Chapter 4: Techniques of Circuit Analysis

Open practice · AI allowed

Practice sheet: Chapter 4, Part 2, Thévenin Equivalents, Superposition and Maximum Power

This sheet is not marked for correctness, and AI tools are allowed. The Part 2 device-free quiz, at the start of week 10, session 1 (Thu 10 Dec), is built from twins of these questions, so make sure you can do each one on your own.

Two-tier items ask for an answer and a reason. Both must be right.
Predict first: where asked, make your prediction before you calculate.
Confidence: mark each question Sure or Unsure. It is not marked; it shows you what to revise.
Conventions: $v_{ab}$ is the voltage of terminal a measured from terminal b. A Thévenin equivalent is $V_{\text{Th}}$ in series with $R_{\text{Th}}$; a Norton equivalent is $I_N$ in parallel with $R_N = R_{\text{Th}}$. To deactivate a voltage source, replace it with a short circuit; to deactivate a current source, replace it with an open circuit. Never deactivate a dependent source. “Power developed” means power delivered by a source. Check every answer: $V_{\text{Th}}/R_{\text{Th}}$ must equal the short-circuit current, and the absorbed powers (passive sign convention) must add to zero.

Prefer paper? Open the printable PDF.

Which questions practise which skill

1Transform a voltage source with a series resistor into a current source with a parallel resistor, with the arrow the right way, and back S1 S2 S3 S7 S11
2Use superposition to find a current, and explain why powers cannot be added S4 S5
3Find $V_{\text{Th}}$, $i_{\text{sc}}$ and $R_{\text{Th}}$, and draw the Thévenin and Norton equivalents S6 S7 S8 S9 S10 S11 S13 S14 S15
4Find $R_{\text{Th}}$ by deactivating the independent sources, and with a test source when the circuit has a dependent source S6 S8 S9 S12 S13
5Choose the load for maximum power transfer and find that power and the efficiency S10 S13 S14 S15
6Check an equivalent with $i_{\text{sc}} = V_{\text{Th}}/R_{\text{Th}}$ S12 S16

S1A source transformationTwo-tier

Which current source and resistor are equivalent at terminals a and b? (“Toward a” gives the direction of the source arrow.)

Answer

a)3 A toward b, 4 Ω in parallelb)3 A toward a, 4 Ω in parallelc)48 A toward a, 4 Ω in paralleld)3 A toward a, 4 Ω in series

Reason

i)A current-source arrow points from + to −, the way current flows through a resistor.ii)The resistor must stay in series to limit the current.iii)Both circuits must give the same short-circuit current, 12/4 = 3 A out of terminal a, and the same resistance seen from a and b.iv)The source current is the voltage times the resistance.

Follow-up: with a and b open, the 4 Ω absorbs 0 W in one circuit and 36 W in the other. Why does that not contradict the equivalence?

Work this out on paper, then compare with the explanation.

S2A chain of source transformations

a) Use successive source transformations to find $i_o$.

b) Check your answer with the node-voltage method. With the bottom wire as the reference, the node voltage above the 3 Ω

b) …and the node voltage above the 15 Ω

S3Resistors you may remove, and resistors you may notPredict first

Predict first: does the 50 Ω change $v_o$? Does the 10 Ω?

a)Both change $v_o$b)Only the 50 Ω changes $v_o$c)Only the 10 Ω changes $v_o$d)Neither changes $v_o$

a) Use source transformations to find $v_o$.

b) Find the power developed by the 100 V source.

c) Find the power developed by the 2 A source.

d) Which resistors could you drop in (a), and why do they matter in (b) and (c)?

Work this out on paper, then compare with the explanation.

S4Superposition

As in Nilsson Example 4.22.
a) Find $i_1$ to $i_4$ with the 60 V source alone.
b) Find them with the 6 A source alone.
c) Add the two sets to get the actual currents.
d) Find the power in the 6 Ω resistor. Compare it with the sum of the powers you would get from (a) and (b) separately.

a) 60 V source alone: $i_1$

a) $i_2$

a) $i_3$

a) $i_4$

b) 6 A source alone: $i_1$

b) $i_2$

b) $i_3$

b) $i_4$

c) Actual currents: $i_1$

c) $i_2$

c) $i_3$

c) $i_4$

d) The power in the 6 Ω resistor

d) The sum of the powers in the 6 Ω from (a) and (b) separately

S5Does power add?Two-tier

With only source 1 on, 2 A flows down through a 5 Ω resistor. With only source 2 on, 1 A flows down through it. What power does the resistor absorb with both sources on?

Answer

a)25 Wb)20 Wc)45 Wd)5 W

Reason

i)The powers add, because energy is conserved.ii)Only the larger source sets the power; the smaller one is masked.iii)Currents from two different sources subtract, because they oppose each other.iv)The currents add, with their directions, to one total current, and the power depends on the square of that total.

S6A Thévenin equivalent from $v_{\text{oc}}$ and $i_{\text{sc}}$

As in Nilsson Example 4.14.

a) Find $V_{\text{Th}}$, the open-circuit voltage $v_{ab}$.

b) Short a to b. Find $i_{\text{sc}}$

b) …then $R_{\text{Th}} = V_{\text{Th}}/i_{\text{sc}}$

c) Check $R_{\text{Th}}$ by deactivating both sources: the resistance seen from a and b

d) A 15 Ω load is connected between a and b. Find its current

d) …and its power

S7A Norton equivalent by source transformations

a) Use source transformations to find the Norton equivalent, then the Thévenin equivalent. $I_N$

a) The Norton source arrow points

a)toward ab)toward b

a) $R_N = R_{\text{Th}}$

a) $V_{\text{Th}}$

b) With a and b open, a current still flows around the two branches. Which source delivers power?

a)The 50 V sourceb)The 30 V source

b) How much power does it deliver?

b) How much power does the other source absorb?

S8A Thévenin equivalent with a dependent source

As in Nilsson Example 4.16.

a) Find $V_{\text{Th}}$.

b) Find $i_{\text{sc}}$

b) …and $R_{\text{Th}}$

c) Check $R_{\text{Th}}$ with a test source: deactivate the 24 V source only, connect a 1 A current source from b to a (so 1 A enters the circuit at a) and find $v_{ab}$.

S9Only a dependent source

As in Nilsson Example 4.19.

a) Without calculating, what is $V_{\text{Th}}$? Why?

b) Find $R_{\text{Th}}$ with a 1 A test current source from b to a. (The dependent source delivers $v_x/40$ amperes, with $v_x$ in volts.)

c) A classmate deactivates the dependent source and gets 15 Ω. What went wrong?

Work this out on paper, then compare with the explanation.

S10Maximum power transfer

As in Nilsson Example 4.21.

a) Find the Thévenin equivalent seen by $R_L$: $V_{\text{Th}}$

a) $R_{\text{Th}}$

b) What value of $R_L$ receives maximum power?

b) How much power is it?

c) With $R_L$ at that value, what percentage of the power delivered by the 90 V source reaches $R_L$?

S11Spot the errorSpot the error

The worked solution below finds the Thévenin equivalent. It contains the kind of polarity slip AI chat tools often make.

The 30 V source and 6 Ω become 5 A in parallel with 6 Ω, arrow pointing up, toward the + terminal.
The 6 V source and 3 Ω, + at the bottom, become 2 A in parallel with 3 Ω, arrow pointing down.
Combine: $5 - 2 = 3$ A pointing up, in parallel with $6 \parallel 3 = 2$ Ω.
Transform back: 3 A pointing up with 2 Ω becomes 6 V in series with 2 Ω, + at the bottom.
Add the 4 Ω: $V_{\text{Th}} = -6$ V and $R_{\text{Th}} = 6$ Ω. Check: deactivating the sources gives $4 + 6 \parallel 3 = 6$ Ω, so the answer is right.

b) Correct it and find $V_{\text{Th}}$.

c) Check $V_{\text{Th}}$ with one node equation: the node voltage above the 3 Ω, with b as the reference

d) Line 5 calls itself a check. Why does it prove nothing about $V_{\text{Th}}$?

Work this out on paper, then compare with the explanation.

S12Spot the error in $R_{\text{Th}}$Spot the error

The worked solution below contains an error.

Deactivate the sources: replace the 20 V source and the 2 A source with short circuits.
The short in place of the 2 A source also shorts the 12 Ω, so $R_{\text{Th}} = 5$ Ω.
Open circuit: $(v - 20)/4 + v/12 - 2 = 0$ gives $v = 21$ V, so $V_{\text{Th}} = 21$ V.
So $i_{\text{sc}} = V_{\text{Th}}/R_{\text{Th}} = 4.2$ A.

a) Which is the first wrong line?

a)Line 1b)Line 2c)Line 3d)Line 4

b) Correct it and find $R_{\text{Th}}$.

c) Short a to b in the original circuit and find $i_{\text{sc}}$ directly.

c) Does it agree with $V_{\text{Th}}/R_{\text{Th}}$, using your corrected $R_{\text{Th}}$?

a)Yesb)No

S13Maximum power with a dependent source

a) Find $V_{\text{Th}}$. Hint: with a and b open, no current flows in the 6 Ω or in the dependent source.

b) Find $R_{\text{Th}}$ with a test source.

c) What load receives maximum power?

c) How much power?

S14Work backwards

A circuit is tested with two loads. With 10 Ω across its terminals, the load voltage is 20 V. With 30 Ω, it is 30 V.

a) Find $V_{\text{Th}}$

a) …and $R_{\text{Th}}$

b) What load resistance would receive maximum power?

b) How much?

S15Your generator subscriptionPredict first

With nothing switched on, the socket reads 230 V. With a heater on, it draws 10 A and the socket reads 210 V. Model the generator and its cables as a Thévenin equivalent.

a) Find $V_{\text{Th}}$

a) $R_{\text{Th}}$

a) The heater’s resistance

b) Predict first: with two identical heaters on, will the socket read about 210 V, 193 V or 170 V?

a)About 210 Vb)About 193 Vc)About 170 V

b) Then calculate it.

b) The total heater power

b) Is it twice the power of one heater?

a)Yesb)No

c) Why does the generator owner not match the load to $R_{\text{Th}}$ for maximum power transfer?

Work this out on paper, then compare with the explanation.

S16Check an equivalent without solving again

Two classmates report the Thévenin equivalent. Student 1: $V_{\text{Th}} = 24$ V, $R_{\text{Th}} = 6$ Ω. Student 2: $V_{\text{Th}} = 24$ V, $R_{\text{Th}} = 4$ Ω. Short a to b, find $i_{\text{sc}}$, and use $i_{\text{sc}} = V_{\text{Th}}/R_{\text{Th}}$ to decide which student is right.

$i_{\text{sc}}$

Which student is right?

a)Student 1b)Student 2