In the cicuit shown in fig. C is a parallel plate air capacitor having plate of area `A = 50cm^2` each and distance d = 1mm apart. `R_1 R_2, and R_3` are resistors having resistance `3Omega, 2Omega, and 1Omega` respectively. Two identical sources each of emf V and of emf V and of negligible internal resistance are connected as shown in fig. if a dielectric strength of are is `E_0 = 3xx10^6Vm^(-10)`, calculate the maximum safe value of V.

Explanation

Due to sources, currents flow
through resistance `R_(1),R_(2)` and
`R_(3)` and capacitor gets charged.
Due to charge, an electic field
is established in the capacitor
whose magnitude cannot
exceed dielectric strength `E_(0)`
of air. Maximum safe value of
V corresponds to the maximum
possible charge on capacitor.
Let the Maximum
possibel charge on capacitor be `q_(0)` then the electic field inside
the capacitor is `E_(0) = q_(0)//A epsilon_(0),`
Then the electirc field inside
the capacitor is `E_(0) = q_(0)//A epsilon_(0),`
where `q_(0) A epsilon_(0)E_(0) = 15,000 epsilon_(0) and C = epsilon_(0) A//d = 5epilon_(0)` farad
Since, in steady state, no current flows through the capacitor
current through various parts of the circuit will be as sown in
fig. Now we analyze the cirucit in a steady state. First applying Kirchhoff's voltage law on mesh ABJA, we get
`I_(2)R_(1) + V +R_(2)(I_(1) - I_(2)) = 0 or 2I_(1) -5I_(2) = -V` (i)
For mesh AJDFGA,
`-R_(2)(I_(1) - I_(2))-R_(3)I_(1) +V = 0 or eI_(1) - 2I_(2) = V` (ii)
From Eqs. (i) and (ii): `I_(1) = (7V)/(11)` and `I_(2) = (%V)/(11)`
Now applying Kirchhoff's voltage law on mesh BDJB, we get
`(q)/(C) +I_(1) R_(3) - V = 0 or or q= (4CV)/(11) = (20)/(11) epsilon_(0)V`
But the maximum possible value of q is `q_(0) = 15,000 epsilon_(0).` Therefore, the maximum safe value of V is `11q_(0)//20epsilon_(0) = 8250V = 8.25kV.`