The effective capacitance of two capacitors of capacitances C 1 and C 2 C 2 > C 1 connected in parallel is 25 6 times the effective capacitance when they are connected in series. The ratio C 2 C 1 is

C p a r a l l e l = 25 6 C s e r i e s ⇒ C 1 + C 2 = 25 6 C 1 C 2 C 1 + C 2 ⇒ 6 C 1 + C 2 2 = 25 C 1 C 2 ⇒ 6 C 1 2 + 6 C 2 2 + 12 C 1 C 2 = 25 C 1 C 2 ⇒ 6 C 1 2 C 2 2 + 6 + 12 C 1 C 2 = 25 C 1 C 2 ⇒ 6 C 1 2 C 2 2 - 13 C 1 C 2 + 6 = 0 This is quadratic equation By solving we get C 2 C 1 = 3 2

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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

The effective capacitance of two capacitors of capacitances $C_{1}$ and $C_{2} (C_{2} > C_{1})$ connected in parallel is $\frac{25}{6}$ times the effective capacitance when they are connected in series. The ratio $\frac{C_{2}}{C_{1}}$ is

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Answer:

Solution:

$C_{p a r a l l e l} = \frac{25}{6} C_{s e r i e s}$

$\Rightarrow C_{1} + C_{2} = \frac{25}{6} \frac{C_{1} C_{2}}{C_{1} + C_{2}}$

$\Rightarrow 6 {(C_{1} + C_{2})}^{2} = 25 C_{1} C_{2}$

$\Rightarrow 6 {C_{1}}^{2} + 6 {C_{2}}^{2} + 12 C_{1} C_{2} = 25 C_{1} C_{2}$

$\Rightarrow 6 \frac{{C_{1}}^{2}}{{C_{2}}^{2}} + 6 + 12 \frac{C_{1}}{C_{2}} = 25 \frac{C_{1}}{C_{2}}$

$\Rightarrow 6 \frac{{C_{1}}^{2}}{{C_{2}}^{2}} - 13 \frac{C_{1}}{C_{2}} + 6 = 0$

This is quadratic equation

By solving we get $\frac{C_{2}}{C_{1}} = \frac{3}{2}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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