{"id":244479,"date":"2025-07-05T07:24:23","date_gmt":"2025-07-05T07:24:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=244479"},"modified":"2025-07-05T07:24:25","modified_gmt":"2025-07-05T07:24:25","slug":"for-the-circuit-shown-in-the-figure-figure-1-find-the-current-through-and-the-potential-difference-across-each-resistor","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/05\/for-the-circuit-shown-in-the-figure-figure-1-find-the-current-through-and-the-potential-difference-across-each-resistor\/","title":{"rendered":"For the circuit shown in the figure (Figure 1), find the current through and the potential difference across each resistor."},"content":{"rendered":"\n

For the circuit shown in the figure (Figure 1), find the current through and the potential difference across each resistor. Part G Find the current through 16
resistor. Express your answer using two significant figures. Figure < 1 of 1 >
3.0
16
12 V 4.0
48
Submit Previous Answers Request Answer Incorrect; Try Again; 27 attempts remaining Part H Find the potential difference across 16
resistor. Express your answer using two significant figures.<\/p>\n\n\n\n

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

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Of course. Here are the correct answers and a step-by-step explanation for solving this circuit problem.<\/p>\n\n\n\n

Correct Answers<\/h3>\n\n\n\n

Part G: Find the current through the 16 \u03a9 resistor.<\/strong>
I = 0.56 A<\/strong><\/p>\n\n\n\n

Part H: Find the potential difference across the 16 \u03a9 resistor.<\/strong>
V = 9.0 V<\/strong><\/p>\n\n\n\n

Explanation<\/h3>\n\n\n\n

This circuit cannot be simplified using basic series and parallel rules alone. The arrangement of the 3.0 \u03a9, 4.0 \u03a9, and 16 \u03a9 resistors forms a delta (\u0394) configuration, which requires a more advanced technique like a Delta-Wye (\u0394-Y) transformation or using Kirchhoff’s Laws. The \u0394-Y transformation is often the most straightforward method.<\/p>\n\n\n\n

1. Identify the Nodes and the Delta Configuration<\/strong>
First, let’s label the important nodes (junctions) in the circuit:<\/p>\n\n\n\n

    \n
  • J1:<\/strong>\u00a0The junction connected to the positive terminal of the 12 V battery, the 3.0 \u03a9 resistor, and the 4.0 \u03a9 resistor.<\/li>\n\n\n\n
  • J2:<\/strong>\u00a0The junction between the 3.0 \u03a9 and 16 \u03a9 resistors.<\/li>\n\n\n\n
  • J3:<\/strong>\u00a0The junction between the 4.0 \u03a9, 16 \u03a9, and 48 \u03a9 resistors.<\/li>\n\n\n\n
  • J4:<\/strong>\u00a0The node connected to the negative terminal of the battery. Note that the wire from J2 and the 48 \u03a9 resistor both connect to this node.<\/li>\n<\/ul>\n\n\n\n

    The resistors R\u2081=3.0 \u03a9, R\u2082=4.0 \u03a9, and R\u2083=16 \u03a9 form a delta (\u0394) between nodes J1, J2, and J3. We can convert this delta into an equivalent wye (Y) network.<\/p>\n\n\n\n

    2. Perform the Delta-to-Wye (\u0394-Y) Transformation<\/strong>
    The formulas for the Y-network resistors (Ra, Rb, Rc) connected to a new central node (N) are:<\/p>\n\n\n\n

      \n
    • Sum of delta resistors: R_\u0394 = R\u2081 + R\u2082 + R\u2083 = 3.0 \u03a9 + 4.0 \u03a9 + 16 \u03a9 = 23 \u03a9.<\/li>\n\n\n\n
    • Ra<\/strong>\u00a0(connects J1 to N): Ra = (R\u2081 * R\u2082) \/ R_\u0394 = (3.0 * 4.0) \/ 23 = 12\/23 \u03a9 \u2248 0.522 \u03a9.<\/li>\n\n\n\n
    • Rb<\/strong>\u00a0(connects J2 to N): Rb = (R\u2081 * R\u2083) \/ R_\u0394 = (3.0 * 16) \/ 23 = 48\/23 \u03a9 \u2248 2.09 \u03a9.<\/li>\n\n\n\n
    • Rc<\/strong>\u00a0(connects J3 to N): Rc = (R\u2082 * R\u2083) \/ R_\u0394 = (4.0 * 16) \/ 23 = 64\/23 \u03a9 \u2248 2.78 \u03a9.<\/li>\n<\/ul>\n\n\n\n

      3. Analyze the Simplified Circuit<\/strong>
      After replacing the delta with the wye, the circuit becomes much simpler:<\/p>\n\n\n\n

        \n
      • Resistor Ra is now in series with the rest of the circuit.<\/li>\n\n\n\n
      • From the new node N, we have two parallel branches connected to the negative terminal (J4):\n
          \n
        • Branch 1:<\/strong>\u00a0Resistor Rb is connected between N and J2. Since J2 is wired directly to J4, this branch just contains Rb.<\/li>\n\n\n\n
        • Branch 2:<\/strong>\u00a0Resistor Rc is connected between N and J3. Resistor R\u2084=48 \u03a9 connects J3 to J4. So, this branch has Rc and the 48 \u03a9 resistor in series.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

          4. Calculate the Equivalent Resistance (R_eq)<\/strong><\/p>\n\n\n\n

            \n
          • Resistance of Branch 2: R_branch2 = Rc + 48 \u03a9 = (64\/23) \u03a9 + 48 \u03a9 = 1168\/23 \u03a9 \u2248 50.8 \u03a9.<\/li>\n\n\n\n
          • Equivalent resistance of the two parallel branches (R_p):
            R_p = (Rb * R_branch2) \/ (Rb + R_branch2) = [(48\/23) * (1168\/23)] \/ [(48\/23) + (1168\/23)] = 2.0045 \u03a9.<\/li>\n\n\n\n
          • Total equivalent resistance of the circuit:
            R_eq = Ra + R_p = (12\/23) \u03a9 + 2.0045 \u03a9 \u2248 0.522 \u03a9 + 2.0045 \u03a9 \u2248 2.527 \u03a9.<\/li>\n<\/ul>\n\n\n\n

            5. Find Voltages at Original Nodes<\/strong>
            To find the current and voltage for the 16 \u03a9 resistor, we need the potentials at nodes J2 and J3.<\/p>\n\n\n\n

              \n
            • Total current from the battery: I_total = V \/ R_eq = 12 V \/ 2.527 \u03a9 \u2248 4.75 A.<\/li>\n\n\n\n
            • The voltage at node N is the voltage across the parallel combination: V_N = I_total * R_p \u2248 4.75 A * 2.0045 \u03a9 \u2248 9.52 V.<\/li>\n\n\n\n
            • Potential at J2: Node J2 is connected directly to the negative terminal (J4), so its potential is 0 V.<\/li>\n\n\n\n
            • Potential at J3: This is the voltage across the 48 \u03a9 resistor. First, find the current in Branch 2: I_branch2 = V_N \/ R_branch2 \u2248 9.52 V \/ 50.8 \u03a9 \u2248 0.1875 A.
              Then, V_J3 = I_branch2 * 48 \u03a9 = 0.1875 A * 48 \u03a9 = 9.0 V.<\/li>\n<\/ul>\n\n\n\n

              6. Calculate Final Answers<\/strong><\/p>\n\n\n\n

                \n
              • Potential Difference (Part H):<\/strong>\u00a0The potential difference across the 16 \u03a9 resistor is the difference between V_J3 and V_J2.
                V_16 = |V_J3 – V_J2| = |9.0 V – 0 V| =\u00a09.0 V<\/strong>.<\/li>\n\n\n\n
              • Current (Part G):<\/strong>\u00a0Now use Ohm’s law for the 16 \u03a9 resistor.
                I_16 = V_16 \/ R_16 = 9.0 V \/ 16 \u03a9 = 0.5625 A.
                Rounding to two significant figures gives\u00a00.56 A<\/strong>.<\/li>\n<\/ul>\n\n\n\n
                \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                For the circuit shown in the figure (Figure 1), find the current through and the potential difference across each resistor. Part G Find the current through 16resistor. Express your answer using two significant figures. Figure < 1 of 1 >3.01612 V 4.048Submit Previous Answers Request Answer Incorrect; Try Again; 27 attempts remaining Part H Find […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-244479","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244479","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=244479"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244479\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244479"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244479"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244479"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}