{"id":3075,"date":"2018-06-08T05:20:14","date_gmt":"2018-06-08T12:20:14","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=3075"},"modified":"2018-06-08T05:20:14","modified_gmt":"2018-06-08T12:20:14","slug":"simple-dc-circuits-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/simple-dc-circuits-problems-and-solutions.htm","title":{"rendered":"Simple DC circuits – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. <\/span><\/span>Based on the figure below, determine the electric current<\/a> through <\/span><\/span>R<\/span><\/span>1<\/span><\/sub>.<\/span><\/span><\/p>\n

\"SimpleKnown :<\/span><\/span><\/u><\/p>\n

Resistor 1 (R<\/span><\/span>1<\/span><\/sub>) = 4 \u03a9<\/span><\/span><\/p>\n

Resistor 2 (R<\/span><\/span>2<\/span><\/sub>) = 4 \u03a9<\/span><\/span><\/p>\n

Resistor<\/a> 3 (R<\/span><\/span>3<\/span><\/sub>) = 8 \u03a9<\/span><\/span><\/p>\n

Electric voltage <\/span><\/span><\/a>(V) = 40 Volt<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> Electric current through <\/span><\/span>R<\/span><\/span>1<\/span><\/sub><\/p>\n

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

The electric current flows from high potential to low potential. The direction of electric current in the circuit above is the same as a clockwise direction.<\/span><\/span><\/p>\n

The electric current that flows out of the battery<\/b><\/u><\/span><\/span><\/p>\n

First, calculate the equivalent resistance (<\/span><\/span>R). <\/span><\/span>Thereafter, calculate the electric current using the equation of Ohm’s law<\/a> :<\/span><\/span><\/p>\n

V = I R <\/span><\/span>or<\/span><\/span> I = V \/ R<\/span><\/span><\/p>\n

V = <\/span><\/span>voltage<\/span><\/span>, I = <\/span><\/span>electric current<\/span><\/span>, R = <\/span><\/span>the equivalent resistance<\/span><\/span><\/p>\n

The equivalent resistance :<\/u><\/span><\/span><\/p>\n

Resistor R<\/span><\/span><\/span>1<\/span><\/span><\/sub> and <\/span><\/span><\/span>resistor R<\/span><\/span><\/span>2<\/span><\/span><\/sub> are connected in parallel<\/span><\/span><\/span>. <\/span><\/span><\/span>The equivalent resistance :<\/span><\/span><\/span><\/p>\n

1\/R<\/span><\/span><\/span>12<\/span><\/span><\/sub> = 1\/R<\/span><\/span><\/span>1<\/span><\/span><\/sub> + 1\/R<\/span><\/span><\/span>2 <\/span><\/span><\/sub>= 1\/4 + 1\/4 = 2\/4<\/span><\/span><\/span><\/p>\n

R<\/span><\/span><\/span>12 <\/span><\/span><\/sub>= 4\/2 = 2 \u03a9<\/span><\/span><\/span><\/p>\n

Resistor R<\/span><\/span><\/span>12<\/span><\/span><\/sub> and <\/span><\/span><\/span>resistor R<\/span><\/span><\/span>3<\/span><\/span><\/sub> are connected in series<\/span><\/span><\/span>. <\/span><\/span><\/span>The equivalent resistance :<\/span><\/span><\/span><\/p>\n

R<\/span><\/span><\/span> = R<\/span><\/span><\/span>12<\/span><\/span><\/sub> + R<\/span><\/span><\/span>3<\/span><\/span><\/sub> = 2 + 8 = 10 \u03a9 <\/span><\/span><\/span><\/p>\n

The electric current that flows out of the battery :<\/u><\/span><\/span><\/span><\/p>\n

I = V \/ R = 40 \/ 10 = 4 A<\/span><\/span><\/span><\/p>\n

The electric current that flows out of the battery is <\/span><\/span><\/span>4 Ampere.<\/span><\/span><\/span><\/p>\n

Electric voltage <\/b><\/u><\/span><\/span><\/span> V<\/b><\/u><\/span><\/span><\/span>ab<\/b><\/u><\/span><\/span><\/sub> <\/b><\/u>and <\/b><\/u><\/span><\/span><\/span>V<\/b><\/u><\/span><\/span><\/span>bc<\/b><\/u><\/span><\/span><\/sub><\/p>\n

\"SimpleKirchhoff<\/b><\/i><\/span><\/span><\/span>‘s first rule<\/b><\/i><\/span><\/span><\/span> states that at<\/i> any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction.<\/i><\/span><\/span><\/span><\/p>\n

Based on Kirchhoff’s first rule<\/a>, <\/span><\/span><\/span>concluded that if the electric current flows out of the battery is 4 A then the electric current through a-b is equal to 4 Ampere, so also the electric current through b-c is 4 Ampere.<\/span><\/span><\/span><\/p>\n

Electric voltage <\/u><\/span><\/span><\/span>V<\/u><\/span><\/span><\/span>ab <\/u><\/span><\/span><\/sub>:<\/u><\/span><\/span><\/span> <\/u><\/p>\n

V<\/span><\/span><\/span>ab<\/span><\/span><\/sub> = I<\/span><\/span><\/span>ab<\/span><\/span><\/sub> R<\/span><\/span><\/span>ab<\/span><\/span><\/sub> = (4)(2) = 8 Volt <\/span><\/span><\/span><\/p>\n

Electric voltage <\/u><\/span><\/span><\/span>V<\/u><\/span><\/span><\/span>bc <\/u><\/span><\/span><\/sub>:<\/u><\/span><\/span><\/span> <\/u><\/p>\n

V<\/span><\/span><\/span>bc<\/span><\/span><\/sub> = I<\/span><\/span><\/span>bc<\/span><\/span><\/sub> R<\/span><\/span><\/span>bc<\/span><\/span><\/sub> = (4)(8) = 32 Volt<\/span><\/span><\/span><\/p>\n

The above circuit is connected in series so that the total electrical voltage is <\/span><\/span><\/span>V = V<\/span><\/span><\/span>ab<\/span><\/span><\/sub> + V<\/span><\/span><\/span>bc<\/span><\/span><\/sub> = 8 Volt + 32 Volt = 40 Volt.<\/span><\/span><\/span><\/p>\n

The electric current flows through <\/b><\/u><\/span><\/span><\/span>R<\/b><\/u><\/span><\/span><\/span>1<\/b><\/u><\/span><\/span><\/sub> = 4 \u03a9<\/b><\/u><\/span><\/span><\/span><\/p>\n

I<\/span><\/span><\/span>1<\/span><\/span><\/sub> = V<\/span><\/span><\/span>ab<\/span><\/span><\/sub> \/ R<\/span><\/span><\/span>1<\/span><\/span><\/sub> = 8 Volt \/ 4 Ohm = 2 A<\/span><\/span><\/span><\/p>\n

I<\/span><\/span><\/span>2<\/span><\/span><\/sub> = V<\/span><\/span><\/span>ab<\/span><\/span><\/sub> \/ R<\/span><\/span><\/span>2<\/span><\/span><\/sub> = 8 Volt \/ 4 Ohm = 2 A<\/span><\/span><\/span><\/p>\n

The electric current that flows out of the battery is 4 A. When it arrives at point a the electric current is divided into two, the electric current of 2 Ampere flows through the resistor R<\/span><\/span><\/span>1<\/span><\/span><\/sub> and <\/span><\/span><\/span><\/p>\n

the electric current of 2 A flows through the resistor R<\/span><\/span><\/span>2<\/span><\/span><\/sub>. 2 A + 2 A = 4 A.<\/span><\/span><\/span><\/p>\n

2. <\/span><\/span><\/span>Based on the figure below, determine the electric current that flows through resistor <\/span><\/span><\/span>4-\u03a9.<\/span><\/span><\/p>\n

\"SimpleKnown :<\/span><\/span><\/u><\/p>\n

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

Resistor 2 (R<\/span><\/span>2<\/span><\/sub>) = 4 \u03a9<\/span><\/span><\/p>\n

Resistor 3 (R<\/span><\/span>3<\/span><\/sub>) = 1.6 \u03a9<\/span><\/span><\/p>\n

The electric voltage <\/span><\/span>(V) = 16 Volt<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> The electric current flows through <\/span><\/span>4 \u03a9<\/span><\/span><\/p>\n

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

The electric current flows from high potential<\/a> to low potential. The direction of electric current in the circuit above is the same as a clockwise direction.<\/span><\/span><\/p>\n

The electric current that flows out of the battery<\/b><\/u><\/span><\/span><\/p>\n

The equivalent resistance :<\/u><\/span><\/span><\/p>\n

Resistor R<\/span><\/span><\/span>1<\/span><\/span> <\/sub>and <\/span><\/span><\/span>resistor R<\/span><\/span><\/span>2<\/span><\/span> <\/sub>are connected in parallel. The equivalent resistor :<\/span><\/span><\/span><\/p>\n

1\/R<\/span><\/span><\/span>12<\/span><\/span><\/sub> = 1\/R<\/span><\/span><\/span>1<\/span><\/span><\/sub> + 1\/R<\/span><\/span><\/span>2 <\/span><\/span><\/sub>= 1\/6 + 1\/4 = 2\/12 + 3\/12 = 5\/12 <\/span><\/span><\/span><\/p>\n

R<\/span><\/span><\/span>12 <\/span><\/span><\/sub>= 12\/5 = 2.4 \u03a9<\/span><\/span><\/span><\/p>\n

Resistor R<\/span><\/span><\/span>12<\/span><\/span> <\/sub>and <\/span><\/span><\/span>resistor R<\/span><\/span><\/span>3<\/span><\/span> <\/sub>are connected in series. The equivalent resistor :<\/span><\/span><\/span><\/p>\n

R<\/span><\/span><\/span> = R<\/span><\/span><\/span>12<\/span><\/span><\/sub> + R<\/span><\/span><\/span>3<\/span><\/span><\/sub> = 2.4 + 1.6 = 4 \u03a9 <\/span><\/span><\/span><\/p>\n

The electric current that flows out of the battery :<\/u><\/span><\/span><\/span><\/p>\n

I = V \/ R = 16 \/ 4 = 4 A<\/span><\/span><\/span><\/p>\n

The electric voltage <\/b><\/u><\/span><\/span><\/span>V<\/b><\/u><\/span><\/span><\/span>ab<\/b><\/u><\/span><\/span> <\/b><\/u><\/sub>and <\/b><\/u><\/span><\/span><\/span>V<\/b><\/u><\/span><\/span><\/span>bc<\/b><\/u><\/span><\/span><\/sub><\/p>\n

\"SimpleBased on Kirchhoff’s first rule, <\/span><\/span><\/span>concluded that if the electric current flows out of the battery is 4 A then the electric current through a-b is equal to 4 Ampere, so also the electric current through b-c is 4 Ampere.<\/span><\/span><\/span><\/p>\n

The electric voltage <\/u><\/span><\/span><\/span>V<\/u><\/span><\/span><\/span>ab <\/u><\/span><\/span><\/sub>:<\/u><\/span><\/span><\/span> <\/u><\/p>\n

V<\/span><\/span><\/span>ab<\/span><\/span><\/sub> = I<\/span><\/span><\/span>ab<\/span><\/span><\/sub> R<\/span><\/span><\/span>ab<\/span><\/span><\/sub> = (4)(2.4) = 9.6 Volt <\/span><\/span><\/span><\/p>\n

The electric voltage <\/u><\/span><\/span><\/span>V<\/u><\/span><\/span><\/span>bc <\/u><\/span><\/span><\/sub>:<\/u><\/span><\/span><\/span> <\/u><\/p>\n

V<\/span><\/span><\/span>bc<\/span><\/span><\/sub> = I<\/span><\/span><\/span>bc<\/span><\/span><\/sub> R<\/span><\/span><\/span>bc<\/span><\/span><\/sub> = (4)(1.6) = 6.4 Volt <\/span><\/span><\/span><\/p>\n

The above circuit is connected in series so that the total electric voltage is <\/span><\/span><\/span>V = V<\/span><\/span><\/span>ab<\/span><\/span><\/sub> + V<\/span><\/span><\/span>bc<\/span><\/span><\/sub> = 9.6 Volt + 6.4 Volt = 16 Volt.<\/span><\/span><\/span><\/p>\n

The electric current that flows through <\/b><\/u><\/span><\/span><\/span>R<\/b><\/u><\/span><\/span><\/span>2<\/b><\/u><\/span><\/span><\/sub> = 4 \u03a9<\/b><\/u><\/span><\/span><\/span><\/p>\n

I<\/span><\/span><\/span>1<\/span><\/span><\/sub> = V<\/span><\/span><\/span>ab<\/span><\/span><\/sub> \/ R<\/span><\/span><\/span>1<\/span><\/span><\/sub> = 9.6 Volt \/ 6 Ohm = 1.6 A<\/span><\/span><\/span><\/p>\n

I<\/span><\/span><\/span>2<\/span><\/span><\/sub> = V<\/span><\/span><\/span>ab<\/span><\/span><\/sub> \/ R<\/span><\/span><\/span>2<\/span><\/span><\/sub> = 9.6 Volt \/ 4 Ohm = 2.4 A<\/span><\/span><\/span><\/p>\n

The electric current that flows out of the battery is 4 A. When it arrives at point a the electric current is divided into two, <\/span><\/span><\/span><\/p>\n

the electric current 1.6 A flows through the resistor R<\/span><\/span><\/span>1<\/span><\/span><\/sub><\/span> and the electric current 2.4 A flows through the resistor R<\/span><\/span><\/span>2<\/span><\/span><\/sub><\/span>. 1.6 A + 2.4 A = 4 A.<\/span><\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

1. Based on the figure below, determine the electric current through R1. Known : Resistor 1 (R1) = 4 \u03a9 Resistor 2 (R2) = 4 \u03a9 Resistor 3 (R3) = 8 \u03a9 Electric voltage (V) = 40 Volt<\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","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":"Simple DC circuits - 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-3075","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\/3075","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=3075"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/3075\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=3075"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=3075"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=3075"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}