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An RC circuit consists of a resistor with resistance 1.0 kΩ, a 120-V battery, and two capacitors, C1 and C2, with capacitances of 20.0 μF and 60.0 μF, respectively. Initially, the capacitors are uncharged; and the switch is closed at t = 0 s. How much charge will be stored in eah capacitor after a long time has elapsed (capacitor one 2.4 x 10^-3 C) (capacitor 2 7.2 x 10^-3 C) and what will the total charge on both capacitors two time constants after the switch is closed? (8.3 x 10^-3)

Respuesta :

okpalawalter8

Answer:

[tex]Q_t= 8.3 * 10^3 C[/tex]

Explanation:

From the question we are told that:

Resistor [tex]R=1000ohms[/tex]

Voltage [tex]v=120_V[/tex]

Capacitance of c_1 [tex]c_1=20 \mu F[/tex]

Capacitance of c_2 [tex]c_2=60 \mu F[/tex]

Time [tex]t=0[/tex]

Generally the equation for charges is mathematically given by

[tex]For C_1\\Charge\ on\ C_1 = CV = 20*120 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2.4 x 10^-3 C\\[/tex]

[tex]ForC_2\\Charge on C_2 = 60*120 =7200 μC = 7.2 x 10^-3\\Charge on C_2 = 7.2 x 10^-3[/tex]

Generally the equation for voltage across capacitors is mathematically given by

[tex]V_c(t)=V(1-e^{-t/RC})[/tex]

[tex]C=C_1+C_2=80 \mu f\\t=2RC=>160000s[/tex]

[tex]V_c(t)=120(1-e^{-(160000)/1000*(80)})[/tex]

[tex]V_c(t)=103.7598[/tex]

Generally the equation for charges is mathematically given by

[tex]Q1(t) = C1Vc(t)\\Q1(t) = 20*103.7598\\Q1(t) = 2075.196\\\\Q2(t) = 60*103.7598\\Q2(t) = 6225.6\\[/tex]

Generally the equation for total charges [tex]Q_t[/tex] is mathematically given by

[tex]Q_t=Q1(t)+Q2(t)[/tex]

[tex]Q_t= 8.3 * 10^3 C[/tex]

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