# 0.10 Electric circuits  (Page 12/12)

 Page 12 / 12

Equivalent resistance of two parallel resistor, ${R}_{p}$

For $2$ resistors in parallel with resistances ${R}_{1}$ and ${R}_{2}$ , the equivalent resistance is:

${R}_{p}=\frac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}$

## Equivalent parallel resistance

Consider a circuit consisting of a single cell and three resistors that are connected in parallel.

Using what we know about voltage and current in parallel circuits we can define the equivalent resistance of several resistors in parallel as:

Equivalent resistance in a parallel circuit, ${R}_{p}$

For $n$ resistors in parallel, the equivalent resistance is:

$\frac{1}{{R}_{p}}=\left(\frac{1}{{R}_{1}},+,\frac{1}{{R}_{2}},+,\frac{1}{{R}_{3}},+,\cdots ,+,\frac{1}{{R}_{n}}\right)$

Let us apply this formula to the following circuit.

What is the total resistance in the circuit?

$\begin{array}{ccc}\hfill \frac{1}{{R}_{p}}& =& \left(\frac{1}{{R}_{1}},+,\frac{1}{{R}_{2}},+,\frac{1}{{R}_{3}}\right)\hfill \\ & =& \left(\frac{1}{10\phantom{\rule{0.166667em}{0ex}}\Omega },+,\frac{1}{2\phantom{\rule{0.166667em}{0ex}}\Omega },+,\frac{1}{1\phantom{\rule{0.166667em}{0ex}}\Omega }\right)\hfill \\ & =& \left(\frac{1+5+10}{10}\right)\hfill \\ & =& \left(\frac{16}{10}\right)\hfill \\ \hfill \therefore \phantom{\rule{1.em}{0ex}}{R}_{p}& =& 0,625\phantom{\rule{0.166667em}{0ex}}\Omega \hfill \end{array}$

## Aim:

To determine the effect of multiple resistors on current in a circuit

• Battery
• Resistors
• Light bulb
• Wires

## Method:

1. Construct the following circuits
2. Rank the three circuits in terms of the brightness of the bulb.

## Conclusions:

The brightness of the bulb is an indicator of how much current is flowing. If the bulb gets brighter because of a change then more current is flowing. If the bulb gets dimmer less current is flowing. You will find that the more resistors you have the brighter the bulb.

Why is this the case? Why do more resistors make it easier for charge to flow in the circuit? It is because they are in parallel so there are more paths for charge to take to move. You can think of it like a highway with more lanes, or the tube of marbles splitting into multiple parallel tubes. The more branches there are, the easier it is for charge to flow. You will learn more about the total resistance of parallel resistors later but always remember that more resistors in parallel mean more pathways. In series the pathways come one after the other so it does not make it easier for charge to flow.

Two $8\phantom{\rule{2pt}{0ex}}\mathrm{k}\Omega$ resistors are connected in parallel. Calculate the equivalent resistance.

1. Since the resistors are in parallel we can use:

$\frac{1}{{R}_{p}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}$
2. $\begin{array}{ccc}\hfill \frac{1}{{R}_{p}}& =& \frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}\hfill \\ & =& \frac{1}{8\phantom{\rule{0.166667em}{0ex}}\mathrm{k}\phantom{\rule{0.166667em}{0ex}}\Omega }+\frac{1}{10\phantom{\rule{0.166667em}{0ex}}\mathrm{k}\phantom{\rule{0.166667em}{0ex}}\Omega }\hfill \\ \hfill {R}_{p}& =& \frac{2}{8}\hfill \\ & =& 4\phantom{\rule{0.166667em}{0ex}}\mathrm{k}\phantom{\rule{0.166667em}{0ex}}\Omega \hfill \end{array}$
3. The equivalent resistance of two $8\phantom{\rule{2pt}{0ex}}\mathrm{k}\Omega$ resistors connected in parallel is $4\phantom{\rule{2pt}{0ex}}\mathrm{k}\Omega$ .

Two resistors are connected in parallel. The equivalent resistance is $100\phantom{\rule{2pt}{0ex}}\Omega$ . If one resistor is $150\phantom{\rule{2pt}{0ex}}\Omega$ , calculate the value of the second resistor.

1. Since the resistors are in parallel we can use:

$\frac{1}{{R}_{p}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}$

We are given the value of ${R}_{p}$ and ${R}_{1}$ .

2. $\begin{array}{ccc}\hfill \frac{1}{{R}_{p}}& =& \frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}\hfill \\ \hfill \therefore \phantom{\rule{1.em}{0ex}}\frac{1}{{R}_{2}}& =& \frac{1}{{R}_{p}}-\frac{1}{{R}_{1}}\hfill \\ & =& \frac{1}{100\phantom{\rule{0.166667em}{0ex}}\Omega }-\frac{1}{150\phantom{\rule{0.166667em}{0ex}}\Omega }\hfill \\ & =& \frac{3-2}{300}\hfill \\ & =& \frac{1}{300}\hfill \\ \hfill {R}_{2}& =& 300\phantom{\rule{0.166667em}{0ex}}\Omega \hfill \end{array}$
3. The second resistor has a resistance of $300\phantom{\rule{2pt}{0ex}}\Omega$ .

## Resistance

1. What is the unit of resistance called and what is its symbol?
2. Explain what happens to the total resistance of a circuit when resistors are added in series?
3. Explain what happens to the total resistance of a circuit when resistors are added in parallel?
4. Why do batteries go flat?

The following presentation summarizes the concepts covered in this chapter.

## Exercises - electric circuits

1. Write definitions for each of the following:
1. resistor
2. coulomb
3. voltmeter
2. Draw a circuit diagram which consists of the following components:
1. 2 batteries in parallel
2. an open switch
3. 2 resistors in parallel
4. an ammeter measuring total current
5. a voltmeter measuring potential difference across one of the parallel resistors
3. Complete the table below:
 Quantity Symbol Unit of meaurement Symbol of unit e.g. Distance e.g. d e.g. kilometer e.g. km Resistance Current Potential difference
4. Draw a diagram of a circuit which contains a battery connected to a lightbulb and a resistor all in series.
1. Also include in the diagram where you would place an ammeter if you wanted to measure the current through the lightbulb.
2. Draw where and how you would place a voltmeter in the circuit to measure the potential difference across the resistor.
5. Thandi wants to measure the current through the resistor in the circuit shown below and sets up the circuit as shown below. What is wrong with her circuit setup?
6. [SC 2003/11] The emf of a battery can best be explained as the $\cdots$
1. rate of energy delivered per unit current
2. rate at which charge is delivered
3. rate at which energy is delivered
4. charge per unit of energy delivered by the battery
7. [IEB 2002/11 HG1] Which of the following is the correct definition of the emf of a battery?
1. It is the product of current and the external resistance of the circuit.
2. It is a measure of the cell's ability to conduct an electric current.
3. It is equal to the “lost volts” in the internal resistance of the circuit.
4. It is the power supplied by the battery per unit current passing through the battery.
8. [IEB 2005/11 HG] Three identical light bulbs A, B and C are connected in an electric circuit as shown in the diagram below.
1. How bright is bulb A compared to B and C?
2. How bright are the bulbs after switch S has been opened?
3. How do the currents in bulbs A and B change when switch S is opened?
 Current in A Current in B (a) decreases increases (b) decreases decreases (c) increases increases (d) increases decreases
9. [IEB 2004/11 HG1] When a current $I$ is maintained in a conductor for a time of $t$ , how many electrons with charge e pass any cross-section of the conductor per second?
1. It
2. It/e
3. Ite
4. e/It

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