{"id":2426,"date":"2018-05-11T12:57:47","date_gmt":"2018-05-11T04:57:47","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2426"},"modified":"2023-08-06T14:55:22","modified_gmt":"2023-08-06T14:55:22","slug":"kirchhoffs-rules-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/kirchhoffs-rules-problems-and-solutions.htm","title":{"rendered":"Kirchhoff’s rules \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Kirchhoff’s rules \u2013 problems and solutions<\/p>\n

1. What is the terminal voltage<\/a> of the battery in the circuit<\/a> below?<\/span><\/span><\/p>\n

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

emf = electromotive force = the potential difference between the terminals when no current<\/a> flow to an external circuit.\"Kirchhoff's<\/span><\/span><\/p>\n

The terminal voltage (V) = the potential difference between the terminals when a current flows from the battery.<\/span><\/span><\/p>\n

If no current is drawn from the battery, the terminal voltage equals the emf.<\/span><\/span><\/p>\n

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

Resistor<\/a> 1 (R<\/span><\/span>1<\/span><\/sub>) <\/span><\/span>= 2 \u03a9 <\/span><\/span><\/p>\n

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

Resistor 3 (R<\/span><\/span>3<\/span><\/sub>) = 4 <\/span><\/span>\u03a9 <\/span><\/span><\/p>\n

emf <\/span><\/span>1 (E<\/span><\/span>1<\/span><\/sub>) = 20 Volt<\/span><\/span><\/p>\n

emf 2 (E<\/span><\/span>2<\/span><\/sub>) = 15 Volt<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> The terminal voltage (V)<\/span><\/span><\/p>\n

Solution 1 :<\/u><\/span><\/span><\/p>\n

The terminal voltage :<\/span><\/span><\/p>\n

V = E<\/span><\/span>1<\/span><\/sub> \u2013 E<\/span><\/span>2 <\/span><\/sub>= 20 \u2013 15 = 5 Volt <\/span><\/span><\/p>\n

Solution 2 :<\/u><\/span><\/span><\/p>\n

Calculate the current flows in the circuit (I) <\/b><\/span><\/span><\/p>\n

First<\/b><\/i><\/span><\/span>, Choose the direction of each current. The direction can be chosen arbitrarily: if the current is actually in the opposite direction, it will come out with a minus sign in the solution.<\/span><\/span><\/p>\n

Then, starting at any point in the circuit, we imagine traveling around a loop, adding emfs and IR terms as we come to them.<\/span><\/span><\/p>\n

Second<\/b><\/i><\/span><\/span>, When we travel through a source in the direction from – to +, the emf is considered to be positive; when we travel from + to -, the emf is considered to be negative.<\/span><\/span><\/p>\n

Third<\/b><\/i><\/span><\/span>, When we travel through a resistor in the same direction as the assumed current, the IR term is negative because the current goes in the direction of decreasing potential. When we travel through a resistor in the direction opposite to the assumed current, the IR term is positive because this represents a rise of potential.<\/span><\/span><\/p>\n

The direction of current is chosen same as the clockwise direction :<\/span><\/span><\/p>\n

E<\/span><\/span>1<\/span><\/sub> \u2013 I R<\/span><\/span>1<\/span><\/sub> \u2013 I R<\/span><\/span>2<\/span><\/sub> \u2013 I R<\/span><\/span>3<\/span><\/sub> \u2013 E<\/span><\/span>2<\/span><\/sub> = 0<\/span><\/span><\/p>\n

20 \u2013 I (2) \u2013 I (4) \u2013 I (4) \u2013 15 = 0<\/span><\/span><\/p>\n

20 \u2013 15 \u2013 I (2) \u2013 I (4) \u2013 I (4) = 0<\/span><\/span><\/p>\n

5 \u2013 10 I = 0<\/span><\/span><\/p>\n

5 = 10 I <\/span><\/span><\/p>\n

I = 5\/10<\/span><\/span><\/p>\n

I = 0.5 Ampere<\/span><\/span><\/p>\n

The electric current flowing in the circuit is 0.5 Ampere. Electric current signed positive means the direction of electric current is the <\/span><\/span><\/span>same as the clockwise direction.<\/span><\/span><\/span><\/p>\n

Calculate the equivalent resistor (R) :<\/b><\/span><\/span> <\/b><\/p>\n

Resistor 1 (<\/span><\/span>R<\/span><\/span>1<\/span><\/sub>), resistor 2 (<\/span><\/span>R<\/span><\/span>2<\/span><\/sub>) <\/span><\/span>and <\/span><\/span>resistor 3 (R<\/span><\/span>3<\/span><\/sub>) <\/span><\/span>are connected in series<\/span><\/span>. <\/span><\/span>The equivalent resistor :<\/span><\/span><\/p>\n

R = R<\/span><\/span>1<\/span><\/sub> + R<\/span><\/span>2<\/sub> <\/span><\/span>+ R<\/span><\/span>3<\/span><\/sub> = <\/span><\/span>2 \u03a9 + 4 \u03a9 + 4 \u03a9 = 10 \u03a9 <\/span><\/span><\/p>\n

The terminal voltage in resistor <\/b><\/span><\/span>R (V)<\/b><\/span><\/span><\/p>\n

V = I R = (0.5)(10) = 5 Volt<\/span><\/span><\/p>\n

<\/p>\n

2. <\/span><\/span>In the circuit as shown in figure below, find the power dissipated in <\/span><\/span>the <\/span><\/span>3-\u03a9 resistor.<\/span><\/span><\/p>\n

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

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

Resistor 1 <\/span><\/span>(R<\/span><\/span>1<\/span><\/sub>) = 2 <\/span><\/span>\u03a9 \"Kirchhoff's<\/span><\/span><\/p>\n

Resistor 2 <\/span><\/span>(R<\/span><\/span>2<\/span><\/sub>) = 3 <\/span><\/span>\u03a9 <\/span><\/span><\/p>\n

Resistor <\/span><\/span>3 (R<\/span><\/span>3<\/span><\/sub>) = 4 <\/span><\/span>\u03a9 <\/span><\/span><\/p>\n

emf <\/span><\/span>1 (E<\/span><\/span>1<\/span><\/sub>) = 8 Volt<\/span><\/span><\/p>\n

emf <\/span><\/span>2<\/span><\/span> (<\/span><\/span>E<\/span><\/span>2<\/span><\/sub>) = 10 Volt<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> The power dissipated in <\/span><\/span>the <\/span><\/span>3-\u03a9 resistor<\/span><\/span><\/p>\n

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

The power<\/a> dissipated in <\/span><\/span>the <\/span><\/span>3-\u03a9 resisto<\/span><\/span>r :<\/span><\/span><\/p>\n

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

P = <\/i><\/span><\/span>power<\/i><\/span><\/span>, V = <\/i><\/span><\/span>the <\/i><\/span><\/span>voltag<\/i><\/span><\/span>e across <\/i><\/span><\/span>the <\/i><\/span><\/span>3-\u03a9 resistor<\/i><\/span><\/span>, I <\/i><\/span><\/span>= the current passes through the <\/i><\/span><\/span>3-\u03a9 resistor<\/i><\/span><\/span><\/p>\n

Calculate the electric current <\/b><\/span><\/span>(I) <\/b><\/span><\/span>passes through <\/b><\/span><\/span>the <\/b><\/span><\/span>3-\u03a9 resistor<\/b><\/span><\/span><\/p>\n

The direction of current is chosen same as the clockwise direction :<\/span><\/span><\/p>\n

E<\/span><\/span>1<\/span><\/sub> \u2013 I R<\/span><\/span>1<\/span><\/sub> \u2013 I R<\/span><\/span>2<\/span><\/sub> \u2013 I R<\/span><\/span>3<\/span><\/sub> + E<\/span><\/span>2<\/span><\/sub> = 0<\/span><\/span><\/p>\n

8 \u2013 I (2) \u2013 I (3) \u2013 I (4) + 10 = 0<\/span><\/span><\/p>\n

18 \u2013 9 I = 0<\/span><\/span><\/p>\n

18 = 9 I<\/span><\/span><\/p>\n

I = 18 \/ 9<\/span><\/span><\/p>\n

I = 2 Ampere<\/span><\/span><\/p>\n

The electric current flowing in the circuit is <\/span><\/span><\/span>2 <\/span><\/span><\/span>Ampere. Electric current signed positive means the direction of electric current is the <\/span><\/span><\/span>same as the clockwise direction.<\/span><\/span><\/span><\/p>\n

Circuits are connected in series so that the electric current flowing in the circuit = the electric current passes through the <\/span><\/span>3-<\/span><\/span>\u03a9 <\/span><\/span>resistor <\/span><\/span>= 2 Ampere.<\/span><\/span><\/p>\n

Calculate voltage <\/b><\/span><\/span>(V) <\/b><\/span><\/span>across the <\/b><\/span><\/span>3-\u03a9 <\/b><\/span><\/span>resistor<\/b><\/span><\/span><\/p>\n

V = I R<\/span><\/span>2<\/span><\/sub> = (2 A)(3 \u03a9) = 6 Volt<\/span><\/span><\/p>\n

The power dissipated in <\/span><\/span>the <\/span><\/span>3-\u03a9 resistor :<\/span><\/span><\/b><\/p>\n

P = V I = (6 Volt)(2 Ampere) = 12 Volt Ampere = 12 Watt<\/span><\/span><\/p>\n

<\/p>\n

3.<\/span><\/span> Based on the circuit as shown in the figure below, what is the potential difference between point A and B.<\/span><\/span><\/p>\n

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

Resistor <\/span><\/span>1 (R<\/span><\/span>1<\/span><\/sub>) = 2 <\/span><\/span>\u03a9 \"Kirchhoff's<\/span><\/span><\/p>\n

Resistor <\/span><\/span>2 (R<\/span><\/span>2<\/span><\/sub>) = 3 <\/span><\/span>\u03a9 <\/span><\/span><\/p>\n

Resistor <\/span><\/span>3 (R<\/span><\/span>3<\/span><\/sub>) = 4 <\/span><\/span>\u03a9 <\/span><\/span><\/p>\n

emf <\/span><\/span>1 (E<\/span><\/span>1<\/span><\/sub>) = 8 Volt<\/span><\/span><\/p>\n

emf <\/span><\/span>2<\/span><\/span> (E<\/span><\/span>2<\/span><\/sub>) = 10 Volt<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> The potential difference <\/span><\/span>(V) <\/span><\/span>between point A and B<\/span><\/span><\/p>\n

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

Calculate the electric current <\/b><\/span><\/span>(I) <\/b><\/span><\/span>flowing in the <\/b><\/span><\/span>3-<\/b><\/span><\/span>\u03a9 <\/b><\/span><\/span>resistor<\/b><\/span><\/span> <\/b><\/p>\n

The direction of current is chosen same as the clockwise direction :<\/span><\/span><\/p>\n

– E<\/span><\/span>1<\/span><\/sub> \u2013 I R<\/span><\/span>1<\/span><\/sub> \u2013 I R<\/span><\/span>2<\/span><\/sub> – E<\/span><\/span>2<\/span><\/sub> \u2013 I R<\/span><\/span>3 <\/span><\/sub>= 0<\/span><\/span><\/p>\n

– 8 \u2013 I (2) \u2013 I (3) – 10 \u2013 I (4) = 0<\/span><\/span><\/p>\n

– 18 \u2013 9 I = 0<\/span><\/span><\/p>\n

– 18 = 9 I<\/span><\/span><\/p>\n

I = -18 \/ 9<\/span><\/span><\/p>\n

I = – 2 Ampere<\/span><\/span><\/p>\n

The electric current flowing in the circuit is <\/span><\/span><\/span>2 <\/span><\/span><\/span>Ampere. Electric current signed <\/span><\/span><\/span>negative <\/span><\/span><\/span>means the direction of electric current <\/span><\/span><\/span>counter<\/span><\/span><\/span>clockwise direction.<\/span><\/span><\/span><\/p>\n

Circuits are connected in series so that the electric current flowing in the circuit = the electric current passes through the <\/span><\/span>3-<\/span><\/span>\u03a9 <\/span><\/span>resistor <\/span><\/span>= 2 Ampere.<\/span><\/span><\/p>\n

Calculate the potential difference <\/b><\/span><\/span>(V) <\/b><\/span><\/span>across the 3-<\/b><\/span><\/span>\u03a9 <\/b><\/span><\/span>resistor :<\/b><\/span><\/span><\/p>\n

V = I R<\/span><\/span>2<\/span><\/sub> = (2 A)(3 \u03a9) = 6 Volt<\/span><\/span><\/p>\n

<\/p>\n

4. <\/span><\/span>Based on the circuit as shown in the figure below, what is the potential difference across the<\/span><\/span> R<\/span><\/span>3<\/span><\/sub> resistor.<\/span><\/span><\/p>\n

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

Resistor <\/span><\/span>1 (R<\/span><\/span>1<\/span><\/sub>) = 2 <\/span><\/span>\u03a9 \"Kirchhoff's<\/span><\/span><\/p>\n

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

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

emf <\/span><\/span>1 (E<\/span><\/span>1<\/span><\/sub>) = 6 Volt<\/span><\/span><\/p>\n

emf 2 (E<\/span><\/span>2<\/span><\/sub>) = 9 Volt<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> The potential difference across the <\/span><\/span>R<\/span><\/span>3<\/span><\/span><\/sub> resistor<\/span><\/span><\/p>\n

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

Calculate the electric current <\/b><\/span><\/span>(I) <\/b><\/span><\/span>passes through the <\/b><\/span><\/span>R<\/b><\/span><\/span>3<\/b><\/span><\/sub> <\/b>resistor<\/b><\/span><\/span><\/p>\n

The direction of current is chosen same as the clockwise direction :<\/span><\/span><\/p>\n

E<\/span><\/span>1<\/span><\/sub> \u2013 I R<\/span><\/span>1<\/span><\/sub> – E<\/span><\/span>2<\/span><\/sub> \u2013 I R<\/span><\/span>2<\/span><\/sub> \u2013 I R<\/span><\/span>3 <\/span><\/sub>= 0<\/span><\/span><\/p>\n

6 \u2013 2I \u2013 9 \u2013 4I \u2013 3I = 0<\/span><\/span><\/p>\n

6 \u2013 9 \u2013 2I \u2013 4I \u2013 3I = 0<\/span><\/span><\/p>\n

-3 \u2013 9I = 0<\/span><\/span><\/p>\n

-3 = 9I<\/span><\/span><\/p>\n

I = -3\/9<\/span><\/span><\/p>\n

I = -1\/3<\/span><\/span><\/p>\n

The electric current flowing in the circuit is <\/span><\/span><\/span>1\/3 <\/span><\/span><\/span>Ampere. Electric current signed <\/span><\/span><\/span>negative <\/span><\/span><\/span>means the direction of electric current <\/span><\/span><\/span>counter<\/span><\/span><\/span>clockwise direction.<\/span><\/span><\/span><\/p>\n

Circuits are connected in series so that the electric current flowing in the circuit = the electric current passes through the R<\/span><\/span>3 <\/span><\/sub>resistor <\/span><\/span>= <\/span><\/span>1\/3 <\/span><\/span>Ampere.<\/span><\/span><\/p>\n

Calculate the potential difference <\/b><\/span><\/span>(V) <\/b><\/span><\/span>across the <\/b><\/span><\/span>R<\/b><\/span><\/span>3<\/b><\/span> <\/b><\/sub>resistor<\/b><\/span><\/span> <\/b><\/p>\n

V = I R<\/span><\/span>3<\/span><\/sub> = (1\/3)(3) = 1 Volt<\/span><\/span><\/p>\n

<\/p>\n

5. <\/span><\/span>The potential electric between point C and D = 4 Volt, find R!<\/span><\/span><\/p>\n

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

Resistor <\/span><\/span>1 (R<\/span><\/span>1<\/span><\/sub>) = 2 <\/span><\/span>\u03a9 \"Kirchhoff's<\/span><\/span><\/p>\n

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

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

emf <\/span><\/span>1 <\/span><\/span>(E<\/span><\/span>1<\/span><\/sub>) = 8 Volt<\/span><\/span><\/p>\n

emf <\/span><\/span>2 <\/span><\/span>(E<\/span><\/span>2<\/span><\/sub>) = 4 Volt<\/span><\/span><\/p>\n

The potential difference between <\/span><\/span>C <\/span><\/span>and <\/span><\/span>D (V<\/span><\/span>CD<\/span><\/sub>) = 4 Volt<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> R<\/span><\/span><\/p>\n

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

V<\/span><\/span>CD<\/span><\/sub> = I R <\/span><\/span><\/p>\n

4 = I R<\/span><\/span><\/p>\n

R = 4 \/ I<\/span><\/span><\/p>\n

\u2026…………<\/i><\/span><\/span><\/p>\n

Calculate the electric current <\/b><\/span><\/span>(I) <\/b><\/span><\/span>passes through the R resistor <\/b><\/span><\/span><\/p>\n

The direction of current is chosen same as the clockwise direction :<\/span><\/span><\/p>\n

– E<\/span><\/span>1<\/span><\/sub> \u2013 2I + E<\/span><\/span>2 <\/span><\/sub>\u2013 2I \u2013 IR = 0<\/span><\/span><\/p>\n

– 8 \u2013 2I + 4 \u2013 2I \u2013 I (4\/I) = 0<\/span><\/span><\/p>\n

– 8 \u2013 2I + 4 \u2013 2I \u2013 4 = 0<\/span><\/span><\/p>\n

– 8 + 4 \u2013 4 \u2013 2I \u2013 2I = 0<\/span><\/span><\/p>\n

– 8 \u2013 4I = 0<\/span><\/span><\/p>\n

– 8 = 4I<\/span><\/span><\/p>\n

I = -8 \/ 4<\/span><\/span><\/p>\n

I = -2 Ampere<\/span><\/span><\/p>\n

The electric current flowing in the circuit is <\/span><\/span><\/span>2 <\/span><\/span><\/span>Ampere. Electric current signed <\/span><\/span><\/span>negative <\/span><\/span><\/span>means the direction of electric current <\/span><\/span><\/span>counter<\/span><\/span><\/span>clockwise direction.<\/span><\/span><\/span><\/p>\n

Circuits are connected in series so that the electric current flowing in the circuit = the electric current passes through the R<\/span><\/span> resistor <\/span><\/span>= <\/span><\/span>2 <\/span><\/span>Ampere.<\/span><\/span><\/p>\n

R :<\/b><\/span><\/span><\/p>\n

R = 4 \/ I = 4 \/ 2 = 2 Ohm<\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

Kirchhoff’s rules \u2013 problems and solutions 1. What is the terminal voltage of the battery in the circuit below? Solution emf = electromotive force = the potential difference between the terminals when no current flow to an external circuit. The terminal voltage (V) = the potential difference between the terminals when a current flows from … 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":"Kirchhoff's rules \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-2426","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\/2426","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=2426"}],"version-history":[{"count":1,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2426\/revisions"}],"predecessor-version":[{"id":8552,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2426\/revisions\/8552"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=2426"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=2426"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=2426"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}