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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Capacitance

JEE Main · Physics · Capacitance

The effective capacitance of two capacitors of capacitances C _ 1...

The effective capacitance of two capacitors of capacitances C1{C}_{1} and C2{C}_{2}\left({C}_{2}> {C}_{1}\right)connectedinparallelisconnected in parallel is\dfrac{25}{6}timestheeffectivecapacitancewhentheyareconnectedinseries.Theratiotimes the effective capacitance when they are connected in series. The ratio\dfrac{{C}_{2}}{{C}_{1}}is

Options

  1. A32\dfrac{3}{2}
  2. B43\dfrac{4}{3}
  3. C53\dfrac{5}{3}
  4. D256\dfrac{25}{6}

Correct answer

A. 32\dfrac{3}{2}

Step-by-step solution

Cparallel=256Cseries{C}_{parallel} = \dfrac{25}{6}{C}_{series}

⇒C1+C2=256C1C2C1+C2⇒ {C}_{1} + {C}_{2} = \dfrac{25}{6}\dfrac{{C}_{1}{C}_{2}}{{C}_{1} + {C}_{2}}

⇒6(C1+C2)2=25C1C2⇒ 6{\left({C}_{1} + {C}_{2}\right)}^{2} = 25{C}_{1}{C}_{2}

⇒6C12+6C22+12C1C2=25C1C2⇒ 6{{C}_{1}}^{2} + 6{{C}_{2}}^{2} + 12{C}_{1}{C}_{2} = 25{C}_{1}{C}_{2}

⇒6C12C22+6+12C1C2=25C1C2⇒ 6\dfrac{{{C}_{1}}^{2}}{{{C}_{2}}^{2}} + 6 + 12\dfrac{{C}_{1}}{{C}_{2}} = 25\dfrac{{C}_{1}}{{C}_{2}}

⇒6C12C22−13C1C2+6=0⇒ 6\dfrac{{{C}_{1}}^{2}}{{{C}_{2}}^{2}} - 13\dfrac{{C}_{1}}{{C}_{2}} + 6 = 0

This is quadratic equation

By solving we get C2C1=32\dfrac{{C}_{2}}{{C}_{1}} = \dfrac{3}{2}
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