EE301 6 Week Exam:�Review

Jeopardy!

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• What would be the parity bits for the LRC? Use even parity. What is the code rate?

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1 1 0 0 _

0 1 1 1 _

1 0 1 0 _

_ _ _ _ _

The Law’s $100

Ω(not Omega) $100

Find the output voltage V_ab across the R₂ resistor.

Source Conversion $100

Use source transformation to determine v_o.

Solve for node voltage V_b

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Nodal Nonsense $100

Find the Thévenin equivalent circuit external to R_LD. Determine I_LD and V_LD when R_LD = 2.5 Ω.

Thevenin Pain $100

How’s it going Cap (& Inductors) $100

The circuit below will not have the same charging equation due to the voltage divider in the system.

After the switch is shut: (CHARGING)

a. Transform the circuit into the Thèvenin equivalent as seen by the capacitor.

b. Determine the charging constant.

c. Write the charging equation for v_c(t).

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Calculate the power to the heater using all three electrical formulas.

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The Law’s $200

Ω(not Omega) $200

Determine the R_bc of this network:

Source Conversion $200

Use source transformation to determine v_o.

Find the voltage at node c and the unknown current.

Nodal Nonsense $200

Find the Thévenin equivalent circuit external to R_LD and determine I_LD.

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Thevenin Pain $200

In the circuit below, the switch is initially open and conditions are at steady-state.

After the switch is shut, determine:

1. how long it will take for the inductor to reach a steady-state condition (>99% of final current).
2. Write the equation for the V_L(t) & i_L(t).
3. Find the Energy stored in the Inductor.

How’s it going Cap (& Inductors) $200

The Law’s $300

Suppose you are at home and use 3 100-W lamps for 3 hours and an Xbox 500W for 2.5 hours. The TV consumes 180 W for 2.5 hours. At $0.148 per kilowatt-hour, how much will this cost you?

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Determine the R_T of this network:

Ω(not Omega) $300

1^st: Label Nodes

2^nd: Redraw Circuit

Since both ends of the 7Ω resistor connect to node a, the 7Ω resistor doesn’t contribute to RT

3^rd: Solve for RT

Use source transformation and parallel current source rule to simplify the circuit and determine v_o.

Source Conversion $300

Nodal Nonsense $300

Solve for node voltage V_b

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Thevenin Pain $300

1. Find the Thévenin equivalent circuit to the left of terminals a-b.
2. Calculate the maximum power transfer to the load if R_LD = R_TH.
3. Determine the power dissipated by R_LD for load resistances of 2Ω and 6Ω.

How’s it going Cap(&Inductors) $300

a

b

6 Ω

3 Ω

2 F

3V

60V

The switch has been in position a for a long time. At time t=0, it moves to position b.

1. Calculate the steady state voltage across the capacitor at position a
2. Calculate the steady state voltage across the capacitor at position b
3. Calculate the time constant (tau) in position b
4. Find the voltage v(t) across the capacitor for t>0
5. Find the energy stored by the capacitor at t=60 secs

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1. V_c at SS = V across R₂. Use VDR understanding Capacitor is an open circuit. (Think Thevenin)

V_c at SS = E x R₂/R_T

R_T = 6 Ω + 3 Ω = 9 Ω

V_c = 3V x (3 Ω / 9 Ω) = 1V

b. Similar to a.

V_c = 60V x (3 Ω / 9 Ω) = 20V

3. Tau = RC. Think Thevenin for R

R = 6Ωll3Ω = 2Ω

Tau = 2Ω x 2F = 4sec

4. V_c(t) = V_f + (V_i – V_f)e^-t/tau

V_c(t) = 20+(1-20)e^-t/4sec = 20-19e^-t/4V

5. E = ½ CV²

E = ½ (2F)(20V)² = 400J

The Law’s $400

Ω(not Omega) $400

Determine I₂. Find V_a, V_bc

Source Conversion $400

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DAILY DOUBLE

Nodal Nonsense $400

Using Nodal Analysis,

solve for V_ab.

Thevenin Pain $400

A stereo is rated for max output power of 150W per channel when R_LD = 8Ω

1. Sketch the Thévenin Equivalent circuit.
2. What would the output power be with two 8Ω speakers as the load and are connected in parallel to one of the channels?

How’s it going Cap(&Inductors) $400

The circuit shown below has been in operation with the switch shut for a long time. The switch opens at time t = 0, determine:

1. How long it will take for the inductor to discharge.
2. Write the discharge equation for the V_L, i_L,.

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Source Conversion $400

Determine I₂ and V₃.

I₂ = -2A due to direction of current flow

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V₁ = V₂ = V₃ = -40V due to the polarity of voltage drop