Respuesta :
The charge on the capacitor after 2 s is 0.43 C.
Explanation:
The formula for finding the voltage while charging a capacitor is
[tex]V = V_{0}(1-e^{\frac{-t}{RC} })[/tex]
Here , V₀ is the initial potential before charging and t is the time at which we have to determine the voltage, R is the resistance and C is the capacitance for the given circuit.
The given problem have given us the values for V₀ = 50 V and maximum current I is given as 500 mA.
Then, resistance can be determined using Ohm's law: [tex]R = \frac{V}{I} =\frac{50}{500*10^{-3} } = 100 Ohm[/tex]
The capacitance is defined as the ratio of charge to the unit voltage.
[tex]C = \frac{Q}{V} = \frac{IT}{IR} = \frac{T}{R}[/tex]
Here T is the time constant which is given as 1 s and R is found to be 100 ohm, then capacitance will be [tex]\frac{1}{100} = 10 mF[/tex]
So, the values for parameters like V₀ = 50 V, R = 100 Ω, C = 10 mF and t = 2 s.
Then, [tex]V =50*(1-e^{\frac{-2}{100*10*10^{-3} } }) =50*(1-e^{-2}) = 50*(1-0.1353)[/tex]
V= 43 V.
Then, [tex]Q = CV = 10*10^{-3} * 43 V = 0.43 C[/tex]
Thus, the charge on the capacitor after 2 s is 0.43 C.
The unit of charge is Coulomb.
The charge on the capacitor after 2 second is 0.43 Coulomb.
To finding the voltage while charging a capacitor is given as,
[tex]V=V_{0}(1-e^{-\frac{t}{RC} } )[/tex]
Here , V₀ is the initial potential, R is the resistance and C is the capacitance
It is given that V₀ = 50 V and maximum current I is given as 500 mA.
So, [tex]R=\frac{V}{I}=\frac{50}{500*10^{-3} } =100ohm[/tex]
Time constant = RC = 1
So, [tex]C=\frac{1}{R} =\frac{1}{100}F[/tex]
Now, we have V₀ = 50 V, R = 100 Ω, C = 1/100 F and t = 2 s.
[tex]V=50(1-e^{-\frac{2}{100*\frac{1}{100} } } )=50(1-e^{-2} )=50(1-0.1353)=43V[/tex]
We know that, [tex]Q=CV=\frac{1}{100}*43=0.43C[/tex]
Thus, the charge on the capacitor after 2 s is 0.43 C.
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