Explanation:
Given that,
The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.
(a) Capacitance of capacitor 1,
[tex]C_1=\dfrac{Q}{V_1}\\\\C_1=\dfrac{6\times 10^{-8}}{40}\\\\C_1=1.5\times 10^{-9}\ F\\\\C_1=1.5\ nF[/tex]
Capacitance of capacitor 2,
[tex]C_2=\dfrac{Q}{V_2}\\\\C_2=\dfrac{6\times 10^{-8}}{50}\\\\C_2=1.2\times 10^{-9}\ F\\\\C_2=1.2\ nF[/tex]
Capacitance of capacitor 3,
[tex]C_3=\dfrac{Q}{V_3}\\\\C_3=\dfrac{6\times 10^{-8}}{60}\\\\C_3=1\times 10^{-9}\ F\\\\C_3=1\ nF[/tex]
(b) The equivalent capacitance in series combination is :
[tex]\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\\\\\dfrac{1}{C}=\dfrac{1}{1.5}+\dfrac{1}{1.2}+\dfrac{1}{1}\\\\C=0.4\ nF[/tex]
Hence, this is the required solution.
