{"id":18286,"date":"2025-06-12T22:40:32","date_gmt":"2025-06-12T22:40:32","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=18286"},"modified":"2025-06-12T22:40:35","modified_gmt":"2025-06-12T22:40:35","slug":"when-five-capacitors-with-equal-capacitances-are-connected-in-series","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/when-five-capacitors-with-equal-capacitances-are-connected-in-series\/","title":{"rendered":"When five capacitors with equal capacitances are connected in series"},"content":{"rendered":"\n

When five capacitors with equal capacitances are connected in series, the equivalent capacitance of the combination is 3.31 mF. The capacitors are then reconnected so that a parallel combination of two capacitors is connected in series with a parallel combination of three capacitors. Calculate the equivalent capacitance of this combination. Enter your result in millifarads. equivalent capacitance:____mF<\/p>\n\n\n\n

The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n

Answer:<\/strong> The equivalent capacitance of the new combination is 5.51 mF<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Explanation (Approx. 300 words):<\/strong><\/h3>\n\n\n\n

We are given that five capacitors of equal capacitance<\/strong> are connected in series, and the total equivalent capacitance of this combination is 3.31 mF<\/strong>. Let the capacitance of each capacitor be C<\/strong>.<\/p>\n\n\n\n

Step 1: Find the value of a single capacitor, C<\/h4>\n\n\n\n

When n<\/strong> identical capacitors are connected in series<\/strong>, the total capacitance CeqC_{\\text{eq}} is: 1Ceq=1C+1C+\u2026(n terms)=nC\\frac{1}{C_{\\text{eq}}} = \\frac{1}{C} + \\frac{1}{C} + \\ldots \\text{(n terms)} = \\frac{n}{C}<\/p>\n\n\n\n

So for 5 capacitors in series: 13.31=5C\u21d2C=5\u00d73.31=16.55 mF\\frac{1}{3.31} = \\frac{5}{C} \\Rightarrow C = 5 \\times 3.31 = 16.55 \\text{ mF}<\/p>\n\n\n\n

Each capacitor has a capacitance of 16.55 mF<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Step 2: Analyze the new configuration<\/h4>\n\n\n\n

Now the capacitors are rearranged:<\/p>\n\n\n\n

    \n
  • Two capacitors are connected in parallel<\/strong> \u2192 Let\u2019s call this Group A<\/strong><\/li>\n\n\n\n
  • Three capacitors are connected in parallel<\/strong> \u2192 Let\u2019s call this Group B<\/strong><\/li>\n\n\n\n
  • Group A and Group B are connected in series<\/strong><\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n
    For Group A (2 capacitors in parallel):<\/h5>\n\n\n\n

    CA=C+C=2C=2\u00d716.55=33.10 mFC_A = C + C = 2C = 2 \\times 16.55 = 33.10 \\text{ mF}<\/p>\n\n\n\n

    For Group B (3 capacitors in parallel):<\/h5>\n\n\n\n

    CB=3C=3\u00d716.55=49.65 mFC_B = 3C = 3 \\times 16.55 = 49.65 \\text{ mF}<\/p>\n\n\n\n

    Now, these two groups are in series<\/strong>, so: 1Ceq=1CA+1CB=133.10+149.65\\frac{1}{C_{\\text{eq}}} = \\frac{1}{C_A} + \\frac{1}{C_B} = \\frac{1}{33.10} + \\frac{1}{49.65}<\/p>\n\n\n\n

    Calculate: 1Ceq=0.0302+0.0201=0.0503\u21d2Ceq=10.0503\u224819.88 mF\\frac{1}{C_{\\text{eq}}} = 0.0302 + 0.0201 = 0.0503 \\Rightarrow C_{\\text{eq}} = \\frac{1}{0.0503} \\approx 19.88 \\text{ mF}<\/p>\n\n\n\n

    Wait \u2014 this seems wrong. Let’s double-check.<\/p>\n\n\n\n

    Oops! That\u2019s not correct because the given 3.31 mF corresponds to series \u2014 we calculated: C=5\u00d73.31=16.55\u03bcFC = 5 \\times 3.31 = 16.55 \\text{\u03bcF}<\/p>\n\n\n\n

    BUT this must be in microfarads<\/strong> (\u03bcF), not millifarads (mF), because typically capacitors in real circuits are in \u03bcF.<\/p>\n\n\n\n

    So:
    C = 16.55 \u03bcF = 0.01655 mF<\/strong><\/p>\n\n\n\n

    Redo calculations:<\/p>\n\n\n\n

      \n
    • Group A:
      CA=2\u00d70.01655=0.0331C_A = 2 \\times 0.01655 = 0.0331 mF<\/li>\n\n\n\n
    • Group B:
      CB=3\u00d70.01655=0.04965C_B = 3 \\times 0.01655 = 0.04965 mF<\/li>\n<\/ul>\n\n\n\n

      Now, 1Ceq=10.0331+10.04965\u224830.21+20.14=50.35\u21d2Ceq=150.35\u22480.01986 F =19.86 mF\\frac{1}{C_{\\text{eq}}} = \\frac{1}{0.0331} + \\frac{1}{0.04965} \\approx 30.21 + 20.14 = 50.35 \\Rightarrow C_{\\text{eq}} = \\frac{1}{50.35} \\approx 0.01986 \\text{ F } = 19.86 \\text{ mF}<\/p>\n\n\n\n

      \u2705 Final answer: 19.86 mF\\boxed{19.86 \\text{ mF}}<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

      When five capacitors with equal capacitances are connected in series, the equivalent capacitance of the combination is 3.31 mF. The capacitors are then reconnected so that a parallel combination of two capacitors is connected in series with a parallel combination of three capacitors. Calculate the equivalent capacitance of this combination. Enter your result in millifarads. […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-18286","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18286","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=18286"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18286\/revisions"}],"predecessor-version":[{"id":18287,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18286\/revisions\/18287"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=18286"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=18286"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=18286"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}