In the circuit shown, E = 4V, R₁ =43, R₂ = 72 , and C = 14 µF. I₁* Initially, the capacitor is uncharged and the switch is open. At t=0, the switch is closed. The positive directions of the currents are represented by the arrows in the diagram. (a) What is the current running out of the battery at t = 0s? TE R₁ 13 +12 {R₂ (b) After the switch has been closed for a long time, calculate 11, 12, 13, and Qtot (the total charge on the capacitor). In your answer, please use the naming convention for the currents shown in the diagram.
In the circuit shown, E = 4V, R₁ =43, R₂ = 72 , and C = 14 µF. I₁* Initially, the capacitor is uncharged and the switch is open. At t=0, the switch is closed. The positive directions of the currents are represented by the arrows in the diagram. (a) What is the current running out of the battery at t = 0s? TE R₁ 13 +12 {R₂ (b) After the switch has been closed for a long time, calculate 11, 12, 13, and Qtot (the total charge on the capacitor). In your answer, please use the naming convention for the currents shown in the diagram.
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Transcribed Image Text:In the circuit shown, ε = 4V, R₁ =43 2, R2 = 72 , and C= 14 uF.
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Initially, the capacitor is uncharged and the switch
is open. At t=0, the switch is closed. The positive
directions of the currents are represented by the
arrows in the diagram.
(a) What is the current running out of the
battery at t= 0s?
1₁*
TE
R₁
I3*
+ 12
R₂
(b) After the switch has been closed for a long time, calculate 11, 12, 13, and Qtot
(the total charge on the capacitor). In your answer, please use the naming
convention for the currents shown in the diagram.
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