A heart pacemaker fires 66 times a minute, each time a 27.0–nF capacitor is charged (by a battery in series with a resistor) to 0.632 of its full voltage. What is the value of the resistance?
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To find the value of the resistance, we can use the equation for the charging of a capacitor through a resistor:
V(t) = V0 * (1 - e^(-t/RC))
In this equation:
- V(t) is the voltage across the capacitor at time t
- V0 is the initial voltage across the capacitor
- R is the resistance
- C is the capacitance
We are given:
- The capacitance, C = 27.0 nF (nano-Farads)
- The time constant, τ = RC
- The voltage at steady state, V(t) = 0.632 * V0
- The firing rate of the pacemaker, which corresponds to the time t for each firing
The time constant τ can be calculated as follows:
τ = 1 / (firing rate) = 1 / (66 times/minute) = 1 / (66/60 seconds) = (60/66) seconds
Using the given voltage at steady state, we have:
V(t) = 0.632 * V0
0.632 * V0 = V0 * (1 - e^(-t/RC))
Since we want the voltage to reach 0.632 of its full voltage, we can set V(t) = 0.632 * V0 and solve for t:
0.632 * V0 = V0 * (1 - e^(-t/RC))
Dividing both sides by V0:
0.632 = 1 - e^(-t/RC)
Rearranging the equation:
e^(-t/RC) = 1 - 0.632
e^(-t/RC) = 0.368
Taking the natural logarithm of both sides:
-ln(0.368) = -t/RC
Simplifying:
t/RC = ln(1/0.368)
Solving for t/RC:
t/RC = ln(2.7174)
Now we can substitute the value of τ = (60/66) seconds:
t/(R * (60/66) * 27.0 * 10^-9) = ln(2.7174)
Simplifying:
t/(R * 0.9091 * (27.0 * 10^-9)) = ln(2.7174)
Calculating the natural logarithm:
t/(R * 0.9091 * (27.0 * 10^-9)) = 0.9977
Multiplying both sides by the denominator:
t = R * 0.9977 * 0.9091 * 27.0 * 10^-9
Rearranging to solve for R:
R = t / (0.9977 * 0.9091 * 27.0 * 10^-9)
Using the given firing rate, t = 60/66 = 0.9091 seconds:
R = (0.9091) / (0.9977 * 0.9091 * 27.0 * 10^-9)
Calculating the value of R:
R ≈ 1207.5 Ω
Therefore, the value of the resistance is approximately 1207.5 ohms.