Open practice · AI allowed

Practice sheet: Chapter 3, Simple Resistive Circuits

This sheet is not marked for correctness, and AI tools are allowed. The device-free quiz at the start of week 5, session 1 (Tue 3 Nov), 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.
Notation: $a \parallel b = ab/(a + b)$. Check every answer: resistors in parallel must have the same voltage, and the powers must add to zero.

Prefer paper? Open the printable PDF.

Which questions practise which skill

1Say which resistors are in series and which are in parallel, and why S1 S2
2Spot a wire that shorts out a resistor S3
3Find the equivalent resistance of a series-parallel network S2 S3 S7 S8 S13 S14
4Use voltage division and current division, and check the result S4 S5 S6 S7 S8 S9 S15 S16
5Explain why a load changes a divider’s output S4 S6
6Convert a Δ to a Y and a Y to a Δ, and use it to solve a bridge S10 S11 S12 S13

S1Series or not?Two-tier

Which pair of resistors is in series?

Answer

a)$R_1$ and $R_2$b)$R_3$ and $R_4$c)$R_2$ and $R_5$d)$R_1$ and $R_3$

Reason

i)Only those two elements meet at the node between them, so they carry the same current.ii)Resistors drawn one after another along a wire are always in series.iii)They connect the same two nodes.iv)They have the same voltage across them.

Follow-up: is $R_2$ in parallel with $R_5$? Explain in one sentence.

a)Yesb)No

S2Equivalent resistance of a ladder

Find $R_{ab}$.

Answer

a)11 Ωb)8 Ωc)51 Ωd)16.5 Ω

S3A wire across a resistorTwo-tier

Find $R_{ab}$.

Answer

a)14 Ωb)22.7 Ωc)10 Ωd)45 Ω

Reason

i)The wire puts 0 V across the 30 Ω resistor, so it carries no current and the 5 Ω resistor connects straight to b.ii)The wire and the 30 Ω resistor are in series, so the 30 Ω stays in the circuit.iii)A wire carries no current, so it can be ignored.iv)The current splits equally between the wire and the 30 Ω resistor.

S4Loading a voltage dividerPredict first

a) With $R_L$ removed, find $v_o$.

b) Predict first: when $R_L = 6$ kΩ is connected, does $v_o$ rise, fall or stay the same?

a)Riseb)Fallc)Stay the same

b) Then find $v_o$ with $R_L = 6$ kΩ.

c) Repeat b) for $R_L = 600$ kΩ: find $v_o$.

c) What rule about $R_L$ and the 6 kΩ resistor does this suggest?

a)$R_L$ should be much smaller than 6 kΩ.b)$R_L$ should equal 6 kΩ.c)$R_L$ should be much larger than 6 kΩ.

S5Current divisionTwo-tier

The current in the 2 kΩ resistor is:

Answer

a)15 mAb)5 mAc)10 mAd)30 mA

Reason

i)$i_j = (R_{eq}/R_j)\,i$ with $R_{eq} = 1$ kΩ, so the smallest resistance takes the largest current.ii)$i_j = (R_j/R_{eq})\,i$, so the largest resistance takes the largest current.iii)The current splits equally between the three resistors.iv)All the current takes the path of least resistance.

Follow-up: the current in the 3 kΩ resistor

Follow-up: the current in the 6 kΩ resistor

S6Design a voltage divider

You need 3 V from a 12 V supply, and the divider may draw at most 1 mA from it.

a) What is the smallest total resistance $R_1 + R_2$ allowed?

kΩ

b) Choose $R_1$ (top) and $R_2$ (bottom). With the smallest total from a), $R_1$ =

kΩ

b) … and $R_2$ =

kΩ

c) The 3 V output now feeds a 3 kΩ load. Find the new output voltage (with $R_1 = 9$ kΩ and $R_2 = 3$ kΩ).

c) Is your design still good?

a)Yesb)No

S7Series-parallel reduction, then division

a) Find the equivalent resistance seen by the source…

a) … and $i_s$.

b) Use current division to find $i_2$…

b) … then voltage division to find $v_o$.

c) Find the power of every element and check the power balance. Power delivered by the 60 V source:

c) Power absorbed by the 6 Ω resistor:

c) Power absorbed by the 12 Ω resistor:

c) Power absorbed by the 2 Ω resistor:

c) Power absorbed by the 4 Ω resistor:

S8Two divisions in a row

As in Nilsson Example 3.7.

a) Find the equivalent resistance seen by the source.

b) Use current division to find $i_o$.

c) Use voltage division to find $v_o$.

S9Spot the errorSpot the error

The worked solution below finds $v_o$. It contains the kind of slip AI chat tools often make with the divider rules.

The 9 Ω and 3 Ω resistors are in series: 12 Ω.
$6 \parallel 12 = 72/18 = 4$ Ω.
$R_{eq} = 4 + 4 = 8$ Ω.
$i_s = 72/8 = 9$ A.
Current division: $i_2 = (12/(6 + 12))(9) = 6$ A.
$v_o = 3i_2 = 18$ V.
KCL at the top node: the 6 Ω resistor carries $9 - 6 = 3$ A, so the currents balance and the answer is correct.

b) Correct it: the correct $i_2$

c) Finish the solution: the correct $v_o$

c) Check it: the voltage across the 6 Ω resistor

d) Line 7 says the currents balance. Why is that not a check?

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

S10Delta to wyeTwo-tier

In the equivalent Y, the resistor connected to terminal a is:

Answer

a)3 Ωb)2 Ωc)6 Ωd)4.5 Ω

Reason

i)Multiply the two Δ resistors that meet at a, and divide by the sum of all three.ii)Take the Δ resistor opposite a and divide it by 3.iii)Divide the sum of the pairwise products by the Δ resistor opposite a.iv)Put the two Δ resistors that meet at a in parallel.

Follow-up: the Y resistor connected to terminal b

Follow-up: the Y resistor connected to terminal c

Check: the resistance between a and b, the same in both circuits

S11Wye to delta

In the equivalent Δ, the resistor between a and b is:

Answer

a)35 Ωb)140 Ωc)70 Ωd)30 Ω

Follow-up: the Δ resistor between b and c

Follow-up: the Δ resistor between c and a

S12A bridge needs a Δ-to-Y step

No two resistors here are in series or in parallel. The 20 Ω, 30 Ω and 50 Ω resistors form a Δ.

a) Replace it with an equivalent Y: the Y resistor at the top node

a) … at the left node

a) … at the right node

b) Find the equivalent resistance seen by the source…

b) … and $i_s$.

c) Go back to the original circuit and find $i_m$ in the 50 Ω resistor.

S13A balanced bridgeTwo-tierPredict first

Find $R_{ab}$.

Predict first: does the 47 Ω resistor change $R_{ab}$?

a)Yesb)No

Answer

a)22.5 Ωb)15.2 Ωc)120 Ωd)69.5 Ω

Reason

i)$10/20 = 30/60$, so both ends of the 47 Ω are at the same voltage and it carries no current.ii)The 47 Ω is in parallel with the whole network.iii)The 47 Ω is in series with the whole network.iv)A resistor across the middle of a bridge never carries current.

S14Work backwards

In the S7 circuit, replace the 12 Ω resistor with an unknown $R$. What value of $R$ makes $i_s = 7.5$ A?

Answer

a)3 Ωb)2 Ωc)6 Ωd)12 Ω

S15The cable from your generatorPredict first

A neighbourhood generator holds 220 V at its terminals. Each conductor of the cable to your home has 0.25 Ω, and the home draws power like a 21.5 Ω resistor.

a) Predict first: does the home receive 220 V?

a)Yesb)No

a) Then find the current…

a) … and $v_L$.

b) Find the power delivered to the home…

b) … and the power lost in the cable.

b) What percentage of the generator’s output is lost?

c) A thicker cable halves each conductor’s resistance. Find $v_L$ now.

S16How a touch screen finds your finger

In a resistive touch screen (Nilsson’s Practical Perspective for this chapter), the grid in the x-direction is a resistance $R_x$ driven by $V_s = 5$ V. A touch splits it into $\alpha R_x$ and $(1 - \alpha)R_x$, and the screen measures $V_x$ across $\alpha R_x$, so $V_x = \alpha V_s$. The screen is $p_x = 1080$ pixels wide, and the pixel column of the touch is $x = (1 - \alpha)p_x$.

a) A touch gives $V_x = 2$ V. Find $\alpha$…

a) … and the pixel column $x$.

b) Which voltage-divider idea makes $V_x = \alpha V_s$?

a)$\alpha R_x$ and $(1 - \alpha)R_x$ are in series across $V_s$, so $V_x$ is a voltage-division fraction of $V_s$.b)$\alpha R_x$ and $(1 - \alpha)R_x$ are in parallel, so the current divides between them.c)The screen loads the divider, so $V_x$ falls in proportion to $\alpha$.

c) Where on the screen is a touch that gives $V_x = 5$ V?

a)At the far left edge, $x = 0$b)In the middle, $x = 540$c)At the far right edge, $x = 1080$