Respuesta :
Answer:
a). The resulting potential difference across each capacitor is 1.8 kV.
(b). The energy lost is zero.
Explanation:
Given that,
First capacitor = 160 pF
Second capacitor = 640 pF
Voltage = 1.8 kV
(a). We need to calculate the initial charge
Using formula of charge
[tex]q=CV[/tex]
Put the value into the formula
[tex]q=160\times10^{-12}\times1.8\times10^{3}[/tex]
[tex]q=2.88\times10^{-7}\ C[/tex]
[tex]q=288\ nC[/tex]
We need to calculate the final charge
Using formula of charge
[tex]q=CV[/tex]
[tex]q=640\times10^{-12}\times1.8\times10^{3}[/tex]
[tex]q=1152\ nC[/tex]
The total charge is
[tex]Q=288+1152[/tex]
[tex]Q=1440\ nC[/tex]
We need to calculate the resulting potential difference across each capacitor
Using formula of potential difference
[tex]V=\dfrac{Q}{C_{1}+C_{2}}}[/tex]
Put the value into the formula
[tex]V=\dfrac{1440\times10^{-9}}{(160+640)\times10^{-12}}[/tex]
[tex]V=1.8\ kV[/tex]
(b). We need to calculate the energy lost when the connections are made
Using formula of energy
[tex]E=\dfrac{1}{2}CV^2-\dfrac{1}{2}CV^2[/tex]
Put the value into the formula
[tex]E=\dfrac{1}{2}\times(160+640)\times10^{-12}\times1800^2-\dfrac{1}{2}\times(160+640)\times10^{-12}\times1800^2[/tex]
[tex]E=0[/tex]
Hence, (a). The resulting potential difference across each capacitor is 1.8 kV.
(b). The energy lost is zero.
