Question #a8e29

1 Answer
ali ergin
May 10, 2016

#"charge of 2"mu F rarr 12*10^-6C#

#"charge of 4"mu F rarr 24*10^-6C#

#"charge of 6"mu F rarr 36*10^-6C#

#"charge of circuit" rarr 36*10^-6C#

Explanation:

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#"first; let's find the total capacitances in circuit. "#

#"for capacitors "4 mu F and 2 mu F," total="4+2=6 mu F"(in parallel)"#

#"for capacitors "6 mu F " and " 6 mu F" " 1/("total")=1/(6)+1/6#

#"total capacitance="6/2=3 mu F#

#C=3*10^-6F" (total capacitance for circuit")#

#C=Q/V#

#Q=C*V#

#Q=3*10^-6*12#

#Q=36*10^-6C" (total charge for circuit)"#

#x+2x=36*10^-6#

#3x=36*10^-6#

#x=12.10^-6C#

#"charge of 2"mu F rarr 12*10^-6C#

#"charge of 4"mu F rarr 24*10^-6C#

#"charge of 6"mu F rarr 36*10^-6C#

#"charge of circuit" rarr 36*10^-6C#

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