question_answer

Two capacitors of 10\[H=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi ni}{r}\] and 20\[i=\frac{4\pi }{{{\mu }_{0}}}\frac{rH}{2\pi n}\] are connected in series with a 30 V battery. The charge on the capacitors will be respectively

A)  100\[=\frac{5\times {{10}^{-2}}\times 0.314\times {{10}^{-4}}}{{{10}^{-6}}\times 2\times 3.14\times 50}\], 100\[\lambda \]               

B)  200\[\overset{0}{\mathop{A}}\,\], 100\[I\]

C)         200\[I\], 200\[(\sqrt{3}+1)/1\]

D)         100\[\sqrt{3}/1\], 200\[(\sqrt{3}+1)/(\sqrt{3}-1)\]

Correct Answer: C

Solution :

The equivalent capacitance, for capacitors in series is \[6\times {{10}^{4}}\] \[6\times {{10}^{-3}}\]   \[6\times {{10}^{-6}}\] Also, \[\frac{Q}{q}\] This charge on the two capacitors in series is same. Hence, \[A=4i+4j-4k\]