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## Homework Statement

2. Homework Equations

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Consider the following circuit where:

$$R_A=330 k\Omega$$

$$R_B=220 k\Omega$$

$$C= 2.2 nF $$

$$v_1(t)=6u(t) V$$

$$ i_2(t)=-10u(t-t_2) \mu A$$

$$t_2 = 12 ms $$

All the currents go clockwise (with the obvious exception of the current source).

The first part of the problem asked for the equations of vc and vb. I had no trouble on getting to the correct answers so I won't post the developed answer:

$$v_C(t) = 0V$$

if $$t<0$$

$$v_C(t) = 6(1-e^{-\frac{t}{1.21 ms}}) V$$

if $$0\leq t <12 ms$$

$$v_C(t) = 2.7+3.3e^{-\frac{t-12ms}{1.21 ms}} V$$

if $$t \geq 12 ms$$

$$v_B(t) = 0V$$

if $$t<0$$

$$v_B(t) = 2.4e^{-\frac{t}{1.21 ms}} V$$

if $$0\leq t <12 ms$$

$$v_B(t) = -1.32e^{-\frac{t-12ms}{1.21 ms}} V$$

if $$t \geq 12 ms$$

Now I have question in the next part of the problem. It asks me the instant of time of the maximum dissipated power by Joule heating, and the value of that power.

3. The Attempt at a Solution

3. The Attempt at a Solution

So we have that the power P

$$P= R_A (i_A)^2 + R_B (i_B)^2$$

The equations for iB are pretty straight forward. We only need to take the equations of vB and divide it by the value of the resistance RB

$$i_B(t) = 0\mu A$$

if $$t<0$$

$$i_B(t) = 10.9e^{-\frac{t}{1.21 ms}} \mu A$$

if $$0\leq t <12 ms$$

$$i_B(t) = -6e^{-\frac{t-12ms}{1.21 ms}} \mu A$$

if $$t \geq 12 ms$$

The equations of iA are less obvious.

The first case is direct:

$$i_A(t) = 0\mu A$$

if $$t<0$$

On the second case the currents through both resistors are the same (since the current source is still not working).

$$i_A(t) = 10.9e^{-\frac{t}{1.21 ms}} \mu A$$

if $$0\leq t <12 ms$$

For the third case, if we take KCL applied to the upper node:

$$ -i_A+i_B-(-10\mu A)=0$$

Using that we get to

$$i_A(t) = 10-6e^{-\frac{t-12ms}{1.21 ms}} \mu A$$

if $$t \geq 12 ms$$

Now looking to this equations and by verifying them in key points (0+, 12-, 12+, and infinity), I concluded that the maximum in both sets is in 0+ and therefore the power is (since in this case both currents are equal):

$$P= (R_a+R_B) \times (i_A^2)=65.35 \mu W $$

However my textbook chooses instead the instant 12+ and says that the power is

$$P= =92.4 \mu W $$

However if, by the results of my equations I try that instant I get to a different answer (13.2 mW).

I already verified my equations three times and I don't find the mistake. Should I be calculating Joule heating in other components (besides the 2 resistors). What am I missing?