Question

Derive an expression for capacitance of a parallel plate capacitor. Three capacitors C1, C2,C3 are

connected i) in series ii) in parallel show that the energy stored in series and parallel combination is

same​

Answer 1

Answer:

അഖ്ഡസ്‌ഫഹ്യ്ക്കൂപ്ലമനഹ്വ്ഫ്സ്സ്സാഖൈഖൈട്ഫിജോല്ലോംകബ്ജെസിഡ്‌സ്സാസ്

Answer 2

Given: Three Capacitors that are $C_{1} , C_{2} , C_{3}$

Find: (i) For series connection expression of capacitance,

(ii) For parallel connection expression of capacitance,

Solution:

(i) For Series Connection,

The electric current is constant, but the voltage is varied and adds to the total voltage,

In capacitor case,

$V = V_{1} + V_{2} + V_{3}$ --- (1)

Now, we know the value of voltage in terms of charge and capacitance.

$V = \dfrac{Q}{C}$

So, we put the value of the in equation (1),

$\dfrac{Q}{C} = \dfrac{Q}{C_{1} } + \dfrac{Q}{C_{2} } + \dfrac{Q}{C_{3} }$

Now, we take common,

$\dfrac{Q}{C} = \dfrac{1}{C_{1} } + \dfrac{1}{C_{2} } + \dfrac{1}{C_{3} }\\\\\\\dfrac{1}{C} = \dfrac{1}{C_{1} } + \dfrac{1}{C_{2} } + \dfrac{1}{C_{3} }$

So, it is the expression of the series combination of the capacitance.

(ii) For Parallel Connection

The voltage is constant, but the electric charge is varied and adds to the total electric charge,

In capacitor case

$Q = Q_{1} + Q_{2} + Q_{3}$    --- (2)

Now, we know the value of electric current in terms of voltage and capacitance.

$Q = CV$

So, we put the value of the in equation (2),

$CV = C_{1} V + C_{2}V+C_{3}V$

Now, we take common

$CV = V(C_{1} +C_{2} +C_{3} )\\\\C = C_{1} +C_{2} +C_{3}$

So, it is the expression of the parallel combination of the capacitance.