Question

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)

Answer 1

Answer:

$Q_t= 8.3 * 10^3 C$

Explanation:

From the question we are told that:

Resistor $R=1000ohms$

Voltage $v=120_V$

Capacitance of c_1 $c_1=20 \mu F$

Capacitance of c_2 $c_2=60 \mu F$

Time $t=0$

Generally the equation for charges is mathematically given by

$For C_1\\Charge\ on\ C_1 = CV = 20*120 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2.4 x 10^-3 C$

$ForC_2\\Charge on C_2 = 60*120 =7200 μC = 7.2 x 10^-3\\Charge on C_2 = 7.2 x 10^-3$

Generally the equation for voltage across capacitors is mathematically given by

$V_c(t)=V(1-e^{-t/RC})$

$C=C_1+C_2=80 \mu f\\t=2RC=>160000s$

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

$V_c(t)=103.7598$

Generally the equation for charges is mathematically given by

$Q1(t) = C1Vc(t)\\Q1(t) = 20*103.7598\\Q1(t) = 2075.196\\\\Q2(t) = 60*103.7598\\Q2(t) = 6225.6$

Generally the equation for total charges $Q_t$ is mathematically given by

$Q_t=Q1(t)+Q2(t)$

$Q_t= 8.3 * 10^3 C$