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21.6 Dc circuits containing resistors and capacitors  (Page 4/10)

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We set this value for the crossing time t size 12{t} {} equal to τ size 12{τ} {} . Therefore,

C = t R = 1 . 6 × 10 − 4 s 10.0 Ω = 16 μF. size 12{C= { {t} over {R} } = { { left (1 "." 6´"10" rSup { size 8{-4} } right )} over {"10"} } ="16" μF} {}

(Note: Capacitance C size 12{C} {} is typically measured in farads, F , defined as Coulombs per volt. From the equation, we see that C size 12{C} {} can also be stated in units of seconds per ohm.)

Discussion

The flash interval of 160 μs size 12{"160" ms} {} (the traverse time of the bullet) is relatively easy to obtain today. Strobe lights have opened up new worlds from science to entertainment. The information from the picture of the apple and bullet was used in the Warren Commission Report on the assassination of President John F. Kennedy in 1963 to confirm that only one bullet was fired.

RC Circuits for timing

RC size 12{ ital "RC"} {} circuits are commonly used for timing purposes. A mundane example of this is found in the ubiquitous intermittent wiper systems of modern cars. The time between wipes is varied by adjusting the resistance in an RC size 12{ ital "RC"} {} circuit. Another example of an RC size 12{ ital "RC"} {} circuit is found in novelty jewelry, Halloween costumes, and various toys that have battery-powered flashing lights. (See [link] for a timing circuit.)

A more crucial use of RC size 12{ ital "RC"} {} circuits for timing purposes is in the artificial pacemaker, used to control heart rate. The heart rate is normally controlled by electrical signals generated by the sino-atrial (SA) node, which is on the wall of the right atrium chamber. This causes the muscles to contract and pump blood. Sometimes the heart rhythm is abnormal and the heartbeat is too high or too low.

The artificial pacemaker is inserted near the heart to provide electrical signals to the heart when needed with the appropriate time constant. Pacemakers have sensors that detect body motion and breathing to increase the heart rate during exercise to meet the body’s increased needs for blood and oxygen.

Part a shows a charging circuit containing cell of e m f script E connected to a resistor R and capacitor C and a closed switch to complete the circuit. The current is shown to flow clockwise through this arm of the circuit alone. A bulb of high resistance R is connected across the capacitor. Part b shows a discharging circuit containing a cell of e m f script E connected to a resistor R and capacitor C and a closed switch to complete the circuit. A bulb of low resistance R is connected across the capacitor. Current flows clockwise through the arm containing the capacitor and the low resistance bulb. Part c is a graph showing variation of voltage verses time for the bulb in above circuit. The voltage is plotted along the vertical axis and the time is plotted along the horizontal axis. The curve has a smooth rise from the origin, reaches a plateau at threshold value of voltage where it begins to drop and rise as a small sawtooth wave with maxima lying along the threshold line.
(a) The lamp in this RC size 12{ ital "RC"} {} circuit ordinarily has a very high resistance, so that the battery charges the capacitor as if the lamp were not there. When the voltage reaches a threshold value, a current flows through the lamp that dramatically reduces its resistance, and the capacitor discharges through the lamp as if the battery and charging resistor were not there. Once discharged, the process starts again, with the flash period determined by the RC size 12{ ital "RC"} {} constant τ size 12{τ} {} . (b) A graph of voltage versus time for this circuit.

Calculating time: RC Circuit in a heart defibrillator

A heart defibrillator is used to resuscitate an accident victim by discharging a capacitor through the trunk of her body. A simplified version of the circuit is seen in [link] . (a) What is the time constant if an 8.00-μF size 12{8 "." "00"-mF} {} capacitor is used and the path resistance through her body is 1.00 × 10 3 Ω ? (b) If the initial voltage is 10.0 kV, how long does it take to decline to 5.00 × 10 2 V ?

Strategy

Since the resistance and capacitance are given, it is straightforward to multiply them to give the time constant asked for in part (a). To find the time for the voltage to decline to 5.00 × 10 2 V , we repeatedly multiply the initial voltage by 0.368 until a voltage less than or equal to 5.00 × 10 2 V is obtained. Each multiplication corresponds to a time of τ size 12{τ} {} seconds.

Solution for (a)

The time constant τ size 12{τ} {} is given by the equation τ = RC size 12{τ= ital "RC"} {} . Entering the given values for resistance and capacitance (and remembering that units for a farad can be expressed as s / Ω size 12{s/ %OMEGA } {} ) gives

τ = RC = ( 1.00 × 10 3 Ω ) ( 8 . 00 μF ) = 8 . 00 ms. size 12{τ= ital "RC"= \( "1000" %OMEGA \) \( 8 "." "00" μF \) =8 "." "00"" ms"} {}

Solution for (b)

In the first 8.00 ms, the voltage (10.0 kV) declines to 0.368 of its initial value. That is:

V = 0 . 368 V 0 = 3.680 × 10 3 V at  t = 8 . 00 ms. size 12{t=8 "." "00"" ms"} {}

(Notice that we carry an extra digit for each intermediate calculation.) After another 8.00 ms, we multiply by 0.368 again, and the voltage is

V ′ = 0.368 V = 0.368 3.680 × 10 3 V = 1.354 × 10 3 V at t = 16.0 ms.

Similarly, after another 8.00 ms, the voltage is

V ′′ = 0.368 V ′ = ( 0.368 ) ( 1.354 × 10 3 V ) = 498 V at t = 24 .0 ms.

Discussion

So after only 24.0 ms, the voltage is down to 498 V, or 4.98% of its original value. Such brief times are useful in heart defibrillation, because the brief but intense current causes a brief but effective contraction of the heart. The actual circuit in a heart defibrillator is slightly more complex than the one in [link] , to compensate for magnetic and AC effects that will be covered in Magnetism .

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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