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प्रश्न
Two capacitors of capacitance 20⋅0 pF and 50⋅0 pF are connected in series with a 6⋅00 V battery. Find (a) the potential difference across each capacitor and (b) the energy stored in each capacitor.
उत्तर
Given :
`C_1 = 20.0 "pF"`
`C_2 = 50.0 "pF"`
When the capacitors are connected in series, their equivalent capacitance is given by `C_(eq) = (C_1C_2)/(C_1+C_2)`
∴ Equivalent capacitance,`C_(eq) = ((50 xx 10^-12) xx (20 xx 10^-12))/((50 xx 10^-12)+(20 xx 10^-12))` = `1.428 xx 10^-11 "F"`
(a) The charge on both capacitors is equal as they are connected in series. It is given by
`q = C_(eq) xx V`
⇒ `q = (1.428 xx 10^-11) xx 6.0 "C"`
Now ,
`V_1 = q/C_1 = ((1.428 xx 10^-11) xx 6.0 "C")/((20 xx 10^-12))`
⇒ `V_1 = 4.29 "V"`
and
`V_2 = (6.00 - 4.29) V = 1.71 "V"`
(b) The energies in the capacitors are given by
`E_1 = q^2/(2C_1)`
= `[(1.428 xx 10^-11) xx 6.0]^2` `xx 1/(2 xx 20 xx 10^-12)`
= `184 "pJ"`
and
`E_2 = q^2/(2C_1)`
= `[(1.428 xx 10^-11) xx 6.0]^2` `xx 1/(2 xx 50 xx 10^-12)`
= `73.5 "pJ"`
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