Question

After how many time constants has the voltage across a discharging capacitor decayed to 0.85{\rm \\%} of its initial value?

Answer 1

Answer:

After $t=4.77\tau$ voltage across the capacitor will be 0.85 % of the initial voltage across the capacitor

Explanation:

Let initially voltage across capacitor is $v_0$

After discharging the voltage across the capacitor is .85% of its initial voltage

So final voltage $v=0.0085v_0$

We know that voltage across capacitor is given by $v=v_0e^{\frac{-t}{\tau }}$

So $0.85v_0=v_0e^{\frac{-t}{\tau }}$

$e^{\frac{-t}{\tau }}=0.0085$

${\frac{-t}{\tau }}=ln0.0085$

$-t=-4.77\tau$

$t=4.77\tau$

So after $t=4.77\tau$ voltage across the capacitor will be 0.85 % of the initial voltage across the capacitor

Time constant of capacitive circuit is equal to the ratio of resistance and capacitance

Answer 2

The number of time constants the voltage across a discharging cap has decayed to 0.85% of its' initial value is;

4.768τ

In RC time constants, the formula for voltage across capacitor when discharging is given by;

v = v₀e^(-t/τ)

where;

v₀ = initial voltage on the capacitor

v = the voltage after time t

t = time in seconds

τ = time constant

We are told that the capacitor decayed to 0.85% of its' initial vale. Thus;

v = 0.0085v₀

Thus;

0.0085v₀ = v₀e^(-t/τ)

v₀ will cancel out to give;

0.0085 = e^(-t/τ)

In 0.0085 = -t/τ

-4.768 = -t/τ

Since we want to find number of time constants, then let us make t the subject to get;

t = 4.768τ

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