- #1
Dell
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- 0
3 capacitators, C1=8μF C2=4μF C3=8μF, are connected in a colum, over a potential difference of 12V, what is the charge of C2??
i know that connecting the capacitators in a colum, ΣC=(Ca*Cb)/(Ca+Cb)
also i know that
C=Q/V, which i think must be the equation i need since i am looking for Q, but i don't think the V i know is the right V, i think its the V for the whole system.
ΣC1,2=(C1*C2)/(C1+C2)=(8*4)/(8+4)=(8/3)μF
ΣC1,2,3=(C3*C1,2)/(C3+C1,2)=(8*(8/3))/(8+(8/3))=2μF
so now i know Vtotal=12V and Ctotal=2μF
C=Q/V
Q=C*V
Qtotal=2*12=24μC
but how do i find Q2 from the Qtotal
i know that connecting the capacitators in a colum, ΣC=(Ca*Cb)/(Ca+Cb)
also i know that
C=Q/V, which i think must be the equation i need since i am looking for Q, but i don't think the V i know is the right V, i think its the V for the whole system.
ΣC1,2=(C1*C2)/(C1+C2)=(8*4)/(8+4)=(8/3)μF
ΣC1,2,3=(C3*C1,2)/(C3+C1,2)=(8*(8/3))/(8+(8/3))=2μF
so now i know Vtotal=12V and Ctotal=2μF
C=Q/V
Q=C*V
Qtotal=2*12=24μC
but how do i find Q2 from the Qtotal