R1 C2 The RC circuit shown above consists of a battery (emf and internal resista

Question

R1 C2 The RC circuit shown above consists of a battery (emf and internal resistance r), two resistors, and two capacitors, which are initially uncharged. Suppose that at time t-0 the switch S is closed, connecting the circuit. (a) what is the time constant of this circuit? (That is, how long until the capacitors are ~63% charged) (hint: consider the total equivalent resistance and capacitance). (b) Find the time it takes to charge the capacitors to 90% of their capacity. (c) What is the current through the battery at t-0? (just after S is closed). (d) What is the current through the battery a long time after the switch is closed?

Explanation / Answer

Given circuit

RC circuit

the resistors are paralle so the resistance is  

R = R1*R2/(R1+R2)

and capacitors are in parallel combination so

the capacitance is C= C1+C2

a) the time constant of the RC circuit is  

T = R*C = (R1*R2)/(R1+R2)*(C1+C2 )

b) charging of a capacitor in RC circuit is  

Q(t) = Q0(1-e^(-t/T))

0.9*Q0 = Q0(1-e^(-t/T))

0.9 = (1-e^(-t/T))

0.1 = e^(-t/T)

ln(0.1) = (-t/T)

t = T(2.30)

that is 2.3 times of the time constant

c) current throught the battery is maximum when t=0 s  

which is equivalent to the I = e/R

i = e/(R+r)

d) long time after the switch is closed is zero because the charges deposition stops when the potential difference between the plat

s is equal to the potential defference between the terminals of the battery

so that the charges may not be in motion so the current is zero

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