Capacitance past IIT problem

6 Answers

1
madhumitha harishankar ·

1
madhumitha harishankar ·

The solution given was faulty..anyone wanna try it?

23
qwerty ·

is the ans

\frac{1}{k}= \frac{1}{k_{1}+k_{2}}+\frac{1}{2k_{3}} ??

1
madhumitha harishankar ·

yes! tats the right answer..can u explain how u went about it?

23
qwerty ·

C = \frac{\varepsilon _{o}S}{d}

where S = area

and C_{1}=C_{2}=\frac{\varepsilon _{o}\frac{S}{2}}{\frac{d}{2}}=C

and C_{3}=\frac{\varepsilon _{o}S}{\frac{d}{2}}=2C

now dilelctrics k1 and k2 are in parallel and k3 is in series with the combination of k1 and k2

\Rightarrow k_{net}C_{net} = \frac{(k_{1}C_{1}+k_{2}C_{2})(k_{3}C_{3}) }{k_{1}C_{1}+k_{2}C_{2}+(k_{3}C_{3} }

\Rightarrow k_{net}C = \frac{(k_{1}C+k_{2}C)(k_{3}(2C)) }{k_{1}C+k_{2}C+k_{3}(2C)}

\Rightarrow k_{net} = \frac{(k_{1}+k_{2})(2k_{3}) }{k_{1}+k_{2}+k_{3}}

\Rightarrow \frac{1}{k_{net}} = \frac{k_{1}+k_{2}+2k_{3}} {(k_{1}+k_{2})(2k_{3}) }

\Rightarrow \frac{1}{k_{net}} = \frac{1}{k_{1}+k_{2}}+\frac{1}{2k_{3}}

1
madhumitha harishankar ·

gotcha
thankyou!

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