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Question 1. A light bulb rated 240 volts and 75 watts. If the voltage decrease by 5 volts and the resistance of the bulb is increased by 8Ω, by how much will the power change. Solution. DATA. Volt;V_1=240v Volt;V_2=240-5=235v Power;P=75watts Resistance;R_1=x?. Resistance;R_2=x+8Ω..............................................................1 Find Power Change?

Power=IV ; I^2 V..............................................................2 P_1=V_1 I ..............................................................3 V_1=IR_1..............................................................4 Initial Voltage is V_1=240v Current I=? Initial Resistance Resistance;R_1=x? V_2=235 Resistance;R_2=x+8Ω V_2=IR_2..............................................................5 I=V_2/R_2 ..............................................................6 Substitute the values of V_2 and R_2 to get the valie of "I" I=235/(x+8Ω)..............................................................7 Recall from equation 3 P_1=V_1 I Substitute the values of P_1 and V_1 to get the value of x in I 75=240× 235/(x+8Ω) (75(x+8Ω))/75=56400/75 (x+8Ω)=752 x=752-8 R_1=x=744Ω R_2=x+8Ω R_2=744+8 R_2=752Ω Lets calculate the current "I" I=V_1/R_1 I=240/744=0.3226A I=V_1/R_1 I=235/752 Recall P_1=V_1 I P_2=V_2 I P_2=235×0.3125 P_2=73.44watts Change in Power difference P_1-P_2 ∆ in Power=75-73.44 ∆P=1.5625watts.

Question 2 〖(d^2 y)/(dx^2 )+4 dy/dx+3y=0 subject to the initial condition y〗_((0))=3;〖y^'〗_((0))=4 where y^'=dy/dx Solution. We will attempt to first solve this differential equation by general method then we will also subject the equation to the given initial conditions let y=e^mx and y^'=me^mx and y^''=m^2 e^mx Where (d^2 y)/(dx^2 )=y^'' and dy/dx=y^' and y=y substitute the values in the above equation m^2 e^mx+4me^mx+3e^mx=0 e^mx (m^2+4m+3)=0 Find the roots of equation m(m+1)+3(m+1)=0 (m+1)(m+3)=0 Therefore m=-1 and m=-3 The general Solution is y=C_1 e^(-x)+C_2 e^(-3x) where C_1 and C_2 are arbitrary constant Now lets subject y^''+4y^'+3y=0 to the initial condition y_((0)) =3;〖y^'〗_((0))=4 we have the following y^2+4y^'+3y=0 where y_((0)) =3;〖y^'〗_((0))=4

y^2+4×4+3×3=0 y^2+16+9=0 y^2+25=0 y^2=-25 y=√(-25) y=-5 Question 3 In a biogas production experiment products increases with temperature as shown in the Table below. Determine the best fits between linear, quadratic and cubic for the observed data x 0 10 20 30 40 50 60 70 80 90 100 y 27.6 31.0 34.0 37 40 42.6 45.5 48.3 51.1 54.0 56.7

[Alexandra Harkai]

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[James Tursa]

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