The diagram shows four \( 6-\mu \mathrm{F} \) capacitors. The capacitance between points \( a \) and \( b \) is: A. \( 6 \mu \mathrm{F} \) B. \( 3 \mu \mathrm{F} \) C. \( 4 \mu \mathrm{F} \) D. \( 9 \mu \mathrm{F} \)
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Six capacitors each of capacitance of $2 \mu \mathrm{F}$ are connected as shown in the figure. The effective capacitance between $A$ and $B$ is (a) $12 \mu \mathrm{F}$ (b) $8 / 3 \mu \mathrm{F}$ (c) $6 \mu \mathrm{F}$ (d) $2 / 3 \mu \mathrm{F}$
In the acconpanying diagram, if $C_{1}=3 \mu \mathrm{F}, C_{2}=6 \mu \mathrm{F}, C_{3}$ $=9 \mu \mathrm{F}, C_{4}=12 \mu \mathrm{F}_{1} C_{5}=15 \mu \mathrm{F}$ and $C_{6}=18 \mu \mathrm{F}$, then the equivalent capacitance between the ends $A$ and $B$ is a. $1.22 \mu \mathrm{F}$ b. $5.16 \mu \mathrm{F}$ c. $2.25 \mu \mathrm{F}$ d. $2.51 \mu \mathrm{F}$
You have three $12 \mu$ F capacitors. Draw diagrams showing how you could arrange all three so that their equivalent capacitance is (a) $4.0 \mu \mathrm{F},$ (b) $8.0 \mu \mathrm{F},$ (c) $18 \mu \mathrm{F},$ and (d) $36 \mu \mathrm{F}$.
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