Types of Current and Domestic Electrification

There are two types of current. They are AC current and DC current. In DC current polarity of current does not change while in Ac current polarity is changed. This note explains about types of current i.e. AC and DC currents.

Summary

There are two types of current. They are AC current and DC current. In DC current polarity of current does not change while in Ac current polarity is changed. This note explains about types of current i.e. AC and DC currents.

Things to Remember

  • There are two types of current. They are AC current and DC current.
  • In DC current the polarity of the current does not change.
  • In AC current the polarity of the current changes according to its frequency.
  • Fuse is the special wire that melts when certain amount of electricity flows through it, Fuses are made according to the maximum amount of current flowing into the circuit.
  • To know what capacity of fuse to use we have to find out how much of current is needed in the circuit. For that we have to know the potential difference (V) of the circuit and total power drawn (P) by the appliances in that circuit.

MCQs

No MCQs found.

Subjective Questions

Q1:

If mean(\(\overline X = 50\)) and \(\sum fx = 750,\) find the number of terms (N).


Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>Mean(\(\overline X\))= 5o<br>\(\sum fx = 750 \)<br>N = ?</p> <p>\begin{align*} Mean\:(\overline {X}) &amp;= \frac{\sum fx}{N}\\ 50 &amp;= \frac{750}{N}\\ or, N &amp;= \frac{750}{50}\\ \therefore N &amp;= 15 \: \: \: \: _{Ans} \end{align*}</p>

Q2:

If \(Mean \: (\overline{X}) = 60 \) and \(\sum fx = 960,\) find the number.


Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>\(Mean \: (\overline{X}) =60 \\ \sum fx = 960 \\ Number \: of \: term \: (N) = ? \)</p> <p>\begin{align*} Mean (\overline {X}) &amp;= \frac{\sum fx}{N}\\ 60 &amp;= \frac{960}{N}\\ or, N &amp;= \frac{960}{60}\\ \therefore N &amp;= 16 \: _{Ans} \end{align*}</p>

Q3:

If \(Mean \: (\overline{X}) = 12, \sum fx = 70 + 10a\) and the number of frequency (N) = 5 + a, find the value of a.


Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>\(Mean \: (\overline{X}) =12 \\ \sum fx = 70 +10a\\ No. \: of \: term \: (N) = 5+a \)</p> <p>\begin{align*}Mean \: (\overline{X}) &amp;= \frac{\sum fx}{N}\\ 12 &amp;= \frac{70 + 10a}{5 + a}\\ or, 60 +12a &amp;=70 + 10a\\ or,12a -10a &amp;= 70 - 60\\ or, 2a &amp;= 10 \\ or, a &amp;= \frac{10}{2}\\ \therefore a &amp;= 5 \: \: _{Ans} \end{align*}</p>

Q4:

Find the value of x, If the mean of 4, 8, 12, x and 25 is 13.


Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>\(Mean \: (\overline{X}) = 13 \\ No. \: of \: terms \: (N) = 5 + a \)</p> <p>\begin{align*} Mean (\overline{X}) &amp;= \frac{\sum X}{N} \\ or, 13 &amp;= \frac{4 + 8+12+x+25}{5}\\ or, 65 &amp;=49 + x\\ or, x &amp;= 65 - 49 \\ x &amp;= 16 \:\: _{Ans} \end{align*}</p>

Q5:

The mean of 4 different number is 7 and the mean of 3 different number is 12. Find the mean of 7 numbers.


Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>First item mean \( (\overline {X_{1}}) = 7 \)<br>Number of first item \(n_{1} = 4 \)<br>Second item mean \( (\overline {X_{2}}) = 12\)<br>No. of Second item \(n_{2}\)= 3<br>Mean of 7 items \(\overline{X}= ?\)</p> <p>\begin{align*} ( \overline {X}) &amp;= \frac{n_1 \overline{X}_1 + n_2 \overline{X}_2}{n_1+n_2}\\ &amp;= \frac{4 \times 7 + 3 \times 12}{4 + 3}\\ &amp;= \frac{28 + 36}{7}\\ &amp;= \frac{64}{7}\\ &amp;= 9.14 \: \: \: _{ans} \end{align*}</p>

Q6:

If the mean of 10, 20, 30, 40, 50, 60 and 30 + k is 40, find the value of k.


Type: Short Difficulty: Easy

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Answer: <p>\(Mean (\overline{X}) = 40 \\ No. \: of \: terms (N)= 7 \\ k = ? \)</p> <p>\begin{align*} Mean(\overline{X}) &amp;= \frac{\sum X}{N} \\ or, 40 &amp;= \frac{10+20+30+40+50+ 60+30+k}{7}\\ or, 280 &amp;= 240 + k\\ or, k &amp;= 280 - 240\\ \therefore k &amp;= 40 \: \: \: \: \: \: \: \: \: _{Ans} \end{align*}</p>

Q7:

The expenditure (in rupees) of 15 days of a man is as follows. Calculate his average expenditure.


Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <table width="286"><tbody><tr><td>Expenditure (in Rs.)x</td> <td>Frequency(f)</td> <td>fx</td> </tr><tr><td>24</td> <td>2</td> <td>48</td> </tr><tr><td>25</td> <td>4</td> <td>100</td> </tr><tr><td>30</td> <td>3</td> <td>90</td> </tr><tr><td>35</td> <td>4</td> <td>140</td> </tr><tr><td>40</td> <td>2</td> <td>80</td> </tr></tbody></table><p>\begin{align*} Mean(\overline{X}) &amp;= \frac{\sum fx}{N}\\ &amp;= \frac{458}{15}\\ &amp;= 30.53 \\ \therefore Average \: expenditure &amp;= Rs \: 30.53 \end{align*}</p> <p></p>

Q8:

The average age of 5 students is 9 years. Out of them the ages of 4 students are 5, 7, 8 and 15 years. What was the age of the remaining students.


Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>Let, the age of remaining students be x.</p> <p>\(average \: age (\overline{X}) = 9 \\ Number \: of \: terms (N) = 5 \)</p> <p>\begin{align*} (\overline{X}) &amp;= \frac{\sum X}{N}\\ or, 9 &amp;= \frac{5+7+8+15+x}{5}\\ or, 45 &amp;= 35 + x \\ \therefore x &amp;=45 -35 = 10 \\ \: \\ \therefore &amp; \text{The age of remaining student is 10} \end{align*}</p>

Q9:

Construct a frequency table of class interval of 10 from the given data and find the mean.
23, 5, 17, 28, 39,  52, 21, 16, 29, 41, 25, 33, 9, 19, 34, 59, 7, 11, 51, 31, 22, 41, 32, 55, 31, 22, 41, 32, 55, 18


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>Calculation of mean</p> <table width="332"><tbody><tr><td>Class interval</td> <td>mid-value (m)</td> <td>frequency (f)</td> <td>fm</td> </tr><tr><td>0-10</td> <td>5</td> <td>3</td> <td>15</td> </tr><tr><td>10-20</td> <td>15</td> <td>5</td> <td>75</td> </tr><tr><td>20-30</td> <td>25</td> <td>6</td> <td>150</td> </tr><tr><td>30-40</td> <td>35</td> <td>5</td> <td>175</td> </tr><tr><td>40-50</td> <td>45</td> <td>2</td> <td>90</td> </tr><tr><td>50-60</td> <td>55</td> <td>4</td> <td>220</td> </tr><tr><td></td> <td></td> <td>N = 25</td> <td></td> </tr></tbody></table><p>We know that,</p> <p>\begin{align*} Mean (\overline{X} ) &amp;= \frac{\sum fm}{N}\\ &amp;= \frac{725}{25}\\ &amp;= 29 \: _{Ans} \end{align*}</p>

Q10:

Find the mean from the following data.

Class-interval 10-20 10-30 10-40 10-50 10-60 10-70 10-80 10-90
Frequency 4 16 56 97 124 137 146 150

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>Calculation of mean</p> <table width="396"><tbody><tr><td>Class interval</td> <td>cf.</td> <td>f</td> <td>mid-value (m)</td> <td>fm</td> </tr><tr><td>10-20</td> <td>4</td> <td>4</td> <td>15</td> <td>60</td> </tr><tr><td>20-30</td> <td>16</td> <td>12</td> <td>25</td> <td>300</td> </tr><tr><td>30-40</td> <td>56</td> <td>40</td> <td>35</td> <td>1400</td> </tr><tr><td>40-50</td> <td>97</td> <td>41</td> <td>45</td> <td>1845</td> </tr><tr><td>50-60</td> <td>124</td> <td>27</td> <td>55</td> <td>1485</td> </tr><tr><td>60-70</td> <td>137</td> <td>13</td> <td>65</td> <td>845</td> </tr><tr><td>70-80</td> <td>146</td> <td>9</td> <td>75</td> <td>675</td> </tr><tr><td>80-90</td> <td>150</td> <td>4</td> <td>85</td> <td>340</td> </tr><tr><td></td> <td></td> <td>N = 150</td> <td></td> <td>\(\sum fm=6950\)</td> </tr></tbody></table><p>\begin{align*}Mean \: (\overline{X})&amp;= \frac{\sum fm}{N}\\ &amp;= \frac{6950}{150}\\ &amp;= 46.33 \: _{Ans} \end{align*}</p>

Q11:

Calculate the average wages from the following data.

Wages 50-60 50-70 50-80 50-90 50-100
No. of people 6 14 26 34 40

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>Calculation of mean</p> <table width="360"><tbody><tr><td>Wages</td> <td>cf</td> <td>f</td> <td>m</td> <td>fm</td> </tr><tr><td>50-60</td> <td>6</td> <td>6</td> <td>55</td> <td>330</td> </tr><tr><td>60-70</td> <td>14</td> <td>8</td> <td>65</td> <td>520</td> </tr><tr><td>70-80</td> <td>26</td> <td>12</td> <td>75</td> <td>900</td> </tr><tr><td>80-90</td> <td>34</td> <td>8</td> <td>85</td> <td>680</td> </tr><tr><td>90-100</td> <td>40</td> <td>6</td> <td>95</td> <td>570</td> </tr><tr><td></td> <td></td> <td>N=40</td> <td></td> <td>\(\sum fm= 3000\)</td> </tr></tbody></table><p>\begin{align*} Mean (\overline {X})&amp;= \frac{\sum fm}{N}\\ &amp;= \frac{3000}{40}\\ &amp;= 75 \: _{Ans} \end{align*}</p>

Q12:

Calculate the average wages from the following data.

Wages 50-60 50-70 50-80 50-90 50-100
No. of people 6 14 26 34 40

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>Calculation of mean</p> <table width="360"><tbody><tr><td>Wages</td> <td>cf</td> <td>f</td> <td>m</td> <td>fm</td> </tr><tr><td>50-60</td> <td>6</td> <td>6</td> <td>55</td> <td>330</td> </tr><tr><td>60-70</td> <td>14</td> <td>8</td> <td>65</td> <td>520</td> </tr><tr><td>70-80</td> <td>26</td> <td>12</td> <td>75</td> <td>900</td> </tr><tr><td>80-90</td> <td>34</td> <td>8</td> <td>85</td> <td>680</td> </tr><tr><td>90-100</td> <td>40</td> <td>6</td> <td>95</td> <td>570</td> </tr><tr><td></td> <td></td> <td>N=40</td> <td></td> <td>\(\sum fm= 3000\)</td> </tr></tbody></table><p>\begin{align*} Mean (\overline {X})&amp;= \frac{\sum fm}{N}\\ &amp;= \frac{3000}{40}\\ &amp;= 75 \: _{Ans} \end{align*}</p>

Q13:

Calculate the mean from the following data.

X 0-10 10-20 20-30 30-40 40-50
f 5 7 8 4 6

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>Calculation of mean</p> <table width="337"><tbody><tr><td>class interval</td> <td>midvalue</td> <td>f</td> <td>fm</td> </tr><tr><td>0-10</td> <td>5</td> <td>5</td> <td>25</td> </tr><tr><td>10-20</td> <td>15</td> <td>7</td> <td>105</td> </tr><tr><td>20-30</td> <td>25</td> <td>8</td> <td>200</td> </tr><tr><td>30-40</td> <td>35</td> <td>4</td> <td>140</td> </tr><tr><td>40-50</td> <td>45</td> <td>6</td> <td>270</td> </tr><tr><td></td> <td></td> <td>N= 30</td> <td>\( \sum fm = 740 \)</td> </tr></tbody></table><p>\begin{align*} Mean (\overline{X}) &amp;= \frac{\sum fm}{N}\\ &amp;= \frac{740}{30}\\ &amp;= 24.6 \: _{ans} \end{align*}</p>

Q14:

Compute the mean from the table given below.

Marks 10-20 20-30 30-40 40-50 50-60 60-70
No. of students 2 5 7 6 3 2

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>Calculation of mean</p> <table width="227"><tbody><tr><td>Marks</td> <td>f</td> <td>m</td> <td>fm</td> </tr><tr><td>10-20</td> <td>2</td> <td>15</td> <td>30</td> </tr><tr><td>20-30</td> <td>5</td> <td>25</td> <td>125</td> </tr><tr><td>30-40</td> <td>7</td> <td>35</td> <td>245</td> </tr><tr><td>40-50</td> <td>6</td> <td>45</td> <td>270</td> </tr><tr><td>50-60</td> <td>3</td> <td>55</td> <td>165</td> </tr><tr><td>60-70</td> <td>2</td> <td>65</td> <td>130</td> </tr><tr><td></td> <td>N=25</td> <td></td> <td>\(\sum fm = 965\)</td> </tr></tbody></table><p>We know that,</p> <p>\begin{align*} Mean (\overline {X}) &amp;= \frac{\sum fm }{N} \\ &amp;= \frac{965}{25}\\ &amp;= 38.6 \: \: _{Ans} \end{align*}</p> <p></p> <p></p> <p></p>

Q15:

If the mean is 32, find the value of p from the following.

x 5-15 15-25 25-35 35-45 45-55 55-65
f 5 8 p 9 7 1

Type: Long Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>To find the value of p.</p> <table width="389"><tbody><tr><td>X</td> <td>f</td> <td>m</td> <td>fm</td> </tr><tr><td>5-15</td> <td>5</td> <td>10</td> <td>50</td> </tr><tr><td>15-25</td> <td>8</td> <td>20</td> <td>160</td> </tr><tr><td>25-35</td> <td>p</td> <td>30</td> <td>30p</td> </tr><tr><td>35-45</td> <td>9</td> <td>40</td> <td>360</td> </tr><tr><td>45-55</td> <td>7</td> <td>50</td> <td>350</td> </tr><tr><td>55-65</td> <td>1</td> <td>60</td> <td>60</td> </tr><tr><td></td> <td>N = p + 30</td> <td></td> <td>\( \sum fm = 30p+ 980 \)</td> </tr></tbody></table><p>We know that,</p> <p>\begin{align*}Mean (\overline{X})&amp;= \frac{\sum fm}{N}\\ 32 &amp;= \frac{30p + 980}{p + 30}\\ or, 32p + 960 &amp;= 30p + 980\\ or, 32p -30p &amp;= 980 - 960 \\ or, p &amp;= \frac{20}{2}\\ \therefore p&amp;=10 \: _{Ans} \end{align*}</p>

Q16:

The mean of the data given below is 36, find the value of p.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>To find the value of p.</p> <p><strong>Solution:</strong></p> <p>To find the value of p.</p> <table width="389"><tbody><tr><td>Age in year</td> <td>No. of teachers (f)</td> <td>m</td> <td>fm</td> </tr><tr><td>10-20</td> <td>3</td> <td>15</td> <td>45</td> </tr><tr><td>20-30</td> <td>8</td> <td>25</td> <td>200</td> </tr><tr><td>30-40</td> <td>15</td> <td>35</td> <td>525</td> </tr><tr><td>40-50</td> <td>p</td> <td>45</td> <td>45p</td> </tr><tr><td>50-60</td> <td>4</td> <td>55</td> <td>220</td> </tr><tr><td></td> <td>N=p+30</td> <td></td> <td>\( \sum fm = 45p + 990\)</td> </tr></tbody></table><p>We know that,</p> <p>\begin{align*} Mean \: (\overline {X}) &amp;= \frac{\sum fm}{N}\\ 36 &amp;= \frac{45p + 990}{p+30}\\ or, 1080 + 36p &amp;= 45p +990\\ or, 45p - 36p &amp;= 1080 -990 \\ or, 9p &amp;= 90 \\ \therefore p &amp;= \frac{90}{9} = 10 \: _{Ans} \end{align*}</p>

Q17:

The mean of the data given below is 36, find the value of p.


Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>To find the value of p.</p> <p><strong>Solution:</strong></p> <p>To find the value of p.</p> <table width="389"><tbody><tr><td>Age in year</td> <td>No. of teachers (f)</td> <td>m</td> <td>fm</td> </tr><tr><td>10-20</td> <td>3</td> <td>15</td> <td>45</td> </tr><tr><td>20-30</td> <td>8</td> <td>25</td> <td>200</td> </tr><tr><td>30-40</td> <td>15</td> <td>35</td> <td>525</td> </tr><tr><td>40-50</td> <td>p</td> <td>45</td> <td>45p</td> </tr><tr><td>50-60</td> <td>4</td> <td>55</td> <td>220</td> </tr><tr><td></td> <td>N=p+30</td> <td></td> <td>\( \sum fm = 45p + 990\)</td> </tr></tbody></table><p>We know that,</p> <p>\begin{align*} Mean \: (\overline {X}) &amp;= \frac{\sum fm}{N}\\ 36 &amp;= \frac{45p + 990}{p+30}\\ or, 1080 + 36p &amp;= 45p +990\\ or, 45p - 36p &amp;= 1080 -990 \\ or, 9p &amp;= 90 \\ \therefore p &amp;= \frac{90}{9} = 10 \: _{Ans} \end{align*}</p>

Q18:

If the mean of the data given below is 39. Calculate the value of m.

Wages 15-25 25-35 35-45 45-55 55-65
No. of workers 4 6 12 m 3

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>To find the value of m.</p> <table width="346"><tbody><tr><td>Wages</td> <td>f</td> <td>m</td> <td>fm</td> </tr><tr><td>15-25</td> <td>4</td> <td>20</td> <td>80</td> </tr><tr><td>25-35</td> <td>6</td> <td>30</td> <td>180</td> </tr><tr><td>35-45</td> <td>12</td> <td>40</td> <td>480</td> </tr><tr><td>45-55</td> <td>m</td> <td>50</td> <td>50m</td> </tr><tr><td>55-65</td> <td>3</td> <td>60</td> <td>180</td> </tr><tr><td></td> <td>N = m + 25</td> <td></td> <td>\( \sum fm = 50m + 920\)</td> </tr></tbody></table><p>\begin{align*} Mean (\overline{X}) &amp;= \frac{\sum fm}{N}\\ 39 &amp;= \frac{50m+920}{m+25}\\ or, 39m + 975 &amp;= 50m + 920 \\ or, 50m - 39m &amp;= 975 - 920 \\ or, 11m &amp;= 55 \\ or, x&amp;=\frac{55}{11}\\ \therefore m &amp;= 5 \: _{Ans} \end{align*}</p> <p></p>

Q19:

If the mean of the following data is 53, find the value k.

Class 0-20 20-40 40-60 60-80 80-100
Frequency 15 k 21 29 17

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>To find the value of k.</p> <table width="245"><tbody><tr><td>Class</td> <td>f</td> <td>m</td> <td>fm</td> </tr><tr><td>0-20</td> <td>15</td> <td>10</td> <td>150</td> </tr><tr><td>20-40</td> <td>k</td> <td>30</td> <td>30k</td> </tr><tr><td>40-60</td> <td>21</td> <td>50</td> <td>1050</td> </tr><tr><td>60-80</td> <td>29</td> <td>70</td> <td>2030</td> </tr><tr><td>80-100</td> <td>17</td> <td>90</td> <td>1530</td> </tr><tr><td></td> <td>N=82+k</td> <td></td> <td>\( \sum fm = 30k + 4760 \)</td> </tr></tbody></table><p></p> <p>\begin{align*} Mean \: (\overline{X} ) &amp;= \frac{\sum fm}{N}\\ 53 &amp;= \frac{30k + 4760}{82 + k}\\ or, 4346 + 53k &amp;= 30k + 4760 \\ or, 53k - 30k &amp;= 4760 - 4346 \\ or, 23k &amp;= 414 \\ or, k &amp;= \frac{414}{23}\\ \therefore k &amp;= 18 \: _{Ans} \end{align*}</p> <p></p>

Q20:

Calculate mean from the following data.

Marks 20-30 30-40 40-50 50-60 60-70
Frequency 12 7 8 3 10

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>Calculating mean</p> <table width="395"><tbody><tr><td>Marks</td> <td>frequency (f)</td> <td>mid value (m)</td> <td>fm</td> </tr><tr><td>20-30</td> <td>12</td> <td>25</td> <td>300</td> </tr><tr><td>30-40</td> <td>7</td> <td>35</td> <td>245</td> </tr><tr><td>40-50</td> <td>8</td> <td>45</td> <td>360</td> </tr><tr><td>50-60</td> <td>3</td> <td>55</td> <td>165</td> </tr><tr><td>60-70</td> <td>10</td> <td>65</td> <td>650</td> </tr><tr><td></td> <td>N = 40</td> <td></td> <td>\(\sum fm = 1720\)</td> </tr></tbody></table><p>\begin{align*} Mean \: (\overline{X}) &amp;= \frac{\sum fm}{N}\\ &amp;= \frac{1720}{40}\\ &amp;= 43 \: _{Ans} \end{align*}</p>

Q21:

The mean and total number of students are 31 and 50 find the missing frequency.

Marks 0-10 10-20 20-30 30-40 40-50 50-60
No. of students 4 x 10 y 6 4

Type: Long Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>To find the value of x and y</p> <table width="320"><tbody><tr><td>Marks</td> <td>f</td> <td>mid value</td> <td>fm</td> </tr><tr><td>0-10</td> <td>4</td> <td>5</td> <td>20 </td> </tr><tr><td>10-20</td> <td>x</td> <td>15</td> <td>15x</td> </tr><tr><td>20-30</td> <td>10</td> <td>25 </td> <td>250</td> </tr><tr><td>30-40</td> <td>y</td> <td>35</td> <td>35y</td> </tr><tr><td>40-50</td> <td>6</td> <td>45</td> <td>270</td> </tr><tr><td>50-60</td> <td>4</td> <td>55 </td> <td>220</td> </tr><tr><td></td> <td>\( N= x+y +24\)</td> <td></td> <td>\(\sum fm= 15x +35y +760 \)</td> </tr></tbody></table><p>\begin{align*} Mean (\overline{X}) &amp;= \frac{\sum fm}{N}\\ 31 &amp;= \frac{15x +35y+760}{50}\\ or, 1150 - 760 &amp;= 15x+35y \\ or, 15x + 35y &amp;= 790 \: ... .. .. . .. (1) \\ x+y+24 &amp;= 50 \\ or, y&amp;=50-24-x \\ y&amp;=26-x ....................(2) \\ Putting \: valu&amp;e \: of \: y \: in\:equation \: ....(1)\\ 15x + 35y&amp;=790\\ or, 15x + 35(26-x) &amp;= 790 \\ or, 15x+910-35x &amp;= 790\\ or, -20x &amp;= 790-910 \\ or, x &amp;= \frac{-120}{-20}\\ \therefore x &amp;= 6 \\ \: \\ putting \: value \: of \: x &amp; \: in \: equation .. (2) \\ y &amp;= 26-6 \\ \therefore y&amp;=20 \end{align*}</p>

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Types of Current and Domestic Electrification

Types of Current and Domestic Electrification

According to the flow of charged particle in a circuit, there are two types of current.

  1. DC Current: In DC current the polarity of the current does not change. The flow of current is linear from the negative terminal to positive terminal. The voltage of direct current is normally the same and small. DC current can be changed to AC current by the use of an inverter.

  2. AC current: In AC current the polarity of the current changes according to its frequency. In Nepal, the distributed electricity has 50Hz of frequency that means the polarity of the current changes 50 times a second. AC current can be changed to dc current by the use of a rectifier.

Fuse:

Fuse is the special wire that melts when a certain amount of electricity flows through it. It happens due to heating effect (link to the heating effect) caused by electricity and we are going to talk about it later on. Fuses are made according to the maximum amount of current flowing into the circuit. If a circuit needs small amperage of current then we can use fuse with low amperage (in household normally 5 to 15A of fuse is used) and if the circuit needs high amount of current then high amperage of fuse is used (in small scale industries up to 30A of fuse is used and in heavy industries fuse use varies greatly). Fuse is made from a special alloy of tin and lead and the amount of lead and tin determines the capacity of the fuse. The small wire is enclosed in a glass case and sealed with small iron plates. But if we are using a fuse box then, the setting is different.

Selection of fuse:

To know what capacity of a fuse to use we have to find out how much of current is needed in the circuit. For that, we have to know the potential difference (V) of the circuit and total power drawn (P) by the appliances in that circuit. Using the formula, I = PV.We know the total current flowing in the circuit. Now if we use a fuse that has less amperage than the current flowing in the circuit the fuse cannot handle the flow of charged particles and breaks. So, we have to use the slightly higher capacity of fuse than that of current in our circuit.

Domestic Electrification

In Nepal, the electricity distributed comes in two wires one live (also called phase) wire and another neutral. The live wire acts as the positive terminal and neutral wire act as the negative terminal. Both the wires are connected to a meter box in our home, also, the live wire has fuse associated with it. All the current flowing in our home flows through it. The normal capacity of that fuse is 15A but it can be increased or decreased according to the household needs. Then from another side of the meter box both (live and neutral wires) are taken out. Those lines are taken into a switch box from where lines for different rooms are panned out. The switch box also has its own fuse for each line taken out from it that regulates current going into each room or distribution board. If distribution board is used then the individual lines to rooms are distributed from here. In our home along with live and neutral wire, an earthing wire is also added. Earthing is a system where excessive electricity is passed through it into the ground. Earthing can be done in various ways. Mostly in earthing a wire is attached to a plate or nail which is inserted into the ground from the main switch box. Now the electric circuit of our home consists of three wires; the live wire that carries current and act as the positive terminal, the neutral wire that act as the negative terminal and earthing wire that carries any excessive electricity from the circuit into the ground.

Things to remember while wiring a home:

  1. A switch must be added the live wire. So that when the switch is off there is no electricity running through the wire.
  2. Fuse if added must be added to the live wire. Because if electrical appliances draw more current than needed fuse will go off and there won't be excessive current in the circuit.
  3. The capacity of fuse must be chosen appropriately to the current requirement. The normal household does not require much of electricity so smaller fuse can be installed but for small scale industries larger fuse than the household is needed and for large industries where heavy machinery are to be operated, a large capacity fuse is required.
  4. Each room or flat must have different fuse associated with them. If one fuse goes off then only that room's (flat) electricity is cut off rather than the entire house.
  5. The places where wires are laid and switches are placed must be prevented from water or moistness. Since water can conduct electricity we can get shocks and loss of electricity can happen.
  6. The line to the power socket and the bulbs must be separate. If there is a short circuit in the power socket, then also, we can light bulbs.

Coloration of Wire:

When we see a wire we actually see copper wire enclosed in a casing of different colors. Those casings are there to insulate the wire and we do not get a shock while touching it. The colors of the wire differ according to their type and country it is found. Normally in Nepal live wire has a red casing to it, neutral has black and earthing has a yellow color. But some might use brown colored wire for live wires, blue for neutral and green for earthing. The coloration is done because it is easier to know which wire is live or neutral or earthing.

Lesson

Electricity and Magnetism

Subject

Science

Grade

Grade 10

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