{"id":1605,"date":"2018-03-16T12:38:41","date_gmt":"2018-03-16T04:38:41","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1605"},"modified":"2023-08-19T01:06:06","modified_gmt":"2023-08-19T01:06:06","slug":"electric-power-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/electric-power-problems-and-solutions.htm","title":{"rendered":"Electric power – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

28 Electric power – problems and solutions<\/span><\/p>\n

1. Calculate the electric current<\/a> of a 10 Watt lamp designed for 12 Volt.<\/span><\/p>\n

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

Power <\/a>(P) = 10 Watt<\/span><\/p>\n

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

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

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

Lamp power = 10 Watt = 10 Joule\/second<\/span><\/p>\n

The equation below shows the relation between the electric power (P), electric potential<\/a> (V) and electric current (I) stated in formula below :<\/span><\/p>\n

P = V I<\/span><\/p>\n

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

I = P \/ V = 10 \/ 220 = 1\/22 = 0.045 Ampere<\/span><\/p>\n

2. Calculate the electric power according to the electric circuit below.<\/span><\/p>\n

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

Resistor<\/a> 1 (R1<\/sub>) = 6 \u03a9<\/span><\/p>\n

Resistor 2 (R2<\/sub>) = 6 \u03a9<\/span><\/p>\n

Electric voltage (V) = 24 Volt\"Electric<\/span><\/p>\n

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

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

Equivalent resistor :<\/b><\/span><\/p>\n

R1<\/sub> and R2 <\/sub>are connected in series. The equivalent resistor : <\/span><\/p>\n

R = R1 <\/sub>+ R2 <\/sub>= 6 \u03a9 + 6 \u03a9 = 12 \u03a9<\/span><\/p>\n

The electric current :<\/span><\/p>\n

I = V\/R = 24\/12 = 2 Ampere<\/span><\/p>\n

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

P = V I = (24)(2) = 48 Watt = 48 Joule\/second<\/span><\/p>\n

3. Calculate the electric current of a 3 kiloWatt designed for 150 Volt.<\/span><\/p>\n

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

Electric power (P) = 3 kiloWatt = 3000 Watt<\/span><\/p>\n

Electric voltage (V) = 150 Volt <\/span><\/p>\n

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

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

Relation of the electric power (P), electric potential (V) and electric current (I) stated in formula below :<\/span><\/p>\n

P = V I<\/span><\/p>\n

The electric current :<\/span><\/p>\n

I = P \/ V<\/span><\/p>\n

I = 3000 Watt \/ 150 Volt<\/span><\/p>\n

I = 20 Ampere<\/span><\/p>\n

4. Calculate the power for a \\(10\\,\\Omega\\) resistor with a \\(5\\,A\\) current.<\/span>
\nSolution: \\( P = I^2 R = 5^2 \\times 10 = 250\\,W \\)<\/span><\/p>\n

5. Determine the power for a circuit with \\(12\\,V\\) and \\(4\\,\\Omega\\) resistance.<\/span>
\nSolution: \\( P = \\frac{V^2}{R} = \\frac{12^2}{4} = 36\\,W \\)<\/span><\/p>\n

6. A \\(60\\,W\\) bulb operates for \\(3\\,h\\). Calculate the energy consumed.<\/span>
\nSolution: \\( E = Pt = 60 \\times 3 = 180\\,Wh = 0.18\\,kWh \\)<\/span><\/p>\n

7. Calculate the current for a \\(200\\,W\\) appliance on a \\(100\\,V\\) supply.<\/span>
\nSolution: \\( I = \\frac{P}{V} = \\frac{200}{100} = 2\\,A \\)<\/span><\/p>\n

8. Find the resistance of a heater that uses \\(1200\\,W\\) at \\(240\\,V\\).<\/span>
\nSolution: \\( R = \\frac{V^2}{P} = \\frac{240^2}{1200} = 48\\,\\Omega \\)<\/span><\/p>\n

9. A \\(9\\,V\\) battery is connected to a \\(3\\,\\Omega\\) resistor. Find the power dissipated.<\/span>
\nSolution: \\( P = \\frac{V^2}{R} = \\frac{9^2}{3} = 27\\,W \\)<\/span><\/p>\n

10. Calculate the power in a \\(10\\,\\Omega\\) resistor with a \\(100\\,V\\) supply.<\/span>
\nSolution: \\( P = \\frac{V^2}{R} = \\frac{100^2}{10} = 1000\\,W \\)<\/span><\/p>\n

11. Determine the voltage required to dissipate \\(50\\,W\\) in a \\(10\\,\\Omega\\) resistor.<\/span>
\nSolution: \\( V = \\sqrt{PR} = \\sqrt{50 \\times 10} = 22.36\\,V \\)<\/span><\/p>\n

12. A motor uses \\(1500\\,W\\) and runs for \\(4\\,h\\). Find the energy consumed.<\/span>
\nSolution: \\( E = Pt = 1500 \\times 4 = 6000\\,Wh = 6\\,kWh \\)<\/span><\/p>\n

13. Find the current of a \\(240\\,V\\) supply if the power is \\(1200\\,W\\).<\/span>
\nSolution: \\( I = \\frac{P}{V} = \\frac{1200}{240} = 5\\,A \\)<\/span><\/p>\n

14. Calculate the resistance for a \\(100\\,W\\) bulb operating at \\(200\\,V\\).<\/span>
\nSolution: \\( R = \\frac{V^2}{P} = \\frac{200^2}{100} = 400\\,\\Omega \\)<\/span><\/p>\n

15. Determine the power in a \\(3\\,A\\) circuit with a \\(15\\,\\Omega\\) resistor.<\/span>
\nSolution: \\( P = I^2 R = 3^2 \\times 15 = 135\\,W \\)<\/span><\/p>\n

16. Calculate the voltage needed to dissipate \\(100\\,W\\) in a \\(20\\,\\Omega\\) resistor.<\/span>
\nSolution: \\( V = \\sqrt{PR} = \\sqrt{100 \\times 20} = 44.72\\,V \\)<\/span><\/p>\n

17. Find the energy consumed by a \\(300\\,W\\) fan running for \\(2\\,h\\).<\/span>
\nSolution: \\( E = Pt = 300 \\times 2 = 600\\,Wh = 0.6\\,kWh \\)<\/span><\/p>\n

18. Calculate the current in a circuit with \\(1000\\,W\\) power and \\(250\\,V\\) voltage.<\/span>
\nSolution: \\( I = \\frac{P}{V} = \\frac{1000}{250} = 4\\,A \\)<\/span><\/p>\n

19. Determine the resistance of a \\(400\\,W\\) toaster operating at \\(100\\,V\\).<\/span>
\nSolution: \\( R = \\frac{V^2}{P} = \\frac{100^2}{400} = 25\\,\\Omega \\)<\/span><\/p>\n

20. Find the power for a \\(20\\,V\\) voltage across a \\(5\\,\\Omega\\) resistor.<\/span>
\nSolution: \\( P = \\frac{V^2}{R} = \\frac{20^2}{5} = 80\\,W \\)<\/span><\/p>\n

21. Calculate the voltage needed to provide \\(200\\,W\\) to a \\(25\\,\\Omega\\) resistor.<\/span>
\nSolution: \\( V = \\sqrt{PR} = \\sqrt{200 \\times 25} = 50\\,V \\)<\/span><\/p>\n

22. Determine the energy consumed by a \\(500\\,W\\) machine running for \\(5\\,h\\).<\/span>
\nSolution: \\( E = Pt = 500 \\times 5 = 2500\\,Wh = 2.5\\,kWh \\)<\/span><\/p>\n

23. Find the current for a \\(60\\,W\\) bulb operating at \\(120\\,V\\).<\/span>
\nSolution: \\( I = \\frac{P}{V} = \\frac{60}{120} = 0.5\\,A \\)<\/span><\/p>\n

24. Calculate the resistance for a \\(40\\,W\\) lamp running at \\(20\\,V\\).<\/span>
\nSolution: \\( R = \\frac{V^2}{P} = \\frac{20^2}{40} = 10\\,\\Omega \\)<\/span><\/p>\n

25. Determine the power in a \\(5\\,A\\) circuit with a \\(12\\,\\Omega\\) resistor.<\/span>
\nSolution: \\( P = I^2 R = 5^2 \\times 12 = 300\\,W \\)<\/span><\/p>\n

26. Calculate the voltage required to deliver \\(150\\,W\\) to a \\(15\\,\\Omega\\) resistor.<\/span>
\nSolution: \\( V = \\sqrt{PR} = \\sqrt{150 \\times 15} = 61.24\\,V \\)<\/span><\/p>\n

27. Find the energy consumed by a \\(800\\,W\\) air conditioner running for \\(3\\,h\\).<\/span>
\nSolution: \\( E = Pt = 800 \\times 3 = 2400\\,Wh = 2.4\\,kWh \\)<\/span><\/p>\n

28. Determine the current in a circuit with \\(500\\,W\\) power and \\(125\\,V\\) voltage.<\/span>
\nSolution: \\( I = \\frac{P}{V} = \\frac{500}{125} = 4\\,A \\)<\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

28 Electric power – problems and solutions 1. Calculate the electric current of a 10 Watt lamp designed for 12 Volt. Known : Power (P) = 10 Watt Voltage (V) = 220 Volt Wanted : Electric current Solution : Lamp power = 10 Watt = 10 Joule\/second The equation below shows the relation between the … 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 power - 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-1605","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\/1605","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=1605"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1605\/revisions"}],"predecessor-version":[{"id":9017,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1605\/revisions\/9017"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1605"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1605"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1605"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}