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, R_L, 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

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

Now when the two capacitors are joined in parallel, their new combined capacitance is

There is a fixed amount of charge (300 µC) so the voltage resulting from the charge rearranging is found from

At t = t₀ + 10 ms, the circuit becomes

The time constant is

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

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This page last updated on July 10, 2019