Consider the circuit shown in the diagram below. Before the switch is closed, both capacitors are uncharged.
Consider the circuit shown in the diagram below. Before the switch is closed, both capacitors are uncharged.
Immediately after the switch is closed, what is the amount of current supplied by the battery?
Assuming the switch remains closed for a long time, which capacitor will be the first to reach 95% of its final charge level?
What is the time constant for charging this capacitor?
Homework Answers
A) i = 9/90 + 9/30 + 9/60 = 0.55A Answer
B) tow of 2micro F capacitor (t1) = 90*2*10^-6 = 180*10^-6
tow of 4.5 micro farad(t2) = 30*4.5*10^-6 = 135*10^-6
=> So 4.5 micro F will charge first.
B) time constant = 135*10^6 sec Answer
A) Immediately after the switch is closed, the capacitors still have a voltage of 0 since they have not had time to charge up. Thus, you can neglect them and calculate the current as usual by summing the resistors and dividing the voltage by it. Req = 1/(1/R1 + 1/R2 + 1/R3) = 16.36 ohm V = 9 V V = IR ----> I = V/R = 9/16.36 = 0.55 A
B) Once the switch is closed for a while, capacitor 2 receives the most direct path of current thus will reach 95 % of it its final charge the fastest.
C) the formula for the time constant is t = RC. ----> t = (R2)(C2) = (30)(4.5E-6) = 135 microFarads
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> FURTHER EXPLANATION:
A) Immediately after the switch is closed, the capacitors still have a voltage of 0 since they have not had time to charge up. Thus, you can neglect them and calculate the current as usual by summing the resistors and dividing the voltage by it.
Req = 1/(1/R1 + 1/R2 + 1/R3) = 16.36 ohm
V = 9 V
V = IR ----> I = V/R = 9/16.36 = 0.55 A
B) Once the switch is closed for a while, capacitor 2 receives the most direct path of current thus will reach 95 % of it its final charge the fastest.
C) the formula for the time constant is t = RC. ----> t = (R2)(C2) = (30)(4.5E-6) = 135 microFarads
ISuckAtPhysics Tue, Feb 22, 2022 6:07 PM