Practice Quiz Review & Reinforcing Build

class: center, middle, inverse, title-slide

.title[
# Practice Quiz Review & Reinforcing Build
]
.subtitle[
## Programmable Electronics — Unit 1
]
.date[
### Class 7
]

---

# Practice Quiz

.q-box[
- 24 questions, covering Classes 1–6.
- This is a **practice run** — not graded, not the real quiz.
- Each slide shows a question. Click once more to reveal the answer.
- Work it out first — in your head or on scratch paper — before you click.
- After the quiz: a quick build reinforcing the voltage divider and potentiometer ideas.
]

---

# Question 1

.q-box[
What three quantities does Ohm's Law relate, and what's the equation solved for **voltage**?
]

.a-box[
.a-label[Answer:] Voltage (`$V$`), current (`$I$`), and resistance (`$R$`).

`$$V = IR$$`
]

---

# Question 2

.q-box[
For current to flow anywhere in a circuit, what has to be true about the path from the positive terminal of the source back to the negative terminal?
]

.a-box[
.a-label[Answer:] It has to be a **closed loop** — one continuous, unbroken conductive path all the way around. Break the path anywhere (a disconnected wire, an open switch) and current stops everywhere in that loop, not just at the break.
]

---

# Question 3

.q-box[
Roughly how many electrons pass a point in one second when a wire carries **1 amp** of current?
]

.a-box[
.a-label[Answer:] About `$6.242 \times 10^{18}$` electrons per second — that's the definition of the amp.
]

---

# Question 4

.q-box[
In the water-tank analogy, what electrical quantity does the **water pressure** represent?
]

.a-box[
.a-label[Answer:] **Voltage** (`$V$`) — the pressure is what pushes electrons through the circuit, the same way pressure pushes water through a hose.
]

---

# Question 5

.q-box[
Write Ohm's Law solved for resistance.
]

.a-box[
.a-label[Answer:]

`$$R = \frac{V}{I}$$`
]

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# Question 6

.q-box[
A red LED (`$V_f \approx 2.0\,\text{V}$`) needs about 20 mA from a 6 V supply. What resistor value do you calculate, **before** rounding to a standard size?
]

.a-box[
.a-label[Answer:]

`$$R = \frac{V_{supply} - V_f}{I_{LED}} = \frac{6\,\text{V} - 2.0\,\text{V}}{0.020\,\text{A}} = 200\,\Omega$$`

Round **up** to the nearest resistor you have — 220 Ω.
]

---

# Question 7

.q-box[
Decode this resistor: **Brown – Black – Orange – Gold**. Give the value and the tolerance.
]

.a-box[
.a-label[Answer:] Brown = 1, Black = 0, Orange = ×1,000, Gold = ±5%.

`$$10 \times 1{,}000 = 10{,}000\,\Omega = 10\,\text{k}\Omega, \pm 5\%$$`
]

---

# Question 8

.q-box[
When your calculated resistor value falls between two standard sizes, why do we round **up** instead of down?
]

.a-box[
.a-label[Answer:] Rounding up gives you *more* resistance than the calculation called for, which means *less* current than calculated — the safe direction. Rounding down would push current (and power) higher than intended, which is how you get an overheating resistor or a burned-out LED.
]

---

# Question 9

.q-box[
Two resistors, R1 and R2, are in series off a battery. How would you connect your multimeter to measure:

1. the **current** through R1?
2. the **voltage** across R2?
]

.a-box[
.a-label[Answer:]

1. **Current** — in *series*: break the circuit and insert the meter into the path, so every electron passes through it.
2. **Voltage** — in *parallel*: touch the probes to both ends of R2 without disconnecting anything.
]

---

# Question 10

.q-box[
A 220 Ω resistor and a 330 Ω resistor are wired in **series**. What is `$R_{total}$`?
]

.a-box[
.a-label[Answer:]

`$$R_{total} = R_1 + R_2 = 220\,\Omega + 330\,\Omega = 550\,\Omega$$`
]

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# Question 11

.q-box[
Those same two resistors (220 Ω and 330 Ω) are now wired in **parallel** instead. Is `$R_{total}$` greater than, less than, or equal to 220 Ω?
]

.a-box[
.a-label[Answer:] **Less than 220 Ω.** Total resistance in a parallel network is always *less* than the smallest branch — more paths for current means less overall resistance to the flow.
]

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# Question 12

.q-box[
Write the power equation in its three equivalent forms.
]

.a-box[
.a-label[Answer:]

`$$P = IV \qquad P = I^2R \qquad P = \frac{V^2}{R}$$`

All three come from substituting Ohm's law (`$V = IR$`) into `$P = IV$`.
]

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# Question 13

.q-box[
A resistor drops 4 V and carries 400 mA. How much power does it dissipate, and how does that compare to a standard ¼ W (0.25 W) rating?
]

.a-box[
.a-label[Answer:]

`$$P = I \times V = 0.4\,\text{A} \times 4\,\text{V} = 1.6\,\text{W}$$`

That's **6.4×** over a 0.25 W rating — not a close call. This is what happened to the 10 Ω resistor: it overheated and visibly browned.
]

---

# Question 14

.q-box[
For the same fixed voltage drop, why does a **smaller** resistor run hotter than a larger one?
]

.a-box[
.a-label[Answer:] Use `$P = \dfrac{V^2}{R}$`. If `$V$` stays roughly fixed (like the leftover voltage after an LED's forward drop), `$R$` sits in the *denominator* — shrink `$R$` and `$P$` goes up sharply. A much smaller resistor dissipates dramatically more heat in a much smaller part.
]

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# Question 15

.q-box[
You want to light a single LED (`$V_f \approx 2\,\text{V}$`, rated for a max of about 25 mA) from a **9 V** supply. What resistor value gives you the standard target current of 20 mA?
]

.a-box[
.a-label[Answer:]

`$$R = \frac{V_{supply} - V_f}{I_{LED}} = \frac{9\,\text{V} - 2\,\text{V}}{0.020\,\text{A}} = 350\,\Omega$$`
]

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# Question 16

.q-box[
Your parts bin only has **10 Ω**, **100 Ω**, and **1000 Ω** resistors on hand — nothing near 350 Ω. Calculate the LED current you'd get with each one.
]

.a-box[
.a-label[Answer:]

`$$10\,\Omega:\ I = \frac{7\,\text{V}}{10\,\Omega} = 700\,\text{mA} \qquad 100\,\Omega:\ I = \frac{7\,\text{V}}{100\,\Omega} = 70\,\text{mA} \qquad 1000\,\Omega:\ I = \frac{7\,\text{V}}{1000\,\Omega} = 7\,\text{mA}$$`
]

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# Question 17

.q-box[
Which of those three resistors would you never use here, and why?
]

.a-box[
.a-label[Answer:] **10 Ω and 100 Ω, both.** 700 mA is 28× the LED's ~25 mA rating — instant burnout. 70 mA is still nearly 3× over rating — enough to overheat and damage the LED, even if it doesn't fail immediately. Any current above the rated max fails the safety check, whether it kills the LED on the spot or just shortens its life.
]

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# Question 18

.q-box[
Given only those three choices, which resistor should you actually use — and what do you give up by using it?
]

.a-box[
.a-label[Answer:] The **1000 Ω**, giving about 7 mA — safely under the LED's rated max. The trade-off is brightness: since brightness roughly tracks current, 7 mA will look noticeably dimmer than the 20 mA target. Safe-but-dim beats bright-but-dead when the ideal part isn't in the bin.
]

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# Question 19

.q-box[
You actually have **two** 1000 Ω resistors in the bin. Wired one way, together they land closer to the ideal ~350 Ω than a single 1000 Ω does. Which configuration, and what resistance does it give?
]

.a-box[
.a-label[Answer:] **Parallel.**

`$$R_{total} = \frac{1000\,\Omega \times 1000\,\Omega}{1000\,\Omega + 1000\,\Omega} = 500\,\Omega$$`

500 Ω is much closer to the ~350 Ω target than either a single 1000 Ω or two in series (which would only make it worse, at 2000 Ω). At 500 Ω, `$I \approx 14\,\text{mA}$` — still safe, and noticeably brighter than the single-resistor option.
]

---

# Question 20

.q-box[
On a breadboard:

1. How many holes in one terminal-strip column are electrically connected to each other?
2. Do the power rails run the full length of the board, or are they split at the center gap like the terminal strip is?
]

.a-box[
.a-label[Answer:]

1. **Five** — every hole in a highlighted column is one single electrical point.
2. The **power rails run the full length** of the board, uninterrupted. Only the terminal-strip columns are split — the columns on one side of the center gap are *not* connected to the columns on the other side.
]

---

# Question 21

.q-box[
Write the voltage divider equation for `$V_{out}$`, measured across `$R_2$` (with `$R_1$` on top, `$R_2$` on the bottom, to ground).
]

.a-box[
.a-label[Answer:]

`$$V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$$`
]

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# Question 22

.q-box[
A potentiometer is wired as a **rheostat** instead of a voltage divider. How many of its three terminals are actually used, and which ones?
]

.a-box[
.a-label[Answer:] **Two** — one end terminal, plus the wiper. The third terminal is left unconnected. This turns the pot into a single adjustable resistor instead of a device that splits a voltage.
]

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# Question 23

.q-box[
Is a photoresistor's resistance **higher** or **lower** in the dark, compared to bright light?
]

.a-box[
.a-label[Answer:] **Higher** in the dark (megaohms), **lower** in bright light (hundreds of ohms). Dark → high resistance; bright → low resistance.
]

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# Question 24

.q-box[
You're building a nightlight: `$V_{out}$` should **rise** as the room gets **darker**, so a tapped LED gets brighter. In the divider `$V_{out} = V_{in} \times \dfrac{R_2}{R_1+R_2}$`, should the photoresistor be `$R_1$` (top) or `$R_2$` (bottom)?
]

.a-box[
.a-label[Answer:] `$R_2$` — the **bottom** position, nearest ground. `$V_{out}$` increases whenever `$R_2$` increases, so putting the LDR there means `$V_{out}$` rises as the room darkens (LDR resistance rises). Putting it on top (`$R_1$`) would make `$V_{out}$` *fall* as it gets darker — backwards for a nightlight.
]

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class: center, middle

# How'd you do?

.hw-box[
- Any question you couldn't answer cold — go back to that class's slides tonight.
- Bring your specific confusions to Studium, not just "I didn't get it."
- The real quiz will follow the same shape as this one.
]

---
class: center, middle

# Build (15 min): potentiometer LED dimmer

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# Quick recap: why these values?

`$$I = \frac{V_{in} - V_f}{R_{fixed} + R_{pot}} \qquad V_{in}=6\,\text{V},\ V_f\approx 2\,\text{V},\ R_{fixed}=220\,\Omega$$`

- **1 kΩ pot:** 18.2 mA (brightest) down to 3.3 mA (dimmest) — visible across the whole turn.
- **10 kΩ / 100 kΩ pot:** still 18.2 mA brightest, but dimmest drops under 0.4 mA — invisible for most of the turn.

That's why Build 1 in Class 6 (and today) uses the **1 kΩ** pot. Full derivation is in the Class 6 deck if anyone wants the details.

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# Build instructions

Same circuit from Class 6, now hands-on with Q21 and Q22 fresh in mind. With your partner, using your breadboard, 6 V battery pack, multimeter, your 1 kΩ potentiometer, an LED, and a 220 Ω resistor:

1. Build a fixed voltage divider with two resistors of your choice. **Predict** `$V_{out}$` first (Q21's equation), then measure it. Do they match?
2. Swap in the potentiometer in place of your two resistors — same two end-terminals across the supply, wiper to `$V_{out}$`. Measure `$V_{out}$` at three positions: fully one way, fully the other way, and roughly in the middle.
3. Now rewire it as a **rheostat** (Q22): one end terminal + the wiper only, in series with your 220 Ω resistor and the LED, across the 6 V supply. Leave the third terminal unconnected.
4. Slowly turn the knob end to end. The LED should dim and brighten smoothly.
5. With your partner, say out loud which of today's questions this circuit connects to — divider equation, rheostat wiring, LED current limiting.

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class: center, middle

# Homework

.hw-box[
- Note anything from today's quiz or build you're still unsure about — bring it to Studium.
- Class 8 picks up with the capacitor and RC timing.
]