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Posted

The battery voltage is constant

the instant the switch is closed the voltage on the capacitor is 0v

thus, the voltage drop across the resistor is equal to the battery voltage

since the cap blocks DC, one plate begins to charge.

When the charge reaches the battery voltage

the voltage drop across the resistor is 0v

Posted

I agree with (e) too ^_^

 

Here are some mathematical aids. See for yourself which one is true ;)

Using the formal definitions introduced in electric circuits:

 

(a)

i = C dV/dt

Cdv = int (i dt)

 

CV = Q

where C = capacitance, V = voltage across capacitor = Vcapacitor, Q = charge on capacitor. Note that Vcapacitor is always less than Vbattery, and so Q/C is always less than Vbattery.

 

i = V/R (Ohm's Law)

i = (Vbattery - Vcapacitor)/R

i = (Vbattery - Q/C)/R

 

(b)

Voltage drop across resistor:

delta V = Vbattery - Vcapacitator

delta V = Vbattery - Q/C

 

©

i = (Vbattery - Q/C)/R

 

(d)

Q = charge on capacitor = charge across capacitor = Qcapacitor

 

(e)

delta V = Vbattery - Q/C

Posted

wait a second there isnt more than one of those options that are correct are there? ie. e is the ONLY correct answer...?? right? cause i think it is...

Posted

Yes ^^ only e is true.

 

Follow the arguments. See if the changing the variables in the equations for a, b, c, and d will make the assumptions in the questions hold true.

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