# Answer in Electric Circuits for Bawe #159159

Explanations & Calculations

1)

2)

• The time constant is given by “small tau=RC scriptsize(s)”: where R is the circuit resistance through which the capacitor either charged or discharged over time.
• Since there are 3 capacitors in this circuit, the equivalent capacitance should be calculated to represent this entire circuit, then the time constant becomes “small tau=RC_{equivalent}”

“qquadqquadnbegin{aligned}nsmall C_{equivalent} &= small frac{5mu Ftimes(2mu F+3mu F)}{5mu F+(2mu F+3mu F)}\n&= small frac{5mu F}{2}\n&= small bold{2.5mu F}nend{aligned}”

• For the discharging path of this circuit though “small S_2,,and,,10Omega”, resistance is “small 10Omega”. Then,

“qquadqquadnbegin{aligned}nsmall tau&= small 10Omega times2.5times10^{-6}F\n&= small 2.5times10^{-5}snend{aligned}”

• The time constant gives an idea of how much time it takes the circuit charge/discharge the capacitor to some amount. Readily it provides information about the time taken (at “small t= RC” ) for the capacitor to charge to 63.2% of the ultimate charge & time taken to discharge to 36.8% from the initially stored charge.
• This gives some idea of the response of the circuit over a change (transient response)

• Half-life is defined as the time taken for a quantity to become half of the initial. By the discharge equation “small Q=Q_0 e^{-frac{t}{RC}}” of this circuit, the time that is taken for the charge to become half can be calculated.

“qquadqquadnbegin{aligned}nsmall Q&tofrac{Q_0}{2}\nsmall frac{Q_0}{2}&= small Q_0e^{-frac{t}{RC}}\nsmall e^{frac{t}{RC}}&= small 2\nsmall t_{frac{1}{2}}&= small RC.ln2\n&=small 10Omegatimes2.5times10^{-6}Ftimes ln2\n&= small bold{1.733times10^{-5}s}nend{aligned}”

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