011
70
At t=0+ i(0+)=i(0-)=0
At t=$$\infty$$ i=2

#### t_n_k

Joined Mar 6, 2009
5,455
And - at t=0+ all of the 2A flows into the left hand branch.

#### p75213

Joined May 24, 2011
70
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.

#### t_n_k

Joined Mar 6, 2009
5,455
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.