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Answer 1 · 2016-05-10T06:54:15

$charge of 2 μ F \to 12 \cdot 10^{- 6} C$ $"charge of 2"mu F rarr 12*10^-6C$

$charge of 4 μ F \to 24 \cdot 10^{- 6} C$ $"charge of 4"mu F rarr 24*10^-6C$

$charge of 6 μ F \to 36 \cdot 10^{- 6} C$ $"charge of 6"mu F rarr 36*10^-6C$

$charge of circuit \to 36 \cdot 10^{- 6} C$ $"charge of circuit" rarr 36*10^-6C$

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

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

$for capacitors 6 μ F and 6 μ F \frac{1}{total} = \frac{1}{6} + \frac{1}{6}$ $"for capacitors "6 mu F " and " 6 mu F" " 1/("total")=1/(6)+1/6$

$total capacitance= \frac{6}{2} = 3 μ F$ $"total capacitance="6/2=3 mu F$

$C = 3 \cdot 10^{- 6} F (total capacitance for circuit)$ $C=3*10^-6F" (total capacitance for circuit")$

$C = \frac{Q}{V}$ $C=Q/V$

$Q = C \cdot V$ $Q=C*V$

$Q = 3 \cdot 10^{- 6} \cdot 12$ $Q=3*10^-6*12$

$Q = 36 \cdot 10^{- 6} C (total charge for circuit)$ $Q=36*10^-6C" (total charge for circuit)"$

$x + 2 x = 36 \cdot 10^{- 6}$ $x+2x=36*10^-6$

$3 x = 36 \cdot 10^{- 6}$ $3x=36*10^-6$

$x = {12.10}^{- 6} C$ $x=12.10^-6C$

$charge of 2 μ F \to 12 \cdot 10^{- 6} C$ $"charge of 2"mu F rarr 12*10^-6C$

$charge of 4 μ F \to 24 \cdot 10^{- 6} C$ $"charge of 4"mu F rarr 24*10^-6C$

$charge of 6 μ F \to 36 \cdot 10^{- 6} C$ $"charge of 6"mu F rarr 36*10^-6C$

$charge of circuit \to 36 \cdot 10^{- 6} C$ $"charge of circuit" rarr 36*10^-6C$