Could someone explain this? I understand that the capacitors act as wires at t=0

Question

Could someone explain this? I understand that the capacitors act as wires at t=0, and I know that for the equivalent resistance (after a long time) the resistances will add as if they are in series since current no longer travels through the capacitors. But why was I instructed to find the voltage across R2 and R3 (for parts b & c)

like so

after a long time

b) Req=19.7ohm

R2=8.4ohm

Voltage of battery = 11.5V

to find R to use in equation below 11.5(8.4ohm/19.7ohm)= 4.9V <---(this is where I'm confused)

Q=RC

9microF • 4.9V= 44.132 microC

I dont understand at all why we multiplied the voltage of the battery times (R2/Req) and then took that value times the capacitance, what equation does this relate to? thanks

In the circuit below, the switch was open for a long time and then closed at t 0 s. The values of the emf, resistons, and capacitors are E 11.5 V, R1 9.2 2, R2 8.4 2, R3 2.1 2, CA 9.0 HF, C 1.1 HF. 1 -El H (a) Immediately after the switch is closed, what is the current through resistor R 1.25 A long time after the switch was closed, what are the charges stored on the two capacitors? (b) on CA: 44.132 (c) on C 1.375 B: v HC

Explanation / Answer

When resistors are in series total voltage is divided

For two resistors R1 and R2 in series

Voltage across any resistor can be obtained by

Vi = Ri*V/(R1+R2)

Which is necessary to obtain charge across capacitors. Qi = CiVi

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