Capacitors

JEE Main level, single correct, medium

44 of 180

Question 44 of 180

Consider the combination of two capacitors C1{C}_{1} and C2{C}_{2}, with C2>C1{C}_{2}\gt {C}_{1}, when connected in parallel, the equivalent capacitance is 1010 times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, C2C1\dfrac{{C}_{2}}{{C}_{1}}.

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Answer: (A) 4+154+\sqrt{15}

When connected in parallel
Ceq=C1+C2{C}_{eq}={C}_{1}+{C}_{2}
When in series
Ceq′=C1C2C1+C2{C}_{eq}^{'}=\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}
C1+C2=10(C1C2C1+C2){C}_{1}+{C}_{2}=10\left(\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}\right)
(C1+C2)2=10C1C2{\left({C}_{1}+{C}_{2}\right)}^{2}=10{C}_{1}{C}_{2}
C12+C22−8C1C2=0{C}_{1}^{2}+{C}_{2}^{2}-8{C}_{1}{C}_{2}=0
dividing by C12{C}_{1}^{2}
(C2C1)2−8C2C1+1=0{\left(\dfrac{{C}_{2}}{{C}_{1}}\right)}^{2}-\dfrac{8{C}_{2}}{{C}_{1}}+1=0
Let C2C1=x\dfrac{{C}_{2}}{{C}_{1}}=x
x2−8x+1=0{x}^{2}-8x+1=0
x=8±64−42=4+15x=\dfrac{8\pm \sqrt{64-4}}{2}=4+\sqrt{15}
This question was Bonus in JEE Main 2021 exam because of incorrect data in original question.

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