64. A heart pacemaker fires 72 times a minute, each time a 25.0-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

Respuesta :

Answer:

=33 .3×10^6Ω

=33.3M Ω

Explanation:

We were told to calculate the Resistance value,

Given the heart pacemaker fires as 72 times a minute, which is the time constant

Then we can convert the pacemaker fires of 72 times a minute to seconds for unit consistency.

1 minutes= 60secs

Then ,Time constant τ=60secs/72=0.8333 seconds

Time constant τ can be calculated using the formula below

τ= RC

Where R= resistance

C = capacitance

Then making RESISTANCE subject of formula we have

R=τ/C

But Capacitance=25.0-nF = 25×10^25F

Substitute the values we have

R=0.833/25×10^25

=33 .3×10^6 Ω

But can still be converted to M Ω= 33.3M Ω

Therefore, the resistance is 33 .3×10^6 Ω or 33.3M Ω

NOTE: 1M= 10^6

The value of the resistance will be "3.33×10⁷ Ω".

Resistance based problem:

According to the question,

Capacitor, C = 25.0 nF

60 sec - 72 fires

now,

Time for 1 fire,

[tex]t = \frac{60}{72}[/tex]

 [tex]= \frac{5}{6} \ sec[/tex]

Now,

⇒            [tex]V = V_0 (1-e^{-\frac{t}{RC} })[/tex]

     [tex]0.632 V_0=V_0(1-e^{\frac{-\frac{5}{6} }{R.25 n C} })[/tex]

[tex]e^{-\frac{1}{R\times 30\times 10^{-9}} } = 1-0.632 = 0.368[/tex]

                  [tex]= 2.72[/tex]

By taking "log" both sides,

⇒ [tex]ln \ e^{\frac{1}{R\times 30\times 10^{-9}} } = ln \ 2.72[/tex]

hence,

The Resistance be:

⇒ [tex]R = \frac{10^9}{30}[/tex]

       [tex]= 3.33\times 10^7 \ \Omega[/tex]

Thus the above approach is right.

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