**E72(a) Review, Solutions**

**Problem 1:**

- R3, R5 and R6 are in series (Rx=R3+R5+R6),

and this is in parallel with R4 (Ry=Rx||R4),

and this is in series with R2 (Rz=R2+Ry),

and this is in parallel with R1,

R_{AB}=Rz||R1=636 Ohm

- R1 is in series with R2 and in parallel with R4 (Rx=(R1+R2)||R4,

and this is in series with R5 and R6 and in parallel with R3

R_{CD}=(Rx+R5+R6)||R3=727 Ohm

- The series combination of R3, R5 and R6 is in parallel with R4,
(Rx=(R3+R4+R5)||R4),

and this is in parallel with the series combination of R1 and R2,

R_{BC}=(R1+R2)||Rx = 545 Ohm

**Problem 2:**

- There are only independent sources so the Thevenin Resistance is 3k Ohm (=1k+1k+2k||2k). The Thevenin Voltage is 4.5 volts (=2.5+2).
- Thevenin Resistance is 1k Ohm (=2k||2k), and the Thevenin voltage is 3.5 volts (=2.5+1).

**Problem 3: **This one is hard, but
relates directly to some transistor concepts.

- Apply a test voltage across nodes A and B, and find resulting current

- Apply a test voltage across nodes C and D, and find resulting current.

**Problem 4: **

- As seen from Rx, the Thevenin equivalent is Vt=9.5 V(=5+4.5), Rt=100 Ohm. Therefore, for max power, Rx=100 Ohm.
- Power dissipated in Rx is equal to the power in the other three resistors. Power=4.75V*47.5mA=.23 W.

**Problem 5: **

For both parts of this problem t=RC and v(t)=v(∞)+(v(0)-v(∞))e

^{-t/t}

- (a) v(0)=0, v(∞)=1. b) v(0)=1, v(∞)=0.
- (a) v(0)=1, v(∞)=-1. b) v(0)=-2, v(∞)=0.

**Problem 6: **

a)

Exact in blue, asymptotic in red- From Bode plot (asymptotic) magnitude is unity (0 dB) and phase is 0 (-6
^{o }on exact plot), so output is equal to input (10sin(100t).- From Bode plot, magnitude is 0.1 (-20 dB) and phase is -90, so output is sin(10000t-90°)=-cos(10000t).
- This is a lowpass filter

b)

Exact in blue, asymptotic in red- From Bode plot, magnitude is 0.1(-20 dB) and phase is 90, so output is sin(100t+90°)=cos(100t).
- From Bode plot, magnitude is unity and phase is 0, so output is 10sin(10000t).
- This is a highpass filter

**Problem 7: **

For both parts of this problem t=RC and v(t)=v(∞)+(v(0)-v(∞))e

^{-t/t}

- When the switch is opened
- the resistance "seen" by the capacitor is R=R3||(R2+R1)=667 Ohm.
- v(0)=V1*R3/(R3+R1)
- v(∞)=V1*R3/(R3+R2+R1)

- When the switch is closed
- the resistance "seen" by the capacitor is R=R3||R1=500 Ohm.
- v(0)=V1*R3/(R3+R2+R1)
- v(∞)=V1*R3/(R3+R1)

**Problem 8: **

For both parts of this problem t=RC and v(t)=v(∞)+(v(0)-v(∞))e

^{-t/t}

- When the switch is opened
- the resistance "seen" by the capacitor is R=R1+R2
- v(0)=(V1)*1/3
- v(∞)=V1

We want to solve for t such that:

- When the switch is closed
- the resistance "seen" by the capacitor is R=R1
- v(0)=(V1)*2/3
- v(∞)=0

We want to solve for t such that

Note that the time constant is different for the two parts, though the form of the solution is not.

**Problem 9: **

**Problem 10: **

Recall that for a generic first order system with a step input that:

so

**Problem 11: **

- At low frequencies, inductor is short, output is zero. At high frequencies, capacitor is short, output is zero. Output must be non-zero at intermediate frequencies, therefore this is a bandpass filter.
- At s=0 and s=∞, the magnitude of the transfer function is zero.
- It decreases.

**Problem 12: **

- At low frequencies, capacitor is open circuit, output=input. At high frequencies, capacitor is short circuit, output=0. Therefore this is a lowpass filter.
- At s=0, Xfer function=1. At s=∞, the Xfer function=0.
- It increases.

**Problem 13: **

- The expression below shows Vo=0 if r=0

- see (i).
- if r<<R, then

**Problem 14: **

- 2.5V+2V=4.5V
- 2.5V+1V=3.5V

**Problem 15: **

We take the positive root since resistance must be positive.