Two resistors, R1 = 8.40Ohm and R2 = 4.20Ohm and two initially uncharged capacit

Question

Two resistors, R1 = 8.40Ohm and R2 = 4.20Ohm and two initially uncharged capacitors, C1 = 0.560uF and C2 = 0.280uF are arranged as shown in the figure below. The switch S is open. The battery is connected across the combination with a potential difference of 21.6V.

A.After waiting for a long time, what is the potential at point a? (Let V = 0 at the negative terminal of the source.)

B.What is the potential at pointbwith the switch left open?

C.The switch is now closed. After waiting for a long time, what is the final potential of pointb?

Two resistors, R1 = 8.40Ohm and R2 = 4.20Ohm and two initially uncharged capacitors, C1 = 0.560uF and C2 = 0.280uF are arranged as shown in the figure below. The switch S is open. The battery is connected across the combination with a potential difference of 21.6V. A.After waiting for a long time, what is the potential at point a? (Let V = 0 at the negative terminal of the source.) B.What is the potential at pointbwith the switch left open? C.The switch is now closed. After waiting for a long time, what is the final potential of pointb?

Explanation / Answer

a) Va=(R2/(R1+R2))*V

b) (V*C1)/(C1+C2)=Vb...for these ones dont enter it as .56E-6 and .28E-6 enter them as .56 and .28

c) It's the same answer as 9 because they have been charging for a long time and when they reach their max capacitance it is the same as point a

Hope this helps.

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