question_answer

Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The [Capacitance and breakdown voltage of the combination will be

A) \[12\Omega {{m}^{-1}}\]             

B)        \[0.6\pi \,\Omega \]      

C)                        \[\frac{C}{3},\,3V\]        

D)        \[6\pi \,\Omega \]

Correct Answer: C

Solution :

In series arrangement charge on each plate of each capacitor has same magnitude. The potential difference is distributed inversely in the ratio of capacitors ie, \[y=(0.02)m\sin (15.7x-2010t)\]Here, V = 3V The equivalent capacitance\[y=(0.02)m\sin (15.7x+2010t)\]is given by \[\frac{1}{{{C}_{s}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}+...\] \[x=\frac{a}{2}\]