here are a quick to questions, how'd i do?

my answer:

 

e

 

-Sarah

Explain why.

this one i am not so sure about...or not confident enough to try and explain it

i did for the other questions though

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

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

 

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i = (Vbattery - Q/C)/R

 

(d)

Q = charge on capacitor = charge across capacitor = Qcapacitor

 

(e)

delta V = Vbattery - Q/C

any hints? pwease

any hints? pwease

 

 

Douglas explained it.

ok i think i understand it now, thanks guys

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...

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.

yeah i thought so, just checking thankss mezarashi