# Electronics (Circuit Analysis) Questions

1. Jan 20, 2014

### stefan10

1. The problem statement, all variables and given/known data

Problem A
http://imagizer.imageshack.us/v2/800x600q90/855/6v0c.png [Broken]

Problem B
http://imagizer.imageshack.us/v2/800x600q90/853/1m2c.png [Broken]
http://imagizer.imageshack.us/v2/800x600q90/34/wfe4.png [Broken]

2. Relevant equations

Kirchhoff's Rules/Ohm's Law

Voltage Divider Formula

$$V_{ab} = (V_{0}*R_{2})/(R_{1} + R_{2})$$

3. The attempt at a solution

Problem A:

I'm unsure how much information is sufficient for what the question is asking. I derived these equations, and relations for different currents:

$$I_{1} = I_{AB}=I_{EF}$$
$$I_{3} = I_{BC}=I_{CD} = I_{DE}$$
$$I_{2}=I_{BE}$$

Loop 1 (Clockwise)
$$0 = V_{0} - I_{1}R - I_{2}R - I_{1}R = V_{0} - 2I_{1}R - I_{2}R$$

Loop 2 (Clockwise)
$$0=V_{0} - 1/2 V_{0} - I_{3}R \ \mbox{(used Voltage Divider Eq for the voltage of GH)}$$
$$\Rightarrow 0 = V_{0} - 2I_{3}R$$

Node Equation

$$0 = I_{1} - I_{3} - I_{2}$$

Solving these equations I get

$$3 I_{1} = I_{3}$$
$$I_{2} = -2 I_{1}$$
$$I_{3} = - (3/2) I_{2}$$

a.) V_{GH} = V_{0}/2
b.) I_{GH} = V_{0}/2R
c.) I'm not sure how to find this.

Problem B:

V Thevenin is equal to V_{GH} right? That means V_{th} = V_{GH} = V_{0}/2

R Thevenin is R parallel to R, which can be found with 1/R_{th} = 1/R + 1/R which give us R_{th} = 2/R.

I'm not as confident of Problem B as I am of Problem A.

Last edited by a moderator: May 6, 2017
2. Jan 20, 2014

### Staff: Mentor

This problem is easiest to solve by using series & parallel combinations of resistors.

For a), collapse the circuit from right to left, combining resistors as you go. You will end up with only one resistor -- this tells you the overall current from the voltage supply. Then re-build the original circuit from left to right, calculating currents and voltages as you go.

3. Jan 20, 2014

### BvU

1. In problem 13, did you get something like V_be = V_o/3 ?
If so, don't you think V_gh = Vo/2 is a little strange ?
I do not understand your 1/2 V_o in loop 2.