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The solution is nt n e kt say ? k if we want to put

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The solution is N(t) = N 0 e kt (say λ = -k, if we want to put in the form of the above discussion). 91
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Half-life of C-14 is approximately 5600 years 2 N o = N(5600) or 2 1 N 0 = N 0 e 5600k . By cancelling N 0 and taking logarithm of both sides we get 2 ln 2 1 ln 5600 - = = k or k= - 5600 2 ln = -0.00012378 Therefore N(t) = N 0 e -0.00012378t With N(t) = 0 1000 1 N we have 1000 1 N 0 =N 0 e -0.00012378t -0.00012378t = ln 1000 1 = - ln 1000. Thus yrs 55800 00012378 . 0 1000 ln t = (b) Let N(t)= N 0 e kt where k= -0.00012378 by part (a). 85.5% of C-14 had decayed; that is, N(t) = 0.145 N 0 or N 0 e - 0.00012378t = 0.145 No Taking logarithm of both sides and solving for t, we get t 15,600 years 4.4 Mixture of Two Salt Solutions 92
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Example. 4.8 A tank contains 300 litres of fluid in which 20 grams of salt is dissolved. Brine containing 1 gm of salt per litre is then pumped into the tank at a rate of 4 L/min; the well-mixed solution is pumped out at the same rate. Find the number N(t) of grams of salt in the tank at time t. Solution: By the data given in this example we have P(t ) = N(t) n=4, p=1, m=300 in the model (1.23) 75 4 4 . 300 4 ) ( P P dt t dP - = - = or 4 75 P dt ) t ( dP = + This is a linear differential of first order in P whose integrating factor is t e dt e 75 1 75 1 = (See Section 2.3) Solution is given by c dt t e t e t P + = 75 1 4 75 1 ). ( P(t) = 300 + c t e 75 1 . - Since P(0) = 20 is given we get 20=P(0) = 300+Ce 0 , that is c= -280 Thus P(t) = 300 - 280 t e 75 1 - 93
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4.5 Series Circuits Let a series circuit contain only a resistor and an inductor as shown in Figure 4.2 Figure 4.2 LR Series circuit By Kirchhoff’s second law the sum of the voltage drop across the inductor Ι dt di and the voltage drop across the resistor (iR) is the same as the impressed voltage (E(t)) on the circuit. Current at time t, i(t), is the solution of the differential equation. ) ( t E Ri dt di = + Ι (4.10) where Ι and R are constants known as the inductance and the resistance respectively. The voltage drop across a capacitor with capacitance C is given by C ) t ( q , where q is the charge on the capacitor. Hence, for the series circuit shown in Figure 4.3 we get the following equation by applying Kirchhoff’s second law ) ( 1 t E q C Ri = + (4.11) 94
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Figure 4.3 RC Series Circuit Since dt dq i = , (4.11) can be written as ) t ( E q C 1 dt dq R = + (4.12) Example 4.9 Find the current in a series RL circuit in which the resistance, inductance, and voltage are constant. Assume that i(o)=0; that is initial current is zero. Solution: It is modeled by (4.10) ) t ( E Ri dt di = + Ι or Ι = Ι + ) t ( E i R dt di (4.13) Since Ι , R and E are constant (4.13) is linear equation of first-order in i with integrating factor Ι = Ι t R e dt R e The solution of (4.13) is c R t R e E t R e ) t ( i or dt t R e E t R e ) t ( i + Ι Ι Ι = Ι Ι Ι = Ι or i(t) = t R ce R E Ι - + (4.14) 95
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Since i(0) = 0, c = - R E Putting this value of c in (4.14) we get ) 1 ( ) ( t R e R E
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