Answer to Question #268221 in Differential Equations for jay

Question

1. Given an RC series circuit that has an emf source of 100 volts, a resistance of 30 kilo ohms, a capacitance of 5 microfarad and the initial charge of the capacitor is 1 coulomb. What is the current in the circuit at the end of 0.03 seconds? Ans. -5.455 A
2. Consider that an object weighing 50 lb is dropped from a height of 1000 ft with zero initial velocity. Assume that the air resistance is proportional to the velocity of the body. If the limiting velocity is known to be 200 ft/sec, find the time it would take for the object to reach the ground? Ans. 9.985 sec.

Answer 1

1.

$$Rq'+\dfrac{1}{C}q=V$$

$$3\cdot10^{4}q'+\dfrac{1}{5\cdot10^{-6}}q=100$$

$$q'+\dfrac{20}{3}q=\dfrac{0.01}{3}$$

$$q'=-\dfrac{20}{3}(q-0.0005)$$

$$\dfrac{dq}{q-0.0005}=-\dfrac{20}{3}dt$$

Integrate

$$q=0.0005+c_1e^{-20t/3}$$

Given $q(0)=1\ C$

$$1=0.0005+c_1$$

$$c_1=0.9995$$

$$q(t)=0.0005+0.9995e^{-20t/3}$$

$$i(t)=q'(t)=0.9995(-\dfrac{20}{3})e^{-20t/3}$$

$$=\dfrac{19.99}{3}e^{-20t/3}$$

$$i(0.03)=-\dfrac{19.99}{3}e^{-20(0.03)/3}=-5.455(A)$$

$-5.455\ A$

1. The second Newton's Law

$$mv'=mg-kv$$

Given $m=50(0.453592)\ kg=22.6796\ kg,$

$g=32.1740 \ ft/s^2$

$$v'+\dfrac{k}{22.6796}v=32.1740$$

$$v'=-\dfrac{k}{22.6796}(v-\dfrac{729.6934504}{k})$$

$$\dfrac{dv}{v-\dfrac{729.6934504}{k}}=-\dfrac{k}{22.6796}dt$$

integrate

$$v(t)=\dfrac{729.6934504}{k}+c_1e^{-kt/22.6796}$$

$$v(0)=0=>c_1=-\dfrac{729.6934504}{k}$$

$$v(t)=\dfrac{729.6934504}{k}-\dfrac{729.6934504}{k}e^{-kt/22.6796}$$

$$v(t)\leq200=>\dfrac{729.6934504}{k}=200$$

$$k=3.648467252$$

$$v(t)=200-200e^{-0.16087t}$$

$$v(t)=-h'(t)$$

$$h(t)=-\int v dt=-\int(200-200e^{-0.16087t})dt$$

$$=-200t-\dfrac{200}{0.16087}e^{-0.16087t}+c_2$$

$$h(0)=1000=-\dfrac{200}{0.16087}+c_2=>c_2=2243.24$$

$$h(t)=-200t-1243.24e^{-0.16087t}+2243.24$$

$$h(t_1)=0=-200t_1-1243.24e^{-0.16087t_1}+2243.24$$

Solve graphically

$9.965$ sec