nvm that just try this, we know that capacitance is series is added like : 1\ C = 1\C1 + 1\C2, right? so now use that value we got in b) (iii) and after calculating you find the capacitance is decreased to half. Note: don't get confused between VOLTAGE in series and CAPACITANCE in series, they are different, I hope this makes sense lol.
@@akidanis6984 i get that the capacitance halves and i get why it does but i don't get the current part. i am saying that if the capacitance halves then why does the current half shouldn't it be voltage that halves not current
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doesnt voltage divide in series and charge stay the same then how does 1/2c=(1/2Q)/v , shouldnt it be the other way around?
nvm that just try this, we know that capacitance is series is added like : 1\ C = 1\C1 + 1\C2, right? so now use that value we got in b) (iii) and after calculating you find the capacitance is decreased to half.
Note: don't get confused between VOLTAGE in series and CAPACITANCE in series, they are different, I hope this makes sense lol.
@@akidanis6984 i get that the capacitance halves and i get why it does but i don't get the current part. i am saying that if the capacitance halves then why does the current half shouldn't it be voltage that halves not current
Q = C/v
But q = it so
I = c/vt
I would be directly proportional to c so if capacitance decreases by 2, current decreases by 2
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