Question

A capacitor used to provide steady voltages in the power supply of a stereo amplifier charges rapidly to 35 V every 1/60 second. It must then hold that voltage to within 1.0 V for the next 1/60 s while it discharges through the amplifier circuit. If the circuit draws 1.2 A from the 35-V supply, what is its effective resistance (in ohms)? What value of capacitance is needed (in mF)?   

Answer 1

The discharging voltage for a capacitor is:

$V(t)=Vi\cdot e^{\frac{-t}{\tau }}$

where $\tau =RC$

This means that when the capacitor is discharging:

$1 V=35V\cdot e^{\frac{-1}{60\cdot \tau }}$

solving for tau:

$Ln(\frac{1}{35})=\frac{-1}{60\cdot \tau }$

$\tau =\frac{-1}{60\cdot Ln(\frac{1}{35})}=0.0046 \Omega \cdot F$

while the capacitor is charging:

$I(t)=\frac{Vf}{R}\cdot e^{\frac{-t}{\tau }}$

$1.2 A=\frac{35}{R}\cdot e^{\frac{-1/60}{\tau }}$

$R=\frac{35V}{1.2 A}\cdot e^{\frac{-1/60}{\tau }}=0.837 \Omega$

If:

$\tau =RC$

$C=\frac{\tau }{R}$

$C=\frac{0.0046 \Omega \cdot F }{0.837\Omega }=0.0054 F = 5495 mF$