{"id":4606,"date":"2026-04-11T12:05:34","date_gmt":"2026-04-11T12:05:34","guid":{"rendered":"https:\/\/ksquareinstitute.in\/blog\/?p=4606"},"modified":"2026-04-11T12:05:35","modified_gmt":"2026-04-11T12:05:35","slug":"physics-current-electricity-pyqs","status":"publish","type":"post","link":"https:\/\/ksquareinstitute.in\/blog\/physics-current-electricity-pyqs\/","title":{"rendered":"Top PYQs from Current Electricity with Concepts & Quick Notes for NEET"},"content":{"rendered":"\n

026Current Electricity is one of the most important and high-weightage chapters in NEET Physics. The questions are mostly numerical and follow standard patterns involving Ohm\u2019s law, resistors, Kirchhoff\u2019s laws, and power dissipation. If you consistently practice Physics Current Electricity PYQs<\/strong>, you will notice that the same concepts are repeated with slight variations. This makes Physics Current Electricity PYQs<\/strong> extremely important for scoring high marks with accuracy.<\/p>\n\n\n\n

This article combines quick notes + top PYQs with detailed solutions<\/strong>, allowing you to revise the entire chapter efficiently. Mastering Physics Current Electricity PYQs<\/strong> ensures faster calculations and strong conceptual clarity.<\/p>\n\n\n\n

\"Physics<\/figure>\n\n\n\n

Physics Current Electricity Quick Notes<\/h1>\n\n\n\n

Before solving Physics Current Electricity PYQs<\/strong>, revise these key formulas:<\/p>\n\n\n\n

    \n
  • Ohm\u2019s Law:<\/li>\n<\/ul>\n\n\n\n

    V<\/mi>=<\/mo>I<\/mi>R<\/mi><\/mrow>V = IR<\/annotation><\/semantics><\/math> V<\/mi>=<\/mo>I<\/mi>R<\/mi><\/mrow>V = IR<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
      \n
    • Power:<\/li>\n<\/ul>\n\n\n\n

      P<\/mi>=<\/mo>V<\/mi>I<\/mi>=<\/mo>I<\/mi>2<\/mn><\/msup>R<\/mi>=<\/mo>V<\/mi>2<\/mn><\/msup>R<\/mi><\/mfrac><\/mrow>P = VI = I^2R = \\frac{V^2}{R}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
        \n
      • Series Combination:<\/li>\n<\/ul>\n\n\n\n

        R<\/mi>=<\/mo>R<\/mi>1<\/mn><\/msub>+<\/mo>R<\/mi>2<\/mn><\/msub>+<\/mo>R<\/mi>3<\/mn><\/msub><\/mrow>R = R_1 + R_2 + R_3<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
          \n
        • Parallel Combination:<\/li>\n<\/ul>\n\n\n\n

          1<\/mn>R<\/mi><\/mfrac>=<\/mo>1<\/mn>R<\/mi>1<\/mn><\/msub><\/mfrac>+<\/mo>1<\/mn>R<\/mi>2<\/mn><\/msub><\/mfrac><\/mrow>\\frac{1}{R} = \\frac{1}{R_1} + \\frac{1}{R_2}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
            \n
          • Drift Velocity:<\/li>\n<\/ul>\n\n\n\n

            v<\/mi>d<\/mi><\/msub>=<\/mo>I<\/mi>n<\/mi>q<\/mi>A<\/mi><\/mrow><\/mfrac><\/mrow>v_d = \\frac{I}{nqA}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
              \n
            • Kirchhoff\u2019s Laws:
              Sum of currents at junction = 0
              Sum of voltages in loop = 0<\/li>\n<\/ul>\n\n\n\n

              These formulas are the backbone of most Physics Current Electricity PYQs<\/strong>.<\/p>\n\n\n\n

              Top Physics Current Electricity PYQs (With Detailed Solutions)<\/h1>\n\n\n\n

              1) Ohm\u2019s law application<\/h2>\n\n\n\n

              A resistor of 10 \u03a9 carries current 2 A. Find voltage.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              V<\/mi>=<\/mo>I<\/mi>R<\/mi>=<\/mo>2<\/mn>\u00d7<\/mo>10<\/mn>=<\/mo>20<\/mn> V<\/mtext><\/mrow>V = IR = 2 \\times 10 = 20\\text{ V}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 20 V<\/strong><\/p>\n\n\n\n

              \n\n\n\n

              2) Power dissipation<\/h2>\n\n\n\n

              A resistor of 5 \u03a9 carries 2 A current.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              P<\/mi>=<\/mo>I<\/mi>2<\/mn><\/msup>R<\/mi>=<\/mo>4<\/mn>\u00d7<\/mo>5<\/mn>=<\/mo>20<\/mn> W<\/mtext><\/mrow>P = I^2R = 4 \\times 5 = 20\\text{ W}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 20 W<\/strong><\/p>\n\n\n\n

              \n\n\n\n

              3) Series resistance<\/h2>\n\n\n\n

              Resistors 2 \u03a9 and 3 \u03a9 are in series.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              R<\/mi>=<\/mo>5<\/mn> \u03a9<\/mtext><\/mrow>R = 5\\text{ \u03a9}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 5 \u03a9<\/strong><\/p>\n\n\n\n

              \n\n\n\n

              4) Parallel resistance<\/h2>\n\n\n\n

              2 \u03a9 and 3 \u03a9 in parallel.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              1<\/mn>R<\/mi><\/mfrac>=<\/mo>1<\/mn>2<\/mn><\/mfrac>+<\/mo>1<\/mn>3<\/mn><\/mfrac>=<\/mo>5<\/mn>6<\/mn><\/mfrac>\u21d2<\/mo>R<\/mi>=<\/mo>6<\/mn>5<\/mn><\/mfrac><\/mrow>\\frac{1}{R} = \\frac{1}{2} + \\frac{1}{3} = \\frac{5}{6} \\Rightarrow R = \\frac{6}{5}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 1.2 \u03a9<\/strong><\/p>\n\n\n\n

              \n\n\n\n

              5) Current division<\/h2>\n\n\n\n

              Two resistors in parallel: 2 \u03a9 and 4 \u03a9. Total current 3 A. Current in 2 \u03a9?<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              Current divides inversely with resistance:I<\/mi>1<\/mn><\/msub>=<\/mo>4<\/mn>6<\/mn><\/mfrac>\u00d7<\/mo>3<\/mn>=<\/mo>2<\/mn> A<\/mtext><\/mrow>I_1 = \\frac{4}{6} \\times 3 = 2\\text{ A}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 2 A<\/strong><\/p>\n\n\n\n

              \n\n\n\n

              6) Kirchhoff loop<\/h2>\n\n\n\n

              Battery 10 V, resistors 2 \u03a9 and 3 \u03a9 in series.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              I<\/mi>=<\/mo>10<\/mn>5<\/mn><\/mfrac>=<\/mo>2<\/mn> A<\/mtext><\/mrow>I = \\frac{10}{5} = 2\\text{ A}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 2 A<\/strong><\/p>\n\n\n\n

              \n\n\n\n

              7) Drift velocity<\/h2>\n\n\n\n

              Current 3 A, area 1 m\u00b2, charge density known.<\/p>\n\n\n\n

              Concept<\/h3>\n\n\n\n

              v<\/mi>d<\/mi><\/msub>=<\/mo>I<\/mi>n<\/mi>q<\/mi>A<\/mi><\/mrow><\/mfrac><\/mrow>v_d = \\frac{I}{nqA}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer:<\/strong> Use formula directly<\/p>\n\n\n\n

              \n\n\n\n

              8) Power comparison<\/h2>\n\n\n\n

              Two resistors same voltage \u2192 power relation?<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              P<\/mi>\u221d<\/mo>1<\/mn>R<\/mi><\/mfrac><\/mrow>P \\propto \\frac{1}{R}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer:<\/strong> Lower resistance \u2192 more power<\/p>\n\n\n\n

              \n\n\n\n

              9) Heating effect<\/h2>\n\n\n\n

              Energy dissipated in 5 s, current 2 A, resistance 5 \u03a9.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              H<\/mi>=<\/mo>I<\/mi>2<\/mn><\/msup>R<\/mi>t<\/mi>=<\/mo>4<\/mn>\u00d7<\/mo>5<\/mn>\u00d7<\/mo>5<\/mn>=<\/mo>100<\/mn> J<\/mtext><\/mrow>H = I^2Rt = 4 \\times 5 \\times 5 = 100\\text{ J}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 100 J<\/strong><\/p>\n\n\n\n

              \n\n\n\n

              10) Combination circuit<\/h2>\n\n\n\n

              Three equal resistors in parallel.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              R<\/mi>=<\/mo>R<\/mi>3<\/mn><\/mfrac><\/mrow>R = \\frac{R}{3}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: R\/3<\/strong><\/p>\n\n\n\n

              \n\n\n\n

              11) Internal resistance<\/h2>\n\n\n\n

              EMF 12 V, current 2 A, external R = 5 \u03a9.<\/p>\n\n\n\n

              Solution<\/h3>\n\n\n\n

              V<\/mi>=<\/mo>I<\/mi>R<\/mi>\u21d2<\/mo>10<\/mn>=<\/mo>12<\/mn>\u2212<\/mo>I<\/mi>r<\/mi>\u21d2<\/mo>r<\/mi>=<\/mo>1<\/mn> \u03a9<\/mtext><\/mrow>V = IR \\Rightarrow 10 = 12 – Ir \\Rightarrow r = 1\\text{ \u03a9}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer: 1 \u03a9<\/strong><\/p>\n\n\n\n

              \n\n\n\n

              12) Potential difference<\/h2>\n\n\n\n

              Work done per charge.<\/p>\n\n\n\n

              Concept<\/h3>\n\n\n\n

              V<\/mi>=<\/mo>W<\/mi>q<\/mi><\/mfrac><\/mrow>V = \\frac{W}{q}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
              \n\n\n\n

              13) Resistivity relation<\/h2>\n\n\n\n

              R<\/mi>=<\/mo>\u03c1<\/mi>l<\/mi>A<\/mi><\/mfrac><\/mrow>R = \\rho \\frac{l}{A}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
              \n\n\n\n

              14) Temperature effect<\/h2>\n\n\n\n

              Resistance increases with temperature.<\/p>\n\n\n\n

              \n\n\n\n

              15) Power maximum condition<\/h2>\n\n\n\n

              R<\/mi>=<\/mo>r<\/mi><\/mrow>R = r<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
              \n\n\n\n

              Quick Revision \u2013 Physics Current Electricity PYQs<\/h1>\n\n\n\n

              To revise Physics Current Electricity PYQs<\/strong>, remember:V<\/mi>=<\/mo>I<\/mi>R<\/mi><\/mrow>V = IR<\/annotation><\/semantics><\/math>V=IR P<\/mi>=<\/mo>I<\/mi>2<\/mn><\/msup>R<\/mi><\/mrow>P = I^2R<\/annotation><\/semantics><\/math>P=I2R R<\/mi>series<\/mtext><\/msub>=<\/mo>R<\/mi>1<\/mn><\/msub>+<\/mo>R<\/mi>2<\/mn><\/msub><\/mrow>R_{\\text{series}} = R_1 + R_2<\/annotation><\/semantics><\/math>Rseries\u200b=R1\u200b+R2\u200b R<\/mi>parallel<\/mtext><\/msub>=<\/mo>R<\/mi>1<\/mn><\/msub>R<\/mi>2<\/mn><\/msub><\/mrow>R<\/mi>1<\/mn><\/msub>+<\/mo>R<\/mi>2<\/mn><\/msub><\/mrow><\/mfrac><\/mrow>R_{\\text{parallel}} = \\frac{R_1R_2}{R_1 + R_2}<\/annotation><\/semantics><\/math>Rparallel\u200b=R1\u200b+R2\u200bR1\u200bR2\u200b\u200b<\/p>\n\n\n\n

              These formulas solve most Physics Current Electricity PYQs<\/strong> instantly.<\/p>\n\n\n\n

              \n\n\n\n

              Most Important NEET Patterns<\/h1>\n\n\n\n

              From repeated trends in Physics Current Electricity PYQs<\/strong>, NEET commonly asks:<\/p>\n\n\n\n

                \n
              • Ohm\u2019s law numericals<\/li>\n\n\n\n
              • Series-parallel circuits<\/li>\n\n\n\n
              • Kirchhoff\u2019s laws<\/li>\n\n\n\n
              • Power dissipation<\/li>\n\n\n\n
              • Internal resistance<\/li>\n<\/ul>\n\n\n\n

                Practicing these Physics Current Electricity PYQs<\/strong> ensures strong command over the chapter.<\/p>\n\n\n\n

                \n\n\n\n

                Final Takeaway<\/h1>\n\n\n\n

                The key to mastering Physics Current Electricity PYQs<\/strong> is circuit simplification and formula application. Most problems become easy if you reduce the circuit step by step.<\/p>\n\n\n\n

                If you revise these Physics Current Electricity PYQs<\/strong> properly, this chapter can become one of your highest scoring areas in NEET.<\/p>\n\n\n\n

                \n\n\n\n

                FAQ \u2013 Physics Current Electricity PYQs<\/h1>\n\n\n\n

                1. Why are Physics Current Electricity PYQs important?<\/h3>\n\n\n\n

                They help identify repeated circuit patterns and improve speed.<\/p>\n\n\n\n

                2. Which topics are most important?<\/h3>\n\n\n\n

                Ohm\u2019s law, Kirchhoff\u2019s laws, and resistance combinations.<\/p>\n\n\n\n

                3. How many questions come in NEET?<\/h3>\n\n\n\n

                Usually 2\u20133.<\/p>\n\n\n\n

                4. Is this chapter scoring?<\/h3>\n\n\n\n

                Yes, one of the highest scoring chapters in NEET Physics.<\/p>\n","protected":false},"excerpt":{"rendered":"

                026Current Electricity is one of the most important and high-weightage chapters in NEET Physics. The questions are mostly numerical and follow standard patterns involving Ohm\u2019s law, resistors, Kirchhoff\u2019s laws, and power dissipation. If you consistently practice Physics Current Electricity PYQs, you will notice that the same concepts are repeated with slight variations. This makes Physics […]<\/p>\n","protected":false},"author":1,"featured_media":4607,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[70,2],"tags":[1111,1108,1113,1117,1114,1115,1110,96,1081,1109,1107,1116,1112],"class_list":["post-4606","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-physics","category-neet","tag-circuit-analysis-neet","tag-current-electricity-neet-questions","tag-current-electricity-quick-notes","tag-electrical-circuits-neet","tag-electricity-numericals-physics","tag-internal-resistance-questions","tag-kirchhoff-law-problems","tag-neet-physics-preparation","tag-neet-physics-pyq-practice","tag-ohms-law-numericals","tag-physics-current-electricity-pyqs","tag-power-dissipation-problems","tag-resistance-combination-questions"],"blocksy_meta":{"page_structure_type":"type-1","styles_descriptor":{"styles":{"desktop":"","tablet":"","mobile":""},"google_fonts":[],"version":6}},"_links":{"self":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/4606","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/comments?post=4606"}],"version-history":[{"count":1,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/4606\/revisions"}],"predecessor-version":[{"id":4608,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/4606\/revisions\/4608"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/media\/4607"}],"wp:attachment":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/media?parent=4606"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/categories?post=4606"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/tags?post=4606"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}