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| # Adventure 1a: Equivalent resistance: Resistors in series | ||
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| ## Goal | ||
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| Understand **equivalent resistance**. | ||
| Simulate a circuit with multiple resistors, connected *in series*. | ||
| Then, build it on a breadboard. | ||
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| ## Simulate circuit | ||
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| 1. Open [Circuit Simulator](https://falstad.com/circuit/circuitjs.html) and | ||
| load the circuit from [](../adv-ohm/adv-ohm.md). | ||
| Here is an example of the finished circuit: [Circuit - Ohm's | ||
| law](https://falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgoqoQFMBaMMAKADcRD8QAWHvTt2wIUUMT1piqMBCwBOggcNF44IZWLDwWAcxAZIPXv32HexQlJYB3EKqob7xgZBuLnpo3xduDXk07eUCxAA). | ||
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| 1. Use your mouse to drag a selection box around your circuit, so you can | ||
| select multiple components. Copy and paste your existing circuit once, so | ||
| you have two independent (sub-)circuits side-by-side, each with its own | ||
| voltage source and connection to ground. | ||
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| 1. In your copy, add a new resistor *in series* with the old one: | ||
| 1. Insert one more resistor. | ||
| 1. Connect it so it sits *between* VCC and the original resistor, **or** | ||
| the original resistor and GND. | ||
| 1. Double click it and set its resistance to $4kΩ$. | ||
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| Your circuit should look something like this: | ||
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|  | ||
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| 1. Click **RUN / Stop** to start the simulation. Hover your mouse over each | ||
| one of the two resistors, notice the *same* current flows through both | ||
| of them. What is its value? How does it compare between the old and the new | ||
| circuit? | ||
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| :::{admonition} Answer | ||
| :class: hint | ||
| :class: dropdown | ||
| It is smaller, $I_{new} = 1mA$, while the current that flows through the | ||
| original circuit is $I_{old} = 5mA$. | ||
| ::: | ||
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| If you would like some help with creating the circuit, see below. | ||
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| :::{hint} | ||
| :class: dropdown | ||
| Here is an example of the finished circuit: [Circuit - Resistors in series](https://falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgoqoQFMBaMMAKAHcQAWbFL4wiijx8BkdiGzYBnflx4jw4hEK5gBy4TNEsA5nN5b9FMLypiAThKmr1KzmqjI4LAG5XpsydIdmutR2bQCK4UGB7qYTYBftS+MMGWCJH2AoQI3qIgaM56hBgGsnm8CCYB4mGaDhUKYhxJ4SBFCqwcTYbVhhaNeJUCHT4gePAsAA6hAlLCaROEwtjRteNR9RKzUOLTq1PpUWJjs3Nr1ZMSCxs9W9296xzHR2spN1dRm49iAB6NkNL2jSjznEBXHAnBAACVWJ8MGBNNhhDgkICIKCwKCwSgWEA). | ||
| ::: | ||
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| 1. Change the value of the second resistor. How does the current change | ||
| depending on the value? What happens if you set the resistance to zero? What | ||
| happens when you set it to a really big value, say $100MΩ$? | ||
| :::{tip} | ||
| Double-click any wire and enable **Show current** so you can see the | ||
| current that flows through the it as you change the resistance. | ||
| ::: | ||
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| 1. Continue changing the value of the second resistor. How does the voltage | ||
| drop across each resistor change depending on the resistance? | ||
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| :::{tip} | ||
| Click **Draw --> Outputs and Labels --> Add Voltmeter/Scope Probe**, | ||
| and add one voltmeter over each resistor. Add wires to each voltmeter so its | ||
| ends connect to the ends of the resistor below it. This way the voltmeter | ||
| will be measuring the voltage drop across each resistor. | ||
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| :::{important} | ||
| Double-click each voltmeter and set its "series resistance" to `0`, i.e., | ||
| infinite. This makes it an *ideal* voltmeter, which does not have any | ||
| effect on the quantity it measures. We will talk more about this in the | ||
| next adventure. | ||
| ::: | ||
| ::: | ||
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| Your circuit should look something like this: | ||
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|  | ||
| ::: | ||
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| 1. Let's name the two resistors and set them to $R_1 = 2kΩ$, $R_2 = 4kΩ$, | ||
| so their voltage drops are $V_1$, $V_2$, respectively. | ||
| What is the value of $V_1$, $V_2$, and what is the sum $V_T = V_1 + | ||
| V_2$? | ||
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| Set the value of $R_2$ to $8kΩ$. What happens to $V_2$, $V_1$, and | ||
| their sum? | ||
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| :::{admonition} Answer | ||
| :class: hint | ||
| :class: dropdown | ||
| **Before:** | ||
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| $$ | ||
| \begin{align*} | ||
| R_1 = 2kΩ \qquad V_1 &= 1.667V \\ | ||
| R_2 = 4kΩ \qquad V_2 &= 3.333V \\ | ||
| V_T = V_1 + V_2 &= 5V | ||
| \end{align*} | ||
| $$ | ||
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| **After:** | ||
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| $$ | ||
| \begin{align*} | ||
| R_1 = 2kΩ \qquad V_1 &= 1V \\ | ||
| R_2 = 8kΩ \qquad V_2 &= 4V \\ | ||
| V_T = V_1 + V_2 &= 5V | ||
| \end{align*} | ||
| $$ | ||
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| So, $V_2$ increases as $R_2$ increases but the sum remains the same. | ||
| ::: | ||
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| 1. Set $R_1 = 2kΩ$, $R_2 = 3kΩ$. Try different values for the single resistor | ||
| in your old circuit so the current that flows through the old circuit | ||
| **matches** the current that flows through the new circuit. How big does the | ||
| single resistor have to be, so it works the same as the two separate | ||
| resistors? | ||
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| :::{admonition} Answer | ||
| :class: hint | ||
| :class: dropdown | ||
| It needs to be $R_T = 5kΩ$. | ||
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| 1. Notice how the single resistor has to be equal to the **sum** of the two | ||
| independent resistors connected in series, to have the same effect. | ||
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| :::{important} | ||
| Two resistors $R_1$, $R_2$ connected in series have a *total* resistance | ||
| which is the sum of their individual values: | ||
| $$R_T = R_1 + R_2$$ | ||
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| In this case the *same* current flows through both resistors: | ||
| $$I_S = I_1 = I_2$$ | ||
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| And the total voltage is *divided* between the two resistors: | ||
| $$V_T = V_1 + V_2$$ | ||
| ::: | ||
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| 1. What current will flow when $R_1 = 2kΩ$, $R_2 = 8kΩ$? | ||
| Use Ohm's law for the equivalent circuit to find out. | ||
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| :::{admonition} Answer | ||
| :class: hint | ||
| :class: dropdown | ||
| We know the equivalent resistance is $R_T = R_1 + R_2 = 2kΩ + 8kΩ = 10kΩ$. | ||
| We can use Ohm's law for the equivalent circuit: | ||
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| $$ | ||
| \begin{align*} | ||
| I &= {V \over R_T} \\ | ||
| \\ | ||
| I &= {5V \over 10kΩ} \\ | ||
| \\ | ||
| Ι &= 0.5mA | ||
| \end{align*} | ||
| $$ | ||
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| 1. What will the voltage drop for each resistance be in this case. | ||
| Use Ohm's law for each one of the resistors. | ||
| What is the ratio $V_1 / V_2$ in this case? | ||
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| :::{admonition} Answer | ||
| :class: hint | ||
| :class: dropdown | ||
| We know the same current passes through both resistors, $I_S = I_1 = I_2 = 0.5mA$. | ||
| We use Ohm's law for each resistor individually: | ||
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| $$ | ||
| \begin{align*} | ||
| V_1 &= I_1 \cdot R_1 & \qquad V_2 &= I_2 \cdot R_2 \\ | ||
| V_1 &= 0.5mA \cdot 2kΩ & \qquad V_2 &= 0.5mA \cdot 8kΩ \\ | ||
| V_1 &= 1V & \qquad V_2 &= 4V | ||
| \end{align*} | ||
| $$ | ||
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| The voltage ratio is: | ||
| $$ | ||
| {V_1 \over V_2} = {1 \over 4} | ||
| $$ | ||
| ::: | ||
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| 1. What is the ratio $V_1 / V_2$ for resistors $R_1$, $R_2$ connected in | ||
| series? | ||
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| :::{important} | ||
| We know Ohm's law still applies to each resistor individually, and the | ||
| *same* current flows through both resistors. | ||
| We can combine these two facts to compute the ratio $V_1 / V_2$. | ||
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| We know that Ohm's law applies to the two resistors individually: | ||
| $$ | ||
| I_1 = {V_1 \over R_1} \qquad I_2 = {V_2 \over R_2} | ||
| $$ | ||
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| And we know the same current flows through both resistors, because they | ||
| are connected in series. So: | ||
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| $$ | ||
| \begin{align*} | ||
| I_1 &= I_2 \\ | ||
| \\ | ||
| {V_1 \over R_1} &= {V_2 \over R_2} \\ | ||
| \\ | ||
| {V_1 \cdot R_2} &= {V_2 \cdot R_1} \\ | ||
| \\ | ||
| {V_1 \over V_2} &= {R_1 \over R_2} | ||
| \end{align*} | ||
| $$ | ||
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| So, the voltage ratio is the resistance ratio, for resistors connected in | ||
| series. | ||
| ::: | ||
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| 1. *Next step:* Add a 1MΩ potentiometer. | ||
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| ## Build circuit | ||
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| Grab your breadboard! Start from the original circuit you had built for | ||
| [](../adv-ohm/adv-ohm.md): | ||
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| 1. Take an 80Ω resistor, and measure it using your multimeter. Write down this | ||
| measurement. | ||
| 1. Add it in series with the original 150Ω resistor. | ||
| 1. Compute the equivalent resistance of the two resistors. | ||
| 1. Use your multimeter to measure the input voltage. | ||
| 1. Compute the current you expect will flow through the circuit, using Ohm's law | ||
| for the equivalent circuit. | ||
| 1. Measure the current that flows through this circuit. | ||
| 1. Compute the voltage drop you expect for each resistor, using Ohm's law. | ||
| 1. Measure the voltage drop across each one of the resistors, and compare with | ||
| your computations above. | ||
| 1. **Next step:** Experiment with a 1MΩ trimmer / potentiometer. | ||
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| ## Celebrate | ||
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| Congratulations! You've simulated and built circuits with resistors in series. | ||
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| Here are more resources to explore: | ||
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| :::{see also} | ||
| * [BBC Bitesize: Resistors in series](https://www.bbc.co.uk/bitesize/guides/z6cvqp3/revision/1) | ||
| * [Wikipedia: Series and parallel circuits](https://en.wikipedia.org/wiki/Series_and_parallel_circuits) | ||
| ::: |
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