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A heart pacemaker fires exactly 69 times a minute, each time a 29.0 nF capacitor is charged (by a battery in series with a resistor) to 0.732 of its full voltage. What is the value of the resistance?

User Nullglob
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Final answer:

To determine the resistance in a heart pacemaker where a 29.0 nF capacitor charges to 0.732 of its full voltage 69 times a minute, the RC time constant can be used along with the formula for capacitor charging.

Step-by-step explanation:

The original question asks what the resistance is in a heart pacemaker circuit where a 29.0 nF capacitor charges to 0.732 of its full voltage 69 times a minute. Since a heart pacemaker fires 69 times a minute, we know that the charging occurs within less than a second per cycle. To find the resistance, we can use the RC time constant τ formula, τ = R * C, where R is resistance, C is capacitance, and τ (tau) is the time constant. The time constant is the time it takes for the capacitor to charge to about 63.2% (−0.632) of its full voltage. Because the capacitor reaches 0.732 of its charge, we must adjust τ to account for this change.

First, we will find τ by using the charging cycle and the fact that 69 cycles occur in a minute, which means one cycle is ≈ 0.8696 seconds. Using the formula V(t) = V0(1 - e^(-t/τ)), where V(t) is the voltage at time t and V0 is the maximum voltage, we solve for τ when V(t)/V0 = 0.732. With the known capacitive value and time per cycle, we can then calculate the resistance in the circuit.

User Paul D Smith
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