In a heart pacemaker, a pulse is delivered to the heart 83 times per minute.?
83/min = 1.383/sec or a period of 0.723 sec So it loses 63.5% in 0.723 s Unfortunately your value for resistance is garbage. I'll call it R voltage on a cap, discharging v = v₀e^(–t/τ) v₀ is the initial voltage on the cap v is the voltage after time t R is resistance in ohms, C is capacitance in farads t is time in seconds RC = τ = time constant v/v₀ = e^(–t/τ) = 0.365 –t/τ = ln 0.365 = –1.008 t/τ = 1.008 t = 0.723 s τ = 0.723 / 1.008 = 0.717 sec RC = 0.717 sec plug in the value for R and solve for C
In a heart pacemaker, a pulse is delivered to the heart 81 times per minute. (concinued below)?
What we know .. R = 1.50^6 Ω t = 60s / 81 = 0.741 s intervals Qt = 0.67.Qo .. Qt / Qo = 0.67 Using the exponential discharge equation .. Qt = Qo.e ^(- t/CR) Qt / Qo = 0.67 = e ^(- t/CR) 0.67 = e ^(- t/CR) Ln 0.67 = - t/CR C = -t /(R.Ln0.67) .. .. - 0.741s / (1.50^6Ω x -0.400) .. .. .. ►C = 1.24^-6 F
In a heart pacemaker, a pulse is delivered to the heart 81 times per minute. The capacitor that controls this ?
Time constant RC = -1.35*ln(1-0.627) = 1.33134 s C = 1.33134/2.2E6 F
Capacitance of a capacitor?
discharge thro RC circuit is Q = Qo exp[-t/RC] at t = T = time period of applied signal, Q/Qo = 0.578 frequency of signal = f = 1/T = 81/60 T = 60/81 = 20/27 ------------------------------ Q/Qo = exp[-t/RC] 0.578 = exp[- T / RC] = 578/1000 exp[+ T / RC] = 1000/578 = 1.73 taking log (natural) T/RC = ln [1.73] = 0.548 C = T/[R*0.548] = 20/(27*2.5*10^6*0.548] C = 0.54*10^-6 Farad C = 0.54 micro farad
RC Circuit Question! Kinda Confused one Which Equation to Use.?
Question: In a heart pacemaker, a pulse is delivered to the heart 81 times per minute. the capacistor that controls this pulsing rate discharges through a resistance of 1.8e6 ohms. one pulse is delivered every time the fully charged capacitor loses 63.2% of its original charge (CONFUSING!). what is the capacitance of the capacitor? My Approach: I went ahead and converted 81 pulses/minute to pulses/second (which is like 1.35 pulses/sec). I am having trouble plugging these numbers into an equation - q = q(initial)*e^(-t/RC). im not sure that's the equation to use. i would appreciate all the help i can get on this problem! even a fully solved problem would be cool. Book Answer: 4.1e-7 F
RC circuits?
I assume that "1.8 106" actually means 1.8×10^6 = 1.8 E6 = 1.8 MΩ. Also, I think that Qnot should be, in fact, Qinit (Q₀), initial value for charge. Qinit exp(-t/RC) gives the charge remaining in the capacitor at any time; and Qinit [1 - exp(-t/RC)] is thus the charge lost (discharged) in the capacitor as a function of time. Rearrange your equation as ΔQ/Q₀ = 1-exp(-t/RC) = 0.643. Then exp(-t/RC) = 1 - 0.643 = 0.357 -t/RC = ln 0.357 = -1.03 RC = -t/-1.03 = 0.9709 t = 0.7374 s C = 0.7374/1.8 E6 = 409.7 E-9 F = 409.7 nF