{"id":1613,"date":"2018-03-23T10:51:47","date_gmt":"2018-03-23T02:51:47","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1613"},"modified":"2023-08-19T00:55:50","modified_gmt":"2023-08-19T00:55:50","slug":"electric-energy-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/electric-energy-problems-and-solutions.htm","title":{"rendered":"Electric energy \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

25 Electric energy \u2013 problems and solutions<\/span><\/p>\n

1. A 220 V – 5 A electric lamp is used for 30 minutes. <\/span>How much energy does it require? <\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Voltage<\/a> (V) = 220 Volt<\/span><\/p>\n

Electric current<\/a> (I) = 5 Ampere<\/span><\/p>\n

Time (t) = 30 minutes = 30 x 60 seconds = 1800 seconds<\/span><\/p>\n

Electric power<\/a> (P) :<\/span><\/p>\n

P = V I = (220 Volt)(5 Ampere) = 1100 Volt Ampere = 1100 Watt = 1100 Joule\/second<\/span><\/p>\n

Electric energy<\/a> = Electric power x time = (1100 Joule\/second)(1800 second) <\/span><\/p>\n

Electric energy = 1,980,000 Joule = 1,980 kiloJoule<\/span><\/p>\n

2. A 220 V \u2013 60 W solder is used for 4 minutes. How much energy does it require.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Power (P) = 60 Watt = 60 Joule\/second<\/span><\/p>\n

Voltage (V) = 220 Volt<\/span><\/p>\n

Time (t) = 4 minutes = 4 x 60 seconds = 240 seconds<\/span><\/p>\n

Wanted:<\/u> Electric power<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

220 Volt – 60 Watt means the electric solder works well if the potential difference or voltage is 220 volts and has a power of 60 Watt = 60 Joule\/second, means that electric solder using the energy of 60 Joules per second.<\/span><\/p>\n

Electric energy = electric power x time interval = (60 Joule\/second)(240 second) = 14,400 Joule.<\/span><\/p>\n

3. The energy used by the iron for 1 minute is 33 kJ, at a voltage of 220 volts. How large the current is in the iron.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Time interval (t) = 1 minute = 60 seconds<\/span><\/p>\n

Energy (W) = 33 kiloJoule = 33,000 Joule <\/span><\/p>\n

Voltage (V) = 220 Volt<\/span><\/p>\n

Wanted :<\/u> Electric current (I)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Electrical power is the electrical energy used during a certain time interval.<\/span><\/span><\/p>\n

P = W \/ t = 33,000 Joule \/ 60 seconds<\/span><\/p>\n

P = 550 Watt<\/span><\/p>\n

Electric current :<\/span><\/p>\n

I = P \/ V = 550 \/ 220 = 2.5 Ampere<\/span><\/p>\n

4. Someone watches TV on average 6 hours each day. The TV is connected to a 220 Volt voltage so that the electric current flows through the TV is 0.5 Amperes. If the electric company charges $0.092 per kWh, then the cost of using electric energy for TV for 1 month (30 days) is…<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Time interval = 6 hours x 30 = 180 hours<\/span><\/p>\n

Voltage (V) = 220 Volt<\/span><\/p>\n

Electric current (I) = 0.5 Ampere<\/span><\/p>\n

Wanted :<\/u> The cost per month<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Power of TV :<\/span><\/p>\n

P = V I = (220 Volt)(0.5 Ampere) = 110 Volt Ampere = 110 Watt<\/span><\/p>\n

Electric energy = electric power x time interval<\/span><\/p>\n

Electric energy of TV = 110 Watt x 180 hours = 19800 Watt hours = 19.8 kilo Watt hours = 19.8 kilo Watt hours = 19.8 kWh<\/span><\/p>\n

The cost of using electric energy for TV during 1 month :<\/span><\/p>\n

19.8 kWh x $ 0.092 \/ kWh = $ 1.8216<\/span><\/p>\n

5. In a house there are 4 lamps 20 Watt, 2 lamps 10 Watt, 3 lamps 40 Watt, are used 5 hours every day. If the electric company charge 0.092 per kWh, then the cost of using electric energy during 1 month (30 days) is ….<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

4 lamps 20 Watt = 4 x 20 Watt = 80 Watt<\/span><\/p>\n

2 lamps 10 Watt = 2 x 10 Watt = 20 Watt<\/span><\/p>\n

3 lamps 40 Watt = 3 x 40 Watt = 120 Watt<\/span><\/p>\n

Total power (W) = 80 Watt + 20 Watt + 120 Watt = 220 Watt<\/span><\/p>\n

Time interval (t) = 5 hours x 30 = 150 hours<\/span><\/p>\n

Wanted :<\/u> The cost of using electric energy during 1 month (30 days) <\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Electric energy = electric power x time interval = 220 Watt x 150 hours = 33,000 Watt hour = 33 kilo Watt hour = 33 kilo Watt hour = 33 kWh<\/span><\/p>\n

The cost of using electric energy during 1 month (30 days) <\/span><\/p>\n

(33 kWh) ( 0.092 \/ kWh) = $ 3.036<\/span><\/p>\n

6. A circuit has a resistance of \\(5\\Omega\\) and a current of \\(2A\\). Find the electric power.<\/span>
\nSolution: \\( P = IV = 2A \\times 5\\Omega = 10W \\)<\/span><\/p>\n

7. A \\(10V\\) battery is connected to a \\(2\\Omega\\) resistor. Calculate the current.<\/span>
\nSolution: \\( I = \\frac{V}{R} = \\frac{10V}{2\\Omega} = 5A \\)<\/span><\/p>\n

8. A light bulb rated at \\(60W\\) operates on a \\(120V\\) supply. Find the resistance.<\/span>
\nSolution: \\( R = \\frac{V^2}{P} = \\frac{120^2}{60} = 240\\Omega \\)<\/span><\/p>\n

9. Calculate the energy consumed by a \\(100W\\) bulb in \\(2\\) hours.<\/span>
\nSolution: \\( E = Pt = 100W \\times 2h = 200Wh = 0.2kWh \\)<\/span><\/p>\n

10. A \\(20\\mu F\\) capacitor is charged to \\(50V\\). Find the stored energy.<\/span>
\nSolution: \\( E = \\frac{1}{2} CV^2 = \\frac{1}{2} \\times 20 \\times 10^{-6}F \\times 50^2 = 25mJ \\)<\/span><\/p>\n

11. Determine the current in a \\(10mH\\) inductor when the rate of change of current is \\(5A\/s\\).<\/span>
\nSolution: \\( V = L\\frac{di}{dt} = 10 \\times 10^{-3}H \\times 5A\/s = 50mV \\)<\/span><\/p>\n

12. Find the power loss in a transmission line with a resistance of \\(8\\Omega\\) and a current of \\(3A\\).<\/span>
\nSolution: \\( P = I^2R = 3^2 \\times 8\\Omega = 72W \\)<\/span><\/p>\n

13. A transformer has a primary voltage of \\(120V\\) and a secondary voltage of \\(240V\\). If the primary current is \\(2A\\), find the secondary current.<\/span>
\nSolution: Secondary current = \\( \\frac{{\\text{{Primary voltage}}}}{{\\text{{Secondary voltage}}}} \\times \\text{{Primary current}} = \\frac{{120V}}{{240V}} \\times 2A = 1A \\)<\/span><\/p>\n

14. Calculate the reactance of a \\(50Hz\\), \\(0.2H\\) inductor.<\/span>
\nSolution: \\( X_L = 2\\pi fL = 2\\pi \\times 50 \\times 0.2 = 62.83\\Omega \\)<\/span><\/p>\n

15. Find the total capacitance of three \\(10\\mu F\\) capacitors connected in series.<\/span>
\nSolution: \\( \\frac{1}{{C_{\\text{{total}}}}} = \\frac{1}{{C_1}} + \\frac{1}{{C_2}} + \\frac{1}{{C_3}} = \\frac{1}{10} + \\frac{1}{10} + \\frac{1}{10} = \\frac{3}{10}, C_{\\text{{total}}} = \\frac{10}{3} \\mu F \\approx 3.33\\mu F \\)<\/span><\/p>\n

16. A \\(0.1F\\) supercapacitor is charged to \\(5V\\). Find the energy stored.<\/span>
\nSolution: \\( E = \\frac{1}{2} CV^2 = \\frac{1}{2} \\times 0.1 \\times 5^2 = 1.25J \\)<\/span><\/p>\n

17. Calculate the power factor of a circuit with a real power of \\(60W\\) and an apparent power of \\(75VA\\).<\/span>
\nSolution: Power factor = \\( \\frac{{\\text{{Real Power}}}}{{\\text{{Apparent Power}}}} = \\frac{{60W}}{{75VA}} \\approx 0.8 \\)<\/span><\/p>\n

18. Find the equivalent resistance of two \\(4\\Omega\\) resistors in parallel.<\/span>
\nSolution: \\( \\frac{1}{{R_{\\text{{eq}}}}} = \\frac{1}{{R_1}} + \\frac{1}{{R_2}} = \\frac{1}{4} + \\frac{1}{4} = \\frac{1}{2}, R_{\\text{{eq}}} = 2\\Omega \\)<\/span><\/p>\n

19. Determine the impedance of a circuit with a resistance of \\(5\\Omega\\) and a reactance of \\(12\\Omega\\).<\/span>
\nSolution: \\( Z = \\sqrt{R^2 + X^2} = \\sqrt{5^2 + 12^2} \\approx 13.34\\Omega \\)<\/span><\/p>\n

15. Calculate the RMS value of a sinusoidal voltage with a peak value of \\(120V\\).<\/span>
\nSolution: \\( V_{\\text{{RMS}}} = \\frac{{V_{\\text{{peak}}}}}{\\sqrt{2}} \\approx 84.85V \\)<\/span><\/p>\n

16. Find the power dissipated in a \\(5\\Omega\\) resistor with a current of \\(4A\\).<\/span>
\nSolution: \\( P = I^2R = 4^2 \\times 5\\Omega = 80W \\)<\/span><\/p>\n

17. Calculate the frequency of a \\(25mH\\) inductor with a reactance of \\(50\\Omega\\).<\/span>
\nSolution: \\( f = \\frac{{X_L}}{{2\\pi L}} = \\frac{50}{{2\\pi \\times 25 \\times 10^{-3}}} \\approx 31.83Hz \\)<\/span><\/p>\n

18. Determine the current in a \\(20\\Omega\\) resistor with a \\(100V\\) supply.<\/span>
\nSolution: \\( I = \\frac{V}{R} = \\frac{100V}{20\\Omega} = 5A \\)<\/span><\/p>\n

19. Find the total inductance of two \\(10mH\\) inductors connected in parallel.<\/span>
\nSolution: \\( \\frac{1}{{L_{\\text{{total}}}}} = \\frac{1}{{L_1}} + \\frac{1}{{L_2}} = \\frac{1}{10} + \\frac{1}{10} = \\frac{1}{5}, L_{\\text{{total}}} = 5mH \\)<\/span><\/p>\n

20. Calculate the energy consumed by a \\(1500W\\) iron used for \\(0.5\\) hours.<\/span>
\nSolution: \\( E = Pt = 1500W \\times 0.5h = 750Wh = 0.75kWh \\)<\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

25 Electric energy \u2013 problems and solutions 1. A 220 V – 5 A electric lamp is used for 30 minutes. How much energy does it require? Solution : Voltage (V) = 220 Volt Electric current (I) = 5 Ampere Time (t) = 30 minutes = 30 x 60 seconds = 1800 seconds Electric power … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Electric energy \u2013 problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-1613","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1613","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=1613"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1613\/revisions"}],"predecessor-version":[{"id":9009,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1613\/revisions\/9009"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1613"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1613"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1613"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}