Electric Load

There are two connection of loads in a circuit. They are parallel connection and series connection. This note has brief introduction on electric loads and how they can be connected to a circuit.

Summary

There are two connection of loads in a circuit. They are parallel connection and series connection. This note has brief introduction on electric loads and how they can be connected to a circuit.

Things to Remember

  • There are two connection of loads in a circuit. They are parallel connection and series connection.
  • The cells are connected in such a way that the potential difference of the circuit is same as the cells used in the circuit but the time of current flowing increases is called parallel connection of loads in a circuit.
  • Loads are said to be connected in the series when they are daisy chained together in a single line.
  • Loads in series have a common current flowing through them as the current that flows through one load must also flow through the others as it can only take one path.

MCQs

No MCQs found.

Subjective Questions

Q1:

Find the upper quartile of the given data:
50, 40, 55, 60, 61, 70,49


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>The given data write in ascending order: 40, 49, 50, 55, 60, 61, 70 where N = 7</p> <p>\begin{align*} Position \: of\: Q_3 &amp;= \frac{3(N + 1)^{th}}{4}term \\ &amp;= \frac{3(7+1)^{th}}{4}\\ &amp;= 6^{th} \: term \\ 6^{th} \: term \: &amp;represent \: 61 \\ \therefore Upper \: quartile \: &amp;(Q_3) = 61 \end{align*}</p> <p></p>

Q2:

Find third quartile (Q3) from the following data: 
45, 30, 31, 37, 42, 43, 40, 48


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>The given data write in ascending order<br>30, 31, 37,40, 42, 43, 45, 48, where N = 8</p> <p>\begin{align*} Position \: of \: Q_3 &amp;= \frac{3(N + 1)^{th}}{4}term \\ &amp;= \frac{3(8 + 1)^{th}}{4} term \\ &amp;= 3 \times 2.25^{th} \: term \\ &amp;= 6.75^{th} \: term \\ \: \\ Q_3 = 6^{th}\:term + &amp;(7^{th}term - 6^{th} term) \times 0.75\\ Q_3 =43 + (45 - &amp;43) \times 0.75 \\ Q_3 = 43 + 1.5 \\ \therefore \: \: Q_3 = 44.5 \: \: _{Ans} \end{align*}</p>

Q3:

2x + 1, 3x - 1, 3x + 5, 5x - 7, 51, 63 and 70are in ascending order. If the first quartile is 20. What will be the value of x.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>N = 7</p> <p>\begin{align*} Position \: of \: Q_1 &amp;= \frac{N + 1^{th}}{4}term \\ &amp;= \frac{7 + 1^{th}}{4}term \\ &amp;= 2^{nd} \: term \\ \: \\ 2^{th}\: term \: repr&amp;esent \: 3x -1 \\ \: \\ Q_1 &amp;= 3x -1 \\ 20 &amp;= 3x -1 \\ or, 3x&amp;= 20 +1 \\ or, x &amp;= \frac{21}{3}\\ \therefore x &amp;= 7 \: _{Ans} \end{align*}</p>

Q4:

Find the lower quartile from the following data 17, 25, 22, 18, 12, 14, 19, 11


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>The given data arranging in ascending order.<br>11, 12, 14, 17, 18, 19, 22, 25 where N = 8.</p> <p>\begin{align*} Position \: of \: first \: quartile \: (Q_1) &amp;= \frac{N+1^{th}}{4}term \\ &amp;= \frac{8+1}{4}term \\ &amp;= 2.25^{th} \: term \\ \: \\ Q_1 = 2^{nd} term + (3^{rd} - &amp;2^{nd}) term \times 0.25 \\ Q_1 = 12 + (14-12) \times &amp;0.25 \\ Q_1 = 12 + 2 \times 0.25\\ \therefore Q_1 = 12.5 \: \: _{Ans} \end{align*}</p>

Q5:

Find the class of lower quartile (Q1) from the following data.

Marks 50 60 70 80 90 100
No. of students 3 4 7 5 2 9

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>Calculating lower quartile (Q<sub>1</sub>)</p> <table width="283"><tbody><tr><td>Marks (X)</td> <td>Frequency (f)</td> <td>Cumulative frequency (cf)</td> </tr><tr><td>50</td> <td>3</td> <td>3</td> </tr><tr><td>60</td> <td>4</td> <td>7</td> </tr><tr><td>70</td> <td>7</td> <td>14</td> </tr><tr><td>80</td> <td>5</td> <td>19</td> </tr><tr><td>90</td> <td>2</td> <td>21</td> </tr><tr><td>100</td> <td>9</td> <td>30</td> </tr><tr><td></td> <td>N = 30</td> <td></td> </tr></tbody></table><p></p> <p>\begin{align*}Position \: of \: lower \: quartile (Q_1) &amp;= \frac{N + 1^{th}}{4}term \\ &amp;= \frac{30 + 1^{th}}{4}term \\ &amp;= 7.75^{th} term \end{align*}</p> <p>7.75<sup>th</sup> term represent cf value 14.</p> <p>\(\therefore\) lower quartile (Q<sub>1</sub>) = 70 \(_{Ans}\)</p> <p></p> <p></p>

Q6:

1, 5, 7, 2x - 4, x + 7, 2x + 1 and 3x + 2 are in ascending order. If the third quartile is 15. What will be the value of x.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>The given data write in ascending order<br>\(1,5,7,2x-4,x+7,2x+1,3x+2\) where N = 7</p> <p>\begin{align*} Position \: of \: Q_3 &amp;= \frac{3(N + 1)^{th}}{4}term \\ &amp;= \frac{3(7+1)^{th}}{4}term \\ &amp;= 3 \times 2^{th} \: term \\ &amp;= 6^{th} term \\ 6^{th} term \: repr&amp;esent \: 2x + 1 \end{align*}</p> <p>\begin{align*} Q_1 &amp;= 2x+1 \\ 2x + 1 &amp;= 15 \\ or, 2x &amp;= 15 -1 \\ \therefore x &amp;= \frac{14}{2} =7 \: _{Ans} \end{align*}</p> <p></p>

Q7:

Calculate the class of third quartile from the given data.

Marks 0 - 10 10 - 20 20 - 30 30 - 40
No. of students 4 8 12 4

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>Calculating the Q<sub>3</sub> class</p> <table width="263"><tbody><tr><td>Marks (x)</td> <td>No. of students(f)</td> <td>Cumulative frequenc (cf)</td> </tr><tr><td>0 - 10</td> <td>4</td> <td>4</td> </tr><tr><td>10 - 20</td> <td>8</td> <td>12</td> </tr><tr><td>20 - 30</td> <td>12</td> <td>24</td> </tr><tr><td>30 - 40</td> <td>4</td> <td>28</td> </tr><tr><td></td> <td>N = 28</td> <td></td> </tr></tbody></table><p>\begin{align*} Position \: of \: Q_3 \: class &amp;= \frac{3N^{th}}{4}term \\ &amp;= \frac{3 \times 28^{th}}{4} term \\ &amp;= 21^{th} \: term \\ Class \: of \: Q_3 &amp;= 20 - 30 \: _{Ans}\end{align*}</p> <p></p>

Q8:

Find the upper quartile from the following data.

Marks 10 20 30 40 50
No. of students 5 4 5 6 7

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>solution:</strong></p> <p>Calculating the Q<sub>3</sub> class</p> <table width="251"><tbody><tr><td>Marks (x)</td> <td>No. of students(f)</td> <td>Cumulative frequenc (cf)</td> </tr><tr><td>10</td> <td>5</td> <td>5</td> </tr><tr><td>20</td> <td>4</td> <td>9</td> </tr><tr><td>30</td> <td>5</td> <td>14</td> </tr><tr><td>40</td> <td>6</td> <td>20</td> </tr><tr><td>50</td> <td>7</td> <td>27</td> </tr><tr><td></td> <td>N = 27</td> <td></td> </tr></tbody></table><p>\begin{align*} Position \: of \: upper\:quartile (Q_3) &amp;= \frac{3(N+1)^{th}}{4}term \\ &amp;= \frac{3(27 + 1)^{th}}{4}term \\ &amp;= 3 \times 7^{th} \: term \\ &amp;=21 ^{th} \: term\\ \: \\ 21^{th} \: term \: represent \: c.f \: 27 \\ \therefore Upper \: quartile \: (Q_3) &amp;= 50 \: \: _{Ans} \end{align*}</p>

Q9:

From the first quartile class from the given graph.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>From graph,<br>Total number of boys (N) = 60</p> <p>\begin{align*} Position \: of \: first \: quartile &amp;= \frac{N^{th}}{4}term \\ &amp;= \frac{60^{th}}{4}term \\ &amp;= 15^{th} term \\ Class \: of \: first \: quartile &amp;= 5 - 10 \: _{Ans} \end{align*}</p>

Videos

Median, Quartiles and Interquartile Range : ExamSolutions
Quartiles and interquartile range
Math Problems : How to Find Quartiles in Math
Electric Load

Electric Load

When a wire is joined to a positive and a negative terminal of a source of electricity that completes a circuit and electrons start to flow and we get an electric current. Then we can add appliances like bulbs or bells or fans etc. to the circuit and use the current from the circuit to do some work. The appliances that use electric energy and convert it to different forms of energy are called electric loads. They pose some resistance to the flow of electricity.

Connection of loads in a circuit:

If we have to join two or more loads in a circuit then it can be done in two ways:

  1. Parallel Connection:

    In this connection, the loads are connected in such a way that electric current divides up among the loads but the potential difference remains the same. To achieve this, the positive terminal of all the loads (that are supposed to be in parallel connections) are connected to the same spot of the positive part of the circuit and the negative terminal of all the loads (that are supposed to be in parallel connections) are connected to the same spot of the negative terminal of the circuit. The added load work as mini circuits and work independently of each other. The brightness of the bulb is not affected due to addition or removal of bulbs in parallel connection because of same potential difference. So this type of connection is done in our house. If any one of the load fails it does not affect other loads (they work fine). In parallel connection, the reciprocal of equivalent resistance is equal to the sum of reciprocal of individual resistances i.e. $$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}$$

  2. Series Connection:

    In this connection, the loads are connected such a way that the electric current flowing through each resistance is equal but the potential difference is different. To achieve this, the loads are connected in series with each other (that's where the name series come from). The current flows continuously through the loads in a straight line. If one of the loads fails, the circuit becomes open and none of the loads would work. The sum of potential difference across the loads gives the total potential difference of the circuit i.e. $$V = V_1 + V_2+ V_3+...+ V_n$$

  3. . The effective resistance is also the sum of the individual resistance of each load i.e. R = R1 + R2+ R+...+Rn. The brightness of the bulb is reduced when we add bulbs into the circuit because the potential difference is divided among the loads.

Lesson

Electricity and Magnetism

Subject

Science

Grade

Grade 10

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