Combination of Cells

The cells which 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 known as parallel combination. This note summarizes parallel and series combination of the cells.

Summary

The cells which 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 known as parallel combination. This note summarizes parallel and series combination of the cells.

Things to Remember

  • To increase the amount of current or potential difference required in a circuit we can add cells to the circuit. This can be achieved in two different ways:
  • The cells which 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 known as parallel combination. (V = V1 = V2 = V3….)
  • In series combination positive terminal of first cell is connected to the negative terminal of the second cell and positive terminal of second cell is connected to the negative terminal of third cell and so on. (V = V1 + V2 + V3 + …. )

MCQs

No MCQs found.

Subjective Questions

Q1:

Find the median of the following data:
60, 64, 65, 70, 72, 80, 85 where N = 7.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>\begin{align*}Position \: of \: median &amp;= \frac{N + 1^{th} }{2}term \\ &amp;= \frac{7+1^{th}}{2}term\\ &amp;= 4^{th} term \\ 4^{th}\: represent \: 70 . &amp; \therefore median &amp;= 70 _{Ans}\end{align*}</p>

Q2:

If the mean of 4, 6, 9, 11, 10 and k is 8, what is its median?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>\(Mean (\overline{X}) = 8 \\ Number \: of \: term (N)= 6 \)</p> <p>\begin{align*} Mean ( \overline {X}) &amp;= \frac{\sum X }{N}\\ or, 8 &amp;= \frac{4 +6+9+11+10+k}{6} \\ or, 48 &amp;= 40 + k \\ or, k &amp;=48 -40 \\ \therefore k &amp;= 8 \end{align*}</p> <p>The given term write in ascending order: 4, 6, 8, 9, 10, 11</p> <p>\begin{align*} Position \: of \: median &amp;= \frac{N+1^{th}}{2}term\\&amp;=\frac{6 + 1^{th}}{2}term\\ &amp;= 3.5^{th} term \\ \: \\ Median &amp;= \frac{8+9}{2}\\ &amp;= \frac{17}{2}\\ &amp;= 8.5 _{Ans} \end{align*}</p>

Q3:

Find the median of first eight multiple of 5.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>First eight multiple of 5 are 5, 10, 15, 20, 25, 30, 35, 40 which are in ascending order. Where N = 8</p> <p>\begin{align*} Position \: of \: Median &amp;= \frac{N+1^{th}}{2}term \\ &amp;= \frac{8+1^{th}}{2}\\ &amp;= 4.5^{th} term \\ \: \\ Median &amp;= \frac{4^{th}term + 5^{th}term}{2}\\ &amp;= \frac{20+25}{2} \\ &amp;= \frac{45}{2}\\ &amp;= 22.5 _{Ans.} \end{align*}</p>

Q4:

In which class interval median marks lines?

Marks 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100
No. of students 10 16 20 30 10

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <table width="266"><tbody><tr><td>Marks</td> <td>No. of students 9 (f)</td> <td>c.f.</td> </tr><tr><td>0-20</td> <td>10</td> <td>10</td> </tr><tr><td>20-40</td> <td>16</td> <td>26</td> </tr><tr><td>40-60</td> <td>20</td> <td>46</td> </tr><tr><td>60-80</td> <td>30</td> <td>76</td> </tr><tr><td>80-100</td> <td>10</td> <td>86</td> </tr><tr><td></td> <td>N = 86</td> <td></td> </tr></tbody></table><p>\begin{align*} Position\:of\:median\:class &amp;= \frac{N^{th}}{2}term\\ &amp;= \frac{86^{th}}{2}term \\ &amp;= 43^{th} \: term \end{align*}</p> <p>The median class lies in 40 - 60 \(_{Ans}\)</p>

Q5:

Find the median class from the following data.

Marks 0-10 10-20 20-30 30-40 40-50
Frequency 5 7 4 3 11

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <table width="349"><tbody><tr><td><strong>Marks</strong></td> <td><strong>Frequency (f)</strong></td> <td><strong>c.f.</strong></td> </tr><tr><td>0-10</td> <td>5</td> <td>5</td> </tr><tr><td>10-20</td> <td>7</td> <td>12</td> </tr><tr><td>20-30</td> <td>4</td> <td>16</td> </tr><tr><td>30-40</td> <td>3</td> <td>19</td> </tr><tr><td>40-50</td> <td>11</td> <td>30</td> </tr><tr><td></td> <td>N =30</td> <td></td> </tr></tbody></table><p>\begin{align*}Position \: of \: median &amp;= \frac{N^{th}}{2}term\\ &amp;= \frac{30^{th}}{2}term\\ &amp;= 15^{th} term. \end{align*}</p> <p>The median class lies in 20 - 30 \(\: _{Ans.}\)</p>

Q6:

6, 8, 10, 12, 2x, 2x + x, 18, 20, 22, 24 are in ascending order. If its median is 15. Find the value of x.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>N = 10<br>Median = 15</p> <p>\begin{align*} Position \: of \: median &amp;= \frac{N + 1^{th}}{2}term \\ &amp;= \frac{10 + 1^{th}}{2}\\ &amp;= 5.5^{th} term \\ \: \\ Median &amp;= \frac{(5^{th} + 6^{th }) term}{2} \\ 15 &amp;= \frac{2x+2x+x}{2}\\ or, 30 &amp;= 4x + 2 \\ 30 - 2 &amp;= 4x\\ or, x &amp;= \frac{28}{4}\\ \therefore x &amp;= 7 \: \: _{Ans} \end{align*}</p>

Q7:

The numbers 2, 3, 4, 4, 2x + 1, 5, 5, 6, 7 have been put in ascending order. If median is 5, find the value of x.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>From the given data we came to know N= 9<br>Median = 5</p> <p>\begin{align*} Position \: of \: median &amp;= \frac{N + 1^{th}}{2}term \\ &amp;= \frac{9 + 1^{th}}{2}term \\ &amp;= 5^{th} term \\ \: \\ 5^{th } term \: &amp; represent \: 2x+1 \\ \therefore Median &amp;= 2x + 1 \\ or, 5 &amp;= 2x + 1 \\ or, 2x &amp;= 5 - 1 \\ or, x &amp;= \frac{4}{2}\\ \therefore x &amp;= 2 \end{align*}</p>

Q8:

Find the median of all even numbers between 60 and 70.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>The even numbers between 60 and 70 are: 62, 64, 66, 68 where N = 4</p> <p>\(Position \: of \: median = \frac{N + 1^{th}}{2}item = \frac{4 + 1^{th}}{2}item = 2.56^{th} \: item \)</p> <p>\begin{align*} Median &amp;= \frac{64+66}{2}\\ &amp;= \frac{130}{2}\\ &amp;= 65 \: \: _{Ans} \end{align*}</p>

Q9:

Study at the given o -give and find the median class.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>From graph paper sum of the frequency (N) = 30</p> <p>\begin{align*} Median \: class &amp;= \frac{N^{th}}{2}item \\ &amp;= \frac{30^{th}}{2}item\\ &amp;= 15^{th}\: item \\ \text{The corresponding } &amp; value \: of \: 15^{th} \: item \: is \: 10-15. \end{align*}</p>

Q10:

From the median class from the given graph.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>From the graph,<br>Total number of students (N) = 60</p> <p>\begin{align*} Position \: of \: Media \: class &amp;= \frac{N^{th}}{2}term \\ &amp;= \frac{60^{th}}{2}term \\ &amp;= 30^{th} \: term \\ \therefore Median \: class &amp;= 20 - 30 \: _{ans} \end{align*}</p>

Q11:

Find the median class from the given Ogive.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>From the graph,<br>Total number of students (N) = 60<br>\begin{align*} Position \: of \: median \: class &amp;= \frac{N^{th}}{2}term \\ &amp;= \frac{100^{th}}{2} term \\ &amp;= 50^{th} term \\ \therefore Median \: class &amp;= 20 - 30 \end{align*}</p>

Q12:

Find the median class and value of median from the adjoining cumulative frequency curve, where 5 - 10 is a class.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>From the graph,<br>Total number of students (N) = 35</p> <p>\begin{align*} Position \: of \: median \: class &amp;= \frac{N^{th}}{4}term\\ &amp;= \frac{35^{th}}{2}term \\ &amp;= 17.5^{th} \: term \end{align*}</p> <p>From the figure</p> <table width="300"><tbody><tr><td>Class</td> <td>Frequency (f)</td> <td>cf</td> </tr><tr><td>5 - 10</td> <td>5</td> <td>5</td> </tr><tr><td>10 - 15</td> <td>5</td> <td>10</td> </tr><tr><td>15 - 20</td> <td>5</td> <td>15</td> </tr><tr><td>20 - 25</td> <td>4</td> <td>19</td> </tr><tr><td>25 - 30</td> <td>6</td> <td>25</td> </tr><tr><td>30 - 35</td> <td>5</td> <td>30</td> </tr><tr><td>35 - 40</td> <td>5</td> <td>35</td> </tr><tr><td></td> <td>N = 35</td> <td></td> </tr></tbody></table><p>Median class = 20 - 25 = 20 \(_{Ans}\)</p>

Q13:

Find the median class from the given graph.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>From graph, the number of men (N) = 60</p> <p>\begin{align*}\text{Position of median class}&amp;=\frac{N^{th}}{2}term \\ &amp;= \frac{60^{th}}{2}term\\ &amp;= 30^{th} \\ median \: class &amp;= 30 - 40 \: _{Ans} \end{align*}</p>

Q14:

From the o-give curve given below determine the class of \(Q_1 \: and \:  Q_3 \)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>From graph,</p> <p>Number of term (N) = 60</p> <p>\begin{align*} Position \: of\:Q_1\: class &amp;= \frac{N^{th}}{4}term \\ &amp;= \frac{60^{th}}{4}term \\ &amp;=15^{th} \:term\\ Q_1 \: class &amp;= 10-20 \: _{Ans} \\ \: \\ Position \: of \: Q_3\:class&amp;= \frac{3N}{4}term\\ &amp;= \frac{3\times60^{th}}{4} \\ &amp;= 45^{th} \: term \\ Q_3 \: class &amp;= 40-50 \:_{Ans} \end{align*}</p>

Q15:

From the given o-give curve, find the first quartile class.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>From graph, Number of students (N) = 20</p> <p>\begin{align*} Position \: of \: Q_1 &amp;= \frac{N^{th}}{4}term \\ &amp;= \frac{20^{th}}{4}term \\ &amp;= 5^{th} \: term \\ class\:of\:Q_1 &amp;= 8 - 12 \: _{Ans} \end{align*}</p>

Q16:

Find the first quartile class from the given graph.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>From graph paper number of the student (N) = 120</p> <p>\begin{align*}Position\:of\:the \: Q_1\: class &amp;= \frac{N^{th}}{4}term \\ &amp;= \frac{120^{th}}{4}term \\ From\:graph\:the\:corres&amp;ponding \: \text{class of } 30^{th}\:term \:is\: 20 - 40 \\ \therefore First \: quartile \: class &amp;= 20-40 \end{align*}</p>

Videos

Finding mean, median, and mode | Descriptive statistics | Probability and Statistics | Khan Academy
Mean, Median and Mode
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Combination of Cells

Combination of Cells

To increase the amount of current or potential difference required in a circuit we can add cells to the circuit. This can be achieved in two different ways:

  1. Parallel Combination:

    In parallel combination, 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. The positive terminal of all the cells are connected to a single point and then connected to the positive terminal of the circuit and similarly, the negative terminal of all the cells are also connected to another single point and then connected to the negative terminal of the circuit. The cells that are used must have same potential difference. Cells with the different potential difference cannot be used in parallel combination. In this type of combination the potential difference between any two points in a circuit is the same and is equal to the individual potential difference of the cells used i.e. V = V1 = V2 = V3. In this combination the current provided by the cells is small and continuous for longer period of time. If we increase the number of cells the current cannot be increased but time of supply is increased.

  2. Series combination:

    In series combination, the positive terminal of the first cell is connected to the negative terminal of the second cell and positive terminal of the second cell is connected to the negative terminal of the third cell and so on. In this combination of a cell, the overall potential difference is the sum of all the individual potential differences of the cells used in the circuit.
    i.e. V = V1 + V2+ V3+ ...+ Vn.

  3. . The current in the resistance in the series combination of cells is number of times the current due to individual cell. So, the current is larger but for shorter period of time. When we increase the number of cells in the series combination potential difference and current both increases.

Lesson

Electricity and Magnetism

Subject

Science

Grade

Grade 10

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