Electrical Systems Modeling (finding current and voltage drops)

[mintsnapple]

Homework Statement

Hello! I've been really unsure of whether my solutions to the first problem and part (b) of the second are right. My book gives few examples and so I've been trying to look on other websites for resources. Sorry if this is a lot, but any and all help would be appreciated.
First problem:

Second problem:

Homework Equations

Kirchoff's Current and Voltage Laqs

The Attempt at a Solution

For the first problem:
I put node 1 right above R_2. I put node 4 to the right of R_6 and above i_in(t). I put node 2 between R_2 and R_3. And I put node 3 above v_in(t) to the left of R_1.
Then, I made a system of equations, summing up the currents at each node.
$$ \sum i_1 = 1(v_1 - v_3) + 1(v_1 - v_2) + v_1 + \frac{1}{\pi}(v_1 - v_4) = 0 $$
$$ \sum i_2 = 1/3(v_2 - 0) + 2/3(v_2 - 0) + 1(v_2 - v_1) = 0 $$
$$ v_3 = v_{in} = 3V,$$, since it is connected to the source voltage.
$$ \sum i_4 = \frac{1}{\pi}(v_4 - v_1) - i_{in} = \frac{1}{\pi}(v_4 - v_1) - 2 = 0 $$
Solving this system, I got $$v_1 = 2V, v_2 = 1V, v_3 = 3V, and v_4 = 2(\pi + 1)V $$

(a). The voltage drop across R_5 is just V_1, which is 2V.
(b). The current through R_4 is related to its voltage by
$$ V_{R4} = R_4(i_{R4} - i_{in})$$. Since
$$ V_{R4} = V_2 = 1V,$$
then
$$ i_{R4} = 2 \frac{2}{3} $$.
(c). The voltage drop across R_2 is just V_1 - V_2, which is 1V.
(d). The current through can be found using Kirchoff's Voltage Law.
$$ V_{in} - R_1(i_{R1}) - V_{R3} - V_{R2} = 0 $$, so
$$ i_{R1} = 3V - 1V - 1V = 1A $$
(e). The voltage drop across the current source is just V_4, which is 2(pi + 1).

Now for part b of the second problem:
(b) We again apply KVL.
$$ v_{in} - R_1(i_R - i_{in}) = 0 $$, so
$$ R_1 i_R = v_{in} + R_1 i_{in} $$
Thank you!

 

[gneill]

Check your part (b). R3 and R4 in parallel yield 1 Ω, so combined with R2 that branch has an equivalent resistance of 2 Ω. But your v1 is 2 V, so the total branch current is 2V / 2 Ω or 1 amp. Thus the current through R4 cannot possibly exceed one ampere.

For part (d), pay attention to the defined direction of the current!

You should move your second question to a new thread; One problem per thread is the general rule.

 

[zoki85]