Answer :
Answer:
The voltage across the capacitor after 8.00 ms is 35.375 V.
Explanation:
Given that,
Capacitor = 11.0 μF
Voltage = 100 V
Resistance = 470 Ω
Time = 8.00 ms
We need to calculate the current
Using formula of current
[tex]I=\dfrac{q}{t}[/tex]
[tex]I=\dfrac{CV}{t}[/tex]
Put the value into the formula
[tex]I=\dfrac{11.0\times10^{-6}\times100}{8.00\times10^{-3}}[/tex]
[tex]I=0.1375\ A[/tex]
We need to calculate the voltage
Using ohm's law
[tex]V=IR[/tex]
[tex]V=0.1375\times470[/tex]
[tex]V=64.625\ V[/tex]
We need to calculate the voltage across the capacitor after 8.00 ms
Using formula of voltage
[tex]V'=100-V[/tex]
[tex]V'=100-64.625[/tex]
[tex]V'=35.375\ V[/tex]
Hence, The voltage across the capacitor after 8.00 ms is 35.375 V.