{"id":88037,"date":"2025-06-01T07:00:11","date_gmt":"2025-06-01T07:00:11","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88037"},"modified":"2025-06-01T07:00:11","modified_gmt":"2025-06-01T07:00:11","slug":"an-electric-circuit-is-given-below-v%e2%82%81-1-v-and-resistance-r-1000-%cf%89","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/an-electric-circuit-is-given-below-v%e2%82%81-1-v-and-resistance-r-1000-%cf%89\/","title":{"rendered":"An electric circuit is given below. V\u2081 = 1 V and Resistance R = 1000 \u03a9"},"content":{"rendered":"

An electric circuit is given below. V\u2081 = 1 V and Resistance R = 1000 \u03a9.
\nThe current through the resistance R is very close to 1 mA and the voltage across point A and B, VAB = 1 V. Now the circuit is changed to :
\nwhere value of V\u2082 = 5 V. The internal resistances of both the batteries are 0.1 \u03a9. The current through the resistance R is about :<\/p>\n

[amp_mcq option1=”1.0 mA” option2=”1.2 mA” option3=”3.0 mA” option4=”5.0 mA” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe first part of the question describes a circuit with a battery V1=1V, resistor R=1000\u03a9, where current is ~1mA and VAB=1V. This indicates R is connected across points A and B, and V1 is the source across A and B. The current is V\/R = 1V\/1000\u03a9 = 1mA, consistent with the description.
\nThe second part describes a changed circuit with V2=5V and internal resistances of both batteries (V1 and V2) are 0.1\u03a9. Assuming A and B are still the points across which R is connected, and both batteries are connected between A and B, the most plausible configuration is that the batteries V1 (1V, 0.1\u03a9) and V2 (5V, 0.1\u03a9) are connected in parallel across A and B, and the resistor R (1000\u03a9) is also connected between A and B.
\nIn a parallel circuit with multiple voltage sources, we can use nodal analysis or source transformation. Using nodal analysis with B as the reference node (0V), the voltage at A is V_A.
\nCurrent through R: I_R = V_A \/ R
\nCurrent from V1 source: I_1 = (V1 – V_A) \/ r1 = (1 – V_A) \/ 0.1
\nCurrent from V2 source: I_2 = (V2 – V_A) \/ r2 = (5 – V_A) \/ 0.1
\nApplying KCL at node A (sum of currents leaving A is zero):
\nI_R + I_1 + I_2 = 0 is incorrect. The sources are supplying current towards A, so currents entering A should be summed to currents leaving A. Let’s assume currents leave A towards R, and into the positive terminals of V1 and V2. This means V_A is higher than the battery terminals connected to A. The positive terminals are connected to A, and negative terminals to B. So current flows from A through R to B, from A through V1 (r1) to B, and from A through V2 (r2) to B.
\nCurrents leaving A: I_R = V_A \/ R, I_1 = V_A – V1 \/ r1? No. This implies V_A is higher than the source terminal.
\nLet’s redraw with proper node labels. A and B are the terminals. R is between A and B. Battery 1 (V1, r1) is between A and B (positive to A, negative to B). Battery 2 (V2, r2) is between A and B (positive to A, negative to B).
\nUsing superposition or combining parallel sources:
\nEquivalent voltage source V_eq = (V1\/r1 + V2\/r2) \/ (1\/r1 + 1\/r2)
\nEquivalent internal resistance r_eq = 1 \/ (1\/r1 + 1\/r2)
\nV_eq = (1V\/0.1\u03a9 + 5V\/0.1\u03a9) \/ (1\/0.1\u03a9 + 1\/0.1\u03a9) = (10 A + 50 A) \/ (10 S + 10 S) = 60 A \/ 20 S = 3 V.
\nr_eq = 1 \/ (10 S + 10 S) = 1 \/ 20 S = 0.05 \u03a9.
\nNow the circuit is a single equivalent source V_eq=3V with internal resistance r_eq=0.05\u03a9 connected across R=1000\u03a9.
\nThe voltage across R is V_AB = V_eq * R \/ (R + r_eq) = 3V * 1000\u03a9 \/ (1000\u03a9 + 0.05\u03a9) = 3V * 1000 \/ 1000.05.
\nV_AB \u2248 3V.
\nThe current through R is I_R = V_AB \/ R = (3V * 1000 \/ 1000.05) \/ 1000\u03a9 = 3V \/ 1000.05\u03a9.
\nI_R \u2248 3 \/ 1000 = 0.003 A = 3 mA.
\nCalculating precisely: 3 \/ 1000.05 \u2248 0.00299985 A \u2248 2.99985 mA.
\nThis is very close to 3.0 mA.<\/p>\n<\/section>\n
\n– The first part of the question establishes the context: R=1000\u03a9, V1=1V providing ~1mA and VAB=1V, suggesting R is connected across the 1V source.
\n– The second part adds V2=5V and internal resistances 0.1\u03a9 for *both* batteries.
\n– Assuming a parallel connection of the two batteries and the resistor R between points A and B is the most likely configuration given the options and the structure implied by the first sentence.
\n– When voltage sources are in parallel, they can be combined into an equivalent source V_eq and r_eq.
\n– Formula for parallel sources: V_eq = (\u03a3 Vi\/ri) \/ (\u03a3 1\/ri) and r_eq = 1 \/ (\u03a3 1\/ri).
\n– Calculate V_eq and r_eq for V1 (1V, 0.1\u03a9) and V2 (5V, 0.1\u03a9).
\n– Calculate the current through R (1000\u03a9) when connected across the equivalent source.
\n– Calculation gives current \u2248 3 mA.
\n<\/section>\n
\nIf the batteries were in series, the total voltage would be 6V (aiding) or 4V (opposing). Total internal resistance would be 0.2\u03a9. Current through 1000\u03a9 would be approximately 6mA or 4mA. Neither of these is a very close match to the options (3.0 mA or 5.0 mA), whereas 3mA is a precise match for the parallel configuration calculation result of ~2.99985 mA. This strongly supports the parallel connection interpretation.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

An electric circuit is given below. V\u2081 = 1 V and Resistance R = 1000 \u03a9. The current through the resistance R is very close to 1 mA and the voltage across point A and B, VAB = 1 V. Now the circuit is changed to : where value of V\u2082 = 5 V. The … <\/p>\n

Detailed SolutionAn electric circuit is given below. V\u2081 = 1 V and Resistance R = 1000 \u03a9<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1093],"tags":[1103,1201,1128],"class_list":["post-88037","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1103","tag-electric-current","tag-physics","no-featured-image-padding"],"yoast_head":"\nAn electric circuit is given below. V\u2081 = 1 V and Resistance R = 1000 \u03a9<\/title>\n<meta name=\"description\" content=\"The first part of the question describes a circuit with a battery V1=1V, resistor R=1000\u03a9, where current is ~1mA and VAB=1V. This indicates R is connected across points A and B, and V1 is the source across A and B. The current is V\/R = 1V\/1000\u03a9 = 1mA, consistent with the description. The second part describes a changed circuit with V2=5V and internal resistances of both batteries (V1 and V2) are 0.1\u03a9. Assuming A and B are still the points across which R is connected, and both batteries are connected between A and B, the most plausible configuration is that the batteries V1 (1V, 0.1\u03a9) and V2 (5V, 0.1\u03a9) are connected in parallel across A and B, and the resistor R (1000\u03a9) is also connected between A and B. In a parallel circuit with multiple voltage sources, we can use nodal analysis or source transformation. Using nodal analysis with B as the reference node (0V), the voltage at A is V_A. Current through R: I_R = V_A \/ R Current from V1 source: I_1 = (V1 - V_A) \/ r1 = (1 - V_A) \/ 0.1 Current from V2 source: I_2 = (V2 - V_A) \/ r2 = (5 - V_A) \/ 0.1 Applying KCL at node A (sum of currents leaving A is zero): I_R + I_1 + I_2 = 0 is incorrect. The sources are supplying current towards A, so currents entering A should be summed to currents leaving A. Let's assume currents leave A towards R, and into the positive terminals of V1 and V2. This means V_A is higher than the battery terminals connected to A. The positive terminals are connected to A, and negative terminals to B. So current flows from A through R to B, from A through V1 (r1) to B, and from A through V2 (r2) to B. Currents leaving A: I_R = V_A \/ R, I_1 = V_A - V1 \/ r1? No. This implies V_A is higher than the source terminal. Let's redraw with proper node labels. A and B are the terminals. R is between A and B. Battery 1 (V1, r1) is between A and B (positive to A, negative to B). Battery 2 (V2, r2) is between A and B (positive to A, negative to B). Using superposition or combining parallel sources: Equivalent voltage source V_eq = (V1\/r1 + V2\/r2) \/ (1\/r1 + 1\/r2) Equivalent internal resistance r_eq = 1 \/ (1\/r1 + 1\/r2) V_eq = (1V\/0.1\u03a9 + 5V\/0.1\u03a9) \/ (1\/0.1\u03a9 + 1\/0.1\u03a9) = (10 A + 50 A) \/ (10 S + 10 S) = 60 A \/ 20 S = 3 V. r_eq = 1 \/ (10 S + 10 S) = 1 \/ 20 S = 0.05 \u03a9. Now the circuit is a single equivalent source V_eq=3V with internal resistance r_eq=0.05\u03a9 connected across R=1000\u03a9. The voltage across R is V_AB = V_eq * R \/ (R + r_eq) = 3V * 1000\u03a9 \/ (1000\u03a9 + 0.05\u03a9) = 3V * 1000 \/ 1000.05. V_AB \u2248 3V. The current through R is I_R = V_AB \/ R = (3V * 1000 \/ 1000.05) \/ 1000\u03a9 = 3V \/ 1000.05\u03a9. I_R \u2248 3 \/ 1000 = 0.003 A = 3 mA. Calculating precisely: 3 \/ 1000.05 \u2248 0.00299985 A \u2248 2.99985 mA. This is very close to 3.0 mA. - The first part of the question establishes the context: R=1000\u03a9, V1=1V providing ~1mA and VAB=1V, suggesting R is connected across the 1V source. - The second part adds V2=5V and internal resistances 0.1\u03a9 for *both* batteries. - Assuming a parallel connection of the two batteries and the resistor R between points A and B is the most likely configuration given the options and the structure implied by the first sentence. - When voltage sources are in parallel, they can be combined into an equivalent source V_eq and r_eq. - Formula for parallel sources: V_eq = (\u03a3 Vi\/ri) \/ (\u03a3 1\/ri) and r_eq = 1 \/ (\u03a3 1\/ri). - Calculate V_eq and r_eq for V1 (1V, 0.1\u03a9) and V2 (5V, 0.1\u03a9). - Calculate the current through R (1000\u03a9) when connected across the equivalent source. - Calculation gives current \u2248 3 mA.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/an-electric-circuit-is-given-below-v\u2081-1-v-and-resistance-r-1000-\u03c9\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"An electric circuit is given below. V\u2081 = 1 V and Resistance R = 1000 \u03a9\" \/>\n<meta property=\"og:description\" content=\"The first part of the question describes a circuit with a battery V1=1V, resistor R=1000\u03a9, where current is ~1mA and VAB=1V. This indicates R is connected across points A and B, and V1 is the source across A and B. The current is V\/R = 1V\/1000\u03a9 = 1mA, consistent with the description. The second part describes a changed circuit with V2=5V and internal resistances of both batteries (V1 and V2) are 0.1\u03a9. Assuming A and B are still the points across which R is connected, and both batteries are connected between A and B, the most plausible configuration is that the batteries V1 (1V, 0.1\u03a9) and V2 (5V, 0.1\u03a9) are connected in parallel across A and B, and the resistor R (1000\u03a9) is also connected between A and B. In a parallel circuit with multiple voltage sources, we can use nodal analysis or source transformation. Using nodal analysis with B as the reference node (0V), the voltage at A is V_A. Current through R: I_R = V_A \/ R Current from V1 source: I_1 = (V1 - V_A) \/ r1 = (1 - V_A) \/ 0.1 Current from V2 source: I_2 = (V2 - V_A) \/ r2 = (5 - V_A) \/ 0.1 Applying KCL at node A (sum of currents leaving A is zero): I_R + I_1 + I_2 = 0 is incorrect. The sources are supplying current towards A, so currents entering A should be summed to currents leaving A. Let's assume currents leave A towards R, and into the positive terminals of V1 and V2. This means V_A is higher than the battery terminals connected to A. The positive terminals are connected to A, and negative terminals to B. So current flows from A through R to B, from A through V1 (r1) to B, and from A through V2 (r2) to B. Currents leaving A: I_R = V_A \/ R, I_1 = V_A - V1 \/ r1? No. This implies V_A is higher than the source terminal. Let's redraw with proper node labels. A and B are the terminals. R is between A and B. Battery 1 (V1, r1) is between A and B (positive to A, negative to B). Battery 2 (V2, r2) is between A and B (positive to A, negative to B). Using superposition or combining parallel sources: Equivalent voltage source V_eq = (V1\/r1 + V2\/r2) \/ (1\/r1 + 1\/r2) Equivalent internal resistance r_eq = 1 \/ (1\/r1 + 1\/r2) V_eq = (1V\/0.1\u03a9 + 5V\/0.1\u03a9) \/ (1\/0.1\u03a9 + 1\/0.1\u03a9) = (10 A + 50 A) \/ (10 S + 10 S) = 60 A \/ 20 S = 3 V. r_eq = 1 \/ (10 S + 10 S) = 1 \/ 20 S = 0.05 \u03a9. Now the circuit is a single equivalent source V_eq=3V with internal resistance r_eq=0.05\u03a9 connected across R=1000\u03a9. The voltage across R is V_AB = V_eq * R \/ (R + r_eq) = 3V * 1000\u03a9 \/ (1000\u03a9 + 0.05\u03a9) = 3V * 1000 \/ 1000.05. V_AB \u2248 3V. The current through R is I_R = V_AB \/ R = (3V * 1000 \/ 1000.05) \/ 1000\u03a9 = 3V \/ 1000.05\u03a9. I_R \u2248 3 \/ 1000 = 0.003 A = 3 mA. Calculating precisely: 3 \/ 1000.05 \u2248 0.00299985 A \u2248 2.99985 mA. This is very close to 3.0 mA. - The first part of the question establishes the context: R=1000\u03a9, V1=1V providing ~1mA and VAB=1V, suggesting R is connected across the 1V source. - The second part adds V2=5V and internal resistances 0.1\u03a9 for *both* batteries. - Assuming a parallel connection of the two batteries and the resistor R between points A and B is the most likely configuration given the options and the structure implied by the first sentence. - When voltage sources are in parallel, they can be combined into an equivalent source V_eq and r_eq. - Formula for parallel sources: V_eq = (\u03a3 Vi\/ri) \/ (\u03a3 1\/ri) and r_eq = 1 \/ (\u03a3 1\/ri). - Calculate V_eq and r_eq for V1 (1V, 0.1\u03a9) and V2 (5V, 0.1\u03a9). - Calculate the current through R (1000\u03a9) when connected across the equivalent source. - Calculation gives current \u2248 3 mA.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/an-electric-circuit-is-given-below-v\u2081-1-v-and-resistance-r-1000-\u03c9\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:00:11+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"4 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"An electric circuit is given below. V\u2081 = 1 V and Resistance R = 1000 \u03a9","description":"The first part of the question describes a circuit with a battery V1=1V, resistor R=1000\u03a9, where current is ~1mA and VAB=1V. This indicates R is connected across points A and B, and V1 is the source across A and B. The current is V\/R = 1V\/1000\u03a9 = 1mA, consistent with the description. The second part describes a changed circuit with V2=5V and internal resistances of both batteries (V1 and V2) are 0.1\u03a9. Assuming A and B are still the points across which R is connected, and both batteries are connected between A and B, the most plausible configuration is that the batteries V1 (1V, 0.1\u03a9) and V2 (5V, 0.1\u03a9) are connected in parallel across A and B, and the resistor R (1000\u03a9) is also connected between A and B. In a parallel circuit with multiple voltage sources, we can use nodal analysis or source transformation. Using nodal analysis with B as the reference node (0V), the voltage at A is V_A. Current through R: I_R = V_A \/ R Current from V1 source: I_1 = (V1 - V_A) \/ r1 = (1 - V_A) \/ 0.1 Current from V2 source: I_2 = (V2 - V_A) \/ r2 = (5 - V_A) \/ 0.1 Applying KCL at node A (sum of currents leaving A is zero): I_R + I_1 + I_2 = 0 is incorrect. The sources are supplying current towards A, so currents entering A should be summed to currents leaving A. Let's assume currents leave A towards R, and into the positive terminals of V1 and V2. This means V_A is higher than the battery terminals connected to A. The positive terminals are connected to A, and negative terminals to B. So current flows from A through R to B, from A through V1 (r1) to B, and from A through V2 (r2) to B. Currents leaving A: I_R = V_A \/ R, I_1 = V_A - V1 \/ r1? No. This implies V_A is higher than the source terminal. Let's redraw with proper node labels. A and B are the terminals. R is between A and B. Battery 1 (V1, r1) is between A and B (positive to A, negative to B). Battery 2 (V2, r2) is between A and B (positive to A, negative to B). Using superposition or combining parallel sources: Equivalent voltage source V_eq = (V1\/r1 + V2\/r2) \/ (1\/r1 + 1\/r2) Equivalent internal resistance r_eq = 1 \/ (1\/r1 + 1\/r2) V_eq = (1V\/0.1\u03a9 + 5V\/0.1\u03a9) \/ (1\/0.1\u03a9 + 1\/0.1\u03a9) = (10 A + 50 A) \/ (10 S + 10 S) = 60 A \/ 20 S = 3 V. r_eq = 1 \/ (10 S + 10 S) = 1 \/ 20 S = 0.05 \u03a9. Now the circuit is a single equivalent source V_eq=3V with internal resistance r_eq=0.05\u03a9 connected across R=1000\u03a9. The voltage across R is V_AB = V_eq * R \/ (R + r_eq) = 3V * 1000\u03a9 \/ (1000\u03a9 + 0.05\u03a9) = 3V * 1000 \/ 1000.05. V_AB \u2248 3V. The current through R is I_R = V_AB \/ R = (3V * 1000 \/ 1000.05) \/ 1000\u03a9 = 3V \/ 1000.05\u03a9. I_R \u2248 3 \/ 1000 = 0.003 A = 3 mA. Calculating precisely: 3 \/ 1000.05 \u2248 0.00299985 A \u2248 2.99985 mA. This is very close to 3.0 mA. - The first part of the question establishes the context: R=1000\u03a9, V1=1V providing ~1mA and VAB=1V, suggesting R is connected across the 1V source. - The second part adds V2=5V and internal resistances 0.1\u03a9 for *both* batteries. - Assuming a parallel connection of the two batteries and the resistor R between points A and B is the most likely configuration given the options and the structure implied by the first sentence. - When voltage sources are in parallel, they can be combined into an equivalent source V_eq and r_eq. - Formula for parallel sources: V_eq = (\u03a3 Vi\/ri) \/ (\u03a3 1\/ri) and r_eq = 1 \/ (\u03a3 1\/ri). - Calculate V_eq and r_eq for V1 (1V, 0.1\u03a9) and V2 (5V, 0.1\u03a9). - Calculate the current through R (1000\u03a9) when connected across the equivalent source. - Calculation gives current \u2248 3 mA.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/an-electric-circuit-is-given-below-v\u2081-1-v-and-resistance-r-1000-\u03c9\/","og_locale":"en_US","og_type":"article","og_title":"An electric circuit is given below. V\u2081 = 1 V and Resistance R = 1000 \u03a9","og_description":"The first part of the question describes a circuit with a battery V1=1V, resistor R=1000\u03a9, where current is ~1mA and VAB=1V. This indicates R is connected across points A and B, and V1 is the source across A and B. The current is V\/R = 1V\/1000\u03a9 = 1mA, consistent with the description. The second part describes a changed circuit with V2=5V and internal resistances of both batteries (V1 and V2) are 0.1\u03a9. Assuming A and B are still the points across which R is connected, and both batteries are connected between A and B, the most plausible configuration is that the batteries V1 (1V, 0.1\u03a9) and V2 (5V, 0.1\u03a9) are connected in parallel across A and B, and the resistor R (1000\u03a9) is also connected between A and B. In a parallel circuit with multiple voltage sources, we can use nodal analysis or source transformation. Using nodal analysis with B as the reference node (0V), the voltage at A is V_A. Current through R: I_R = V_A \/ R Current from V1 source: I_1 = (V1 - V_A) \/ r1 = (1 - V_A) \/ 0.1 Current from V2 source: I_2 = (V2 - V_A) \/ r2 = (5 - V_A) \/ 0.1 Applying KCL at node A (sum of currents leaving A is zero): I_R + I_1 + I_2 = 0 is incorrect. The sources are supplying current towards A, so currents entering A should be summed to currents leaving A. Let's assume currents leave A towards R, and into the positive terminals of V1 and V2. This means V_A is higher than the battery terminals connected to A. The positive terminals are connected to A, and negative terminals to B. So current flows from A through R to B, from A through V1 (r1) to B, and from A through V2 (r2) to B. Currents leaving A: I_R = V_A \/ R, I_1 = V_A - V1 \/ r1? No. This implies V_A is higher than the source terminal. Let's redraw with proper node labels. A and B are the terminals. R is between A and B. Battery 1 (V1, r1) is between A and B (positive to A, negative to B). Battery 2 (V2, r2) is between A and B (positive to A, negative to B). Using superposition or combining parallel sources: Equivalent voltage source V_eq = (V1\/r1 + V2\/r2) \/ (1\/r1 + 1\/r2) Equivalent internal resistance r_eq = 1 \/ (1\/r1 + 1\/r2) V_eq = (1V\/0.1\u03a9 + 5V\/0.1\u03a9) \/ (1\/0.1\u03a9 + 1\/0.1\u03a9) = (10 A + 50 A) \/ (10 S + 10 S) = 60 A \/ 20 S = 3 V. r_eq = 1 \/ (10 S + 10 S) = 1 \/ 20 S = 0.05 \u03a9. Now the circuit is a single equivalent source V_eq=3V with internal resistance r_eq=0.05\u03a9 connected across R=1000\u03a9. The voltage across R is V_AB = V_eq * R \/ (R + r_eq) = 3V * 1000\u03a9 \/ (1000\u03a9 + 0.05\u03a9) = 3V * 1000 \/ 1000.05. V_AB \u2248 3V. The current through R is I_R = V_AB \/ R = (3V * 1000 \/ 1000.05) \/ 1000\u03a9 = 3V \/ 1000.05\u03a9. I_R \u2248 3 \/ 1000 = 0.003 A = 3 mA. Calculating precisely: 3 \/ 1000.05 \u2248 0.00299985 A \u2248 2.99985 mA. This is very close to 3.0 mA. - The first part of the question establishes the context: R=1000\u03a9, V1=1V providing ~1mA and VAB=1V, suggesting R is connected across the 1V source. - The second part adds V2=5V and internal resistances 0.1\u03a9 for *both* batteries. - Assuming a parallel connection of the two batteries and the resistor R between points A and B is the most likely configuration given the options and the structure implied by the first sentence. - When voltage sources are in parallel, they can be combined into an equivalent source V_eq and r_eq. - Formula for parallel sources: V_eq = (\u03a3 Vi\/ri) \/ (\u03a3 1\/ri) and r_eq = 1 \/ (\u03a3 1\/ri). - Calculate V_eq and r_eq for V1 (1V, 0.1\u03a9) and V2 (5V, 0.1\u03a9). - Calculate the current through R (1000\u03a9) when connected across the equivalent source. - Calculation gives current \u2248 3 mA.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/an-electric-circuit-is-given-below-v\u2081-1-v-and-resistance-r-1000-\u03c9\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:00:11+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"4 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/an-electric-circuit-is-given-below-v%e2%82%81-1-v-and-resistance-r-1000-%cf%89\/","url":"https:\/\/exam.pscnotes.com\/mcq\/an-electric-circuit-is-given-below-v%e2%82%81-1-v-and-resistance-r-1000-%cf%89\/","name":"An electric circuit is given below. V\u2081 = 1 V and Resistance R = 1000 \u03a9","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:00:11+00:00","dateModified":"2025-06-01T07:00:11+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The first part of the question describes a circuit with a battery V1=1V, resistor R=1000\u03a9, where current is ~1mA and VAB=1V. This indicates R is connected across points A and B, and V1 is the source across A and B. The current is V\/R = 1V\/1000\u03a9 = 1mA, consistent with the description. The second part describes a changed circuit with V2=5V and internal resistances of both batteries (V1 and V2) are 0.1\u03a9. Assuming A and B are still the points across which R is connected, and both batteries are connected between A and B, the most plausible configuration is that the batteries V1 (1V, 0.1\u03a9) and V2 (5V, 0.1\u03a9) are connected in parallel across A and B, and the resistor R (1000\u03a9) is also connected between A and B. In a parallel circuit with multiple voltage sources, we can use nodal analysis or source transformation. Using nodal analysis with B as the reference node (0V), the voltage at A is V_A. Current through R: I_R = V_A \/ R Current from V1 source: I_1 = (V1 - V_A) \/ r1 = (1 - V_A) \/ 0.1 Current from V2 source: I_2 = (V2 - V_A) \/ r2 = (5 - V_A) \/ 0.1 Applying KCL at node A (sum of currents leaving A is zero): I_R + I_1 + I_2 = 0 is incorrect. The sources are supplying current towards A, so currents entering A should be summed to currents leaving A. Let's assume currents leave A towards R, and into the positive terminals of V1 and V2. This means V_A is higher than the battery terminals connected to A. The positive terminals are connected to A, and negative terminals to B. So current flows from A through R to B, from A through V1 (r1) to B, and from A through V2 (r2) to B. Currents leaving A: I_R = V_A \/ R, I_1 = V_A - V1 \/ r1? No. This implies V_A is higher than the source terminal. Let's redraw with proper node labels. A and B are the terminals. R is between A and B. Battery 1 (V1, r1) is between A and B (positive to A, negative to B). Battery 2 (V2, r2) is between A and B (positive to A, negative to B). Using superposition or combining parallel sources: Equivalent voltage source V_eq = (V1\/r1 + V2\/r2) \/ (1\/r1 + 1\/r2) Equivalent internal resistance r_eq = 1 \/ (1\/r1 + 1\/r2) V_eq = (1V\/0.1\u03a9 + 5V\/0.1\u03a9) \/ (1\/0.1\u03a9 + 1\/0.1\u03a9) = (10 A + 50 A) \/ (10 S + 10 S) = 60 A \/ 20 S = 3 V. r_eq = 1 \/ (10 S + 10 S) = 1 \/ 20 S = 0.05 \u03a9. Now the circuit is a single equivalent source V_eq=3V with internal resistance r_eq=0.05\u03a9 connected across R=1000\u03a9. The voltage across R is V_AB = V_eq * R \/ (R + r_eq) = 3V * 1000\u03a9 \/ (1000\u03a9 + 0.05\u03a9) = 3V * 1000 \/ 1000.05. V_AB \u2248 3V. The current through R is I_R = V_AB \/ R = (3V * 1000 \/ 1000.05) \/ 1000\u03a9 = 3V \/ 1000.05\u03a9. I_R \u2248 3 \/ 1000 = 0.003 A = 3 mA. Calculating precisely: 3 \/ 1000.05 \u2248 0.00299985 A \u2248 2.99985 mA. This is very close to 3.0 mA. - The first part of the question establishes the context: R=1000\u03a9, V1=1V providing ~1mA and VAB=1V, suggesting R is connected across the 1V source. - The second part adds V2=5V and internal resistances 0.1\u03a9 for *both* batteries. - Assuming a parallel connection of the two batteries and the resistor R between points A and B is the most likely configuration given the options and the structure implied by the first sentence. - When voltage sources are in parallel, they can be combined into an equivalent source V_eq and r_eq. - Formula for parallel sources: V_eq = (\u03a3 Vi\/ri) \/ (\u03a3 1\/ri) and r_eq = 1 \/ (\u03a3 1\/ri). - Calculate V_eq and r_eq for V1 (1V, 0.1\u03a9) and V2 (5V, 0.1\u03a9). - Calculate the current through R (1000\u03a9) when connected across the equivalent source. - Calculation gives current \u2248 3 mA.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/an-electric-circuit-is-given-below-v%e2%82%81-1-v-and-resistance-r-1000-%cf%89\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/an-electric-circuit-is-given-below-v%e2%82%81-1-v-and-resistance-r-1000-%cf%89\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/an-electric-circuit-is-given-below-v%e2%82%81-1-v-and-resistance-r-1000-%cf%89\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-1\/"},{"@type":"ListItem","position":3,"name":"An electric circuit is given below. V\u2081 = 1 V and Resistance R = 1000 \u03a9"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88037","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=88037"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88037\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88037"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88037"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88037"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}