P 8.4-3 Cardiac pacemakers are used by people to maintain regular heart rhythm when they
have a damaged heart. The circuit of a pacemaker can be represented as shown in
Figure P 8.4-3. The resistance of the wires, R, can be neglected since R < 1 mΩ. The heart’s
load resistance, RL, is 1 kΩ. The first switch is activated at t = t0, and the second switch is
activated at t1 = t0 + 10 ms. This cycle is repeated every second. Find v(t) for t0 ≤ t ≤ 1. Note
that it is easiest to consider t0 = 0 for this calculation. The cycle repeats by switch 1 returning
to position a and switch 2 returning to its open position.
Hint: Use q = Cv to determine v(0–) for the 100-μF capacitor.
Figure P 8.4-3
Prior to time t0, whether it is 0 or not, if the circuit is in steady state, then it looks like
The capacitor is assumed to be fully charged and acting like an open circuit so there is no
current flowing through R. The voltage across the capacitor must equal the power source
or
𝒗𝑪𝟏 = 𝟑 𝑽
At t = t0 Switch 1 flips and we have the circuit
Now the charge on the 100 µF capacitor when it is charged to 3 V before it is connected to
the 400 µF capacitor can be found from
𝒒𝑪𝟏 = 𝑪𝑪𝟏 𝒗𝑪𝟏 = (𝟏𝟎𝟎 𝝁𝑭)(𝟑 𝑽) = 𝟑𝟎𝟎 𝝁𝑪
Now when the two capacitors are joined in parallel, their new combined capacitance is
𝑪𝒕𝒐𝒕 = 𝑪𝟏𝟎𝟎 + 𝑪𝟒𝟎𝟎 = 𝟏𝟎𝟎 𝝁𝑭 + 𝟒𝟎𝟎 𝝁𝑭 = 𝟓𝟎𝟎 𝝁𝑭
There is a fixed amount of charge (300 µC) so the voltage resulting from the charge
rearranging is found from
𝒗𝒕𝒐𝒕 =
𝒒𝒕𝒐𝒕 𝟑𝟎𝟎 𝝁𝑪
=
= 𝟎. 𝟔𝟎𝟎 𝑽
𝑪𝒕𝒐𝒕 𝟓𝟎𝟎 𝝁𝑭
At t = t0 + 10 ms, the circuit becomes
The time constant is
𝝉 = 𝑹𝑳 𝑪 = (𝟏𝟎𝟎𝟎 𝛀)(𝟓𝟎𝟎 𝝁𝑭) = 𝟓𝟎𝟎 𝒎𝒔 = 𝟎. 𝟓𝟎𝟎 𝒔
So when switch 2 is thrown we have an effective capacitor with a voltage of 0.60 V across it
that is discharging the charges on it through the load resistor with a time constant of 0.50 s.
So we know this will follow just standard capacitor discharge. The solution is
−𝟏 (𝒕−𝒕 )
𝟎
𝒗(𝒕) = 𝟎. 𝟔𝟎 𝑽 𝒆−𝟐 𝒔
𝒇𝒐𝒓 𝒕𝟎 ≤ 𝒕 ≤ 𝟏𝒔
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This page last updated on February 19, 2021