Various Math #37 - Q&A

Question 1 (a) Compute the reactance of a 0.450-H inductor at frequencies of 60.0 Hz and 600 Hz. (b) Compute the reactance of a 2.50 μF capacitor at the same frequencies. (c) At what frequency is the reactance of a 0.450-H inductor equal to that of a 2.50 μF capacitor? Answer 1 Step 1 Given The inductance of the coil is L = 0.45H. Solution (a) For ac source connected with a coil, the inductance reactance X_L in the circuit is given by equation 31.12 in the form X_L = 2πfL This inductance reactance is related to the frequency f and the inductance L of the coil. Now, we want to get X_L at f = 60Hz and 600 Hz. For f = 60Hz: X_L = 2πfL = 2π(60Hz)(0.45H) = 170Ω For f = 600Hz: X_L = 2πfL = 2π(600Hz)(0.45H) = 1700Ω Answer 2 (b) The capacitance of the capacitor is C = 2.5µF. The capacitive reactance X_C in the circuit is given by equation 31.18 in the form X_C = 1/(2πfC) Now, we calcaute X_C or 60 Hz and 600 Hz as next X_C(60) = 1 / (2πfC) = 1 / (2π(60Hz) * (2.5 × 10⁻⁶F)) = 1.06 × 10³Ω X_C(600) = 1 / (2πfC) = 1 / (2π(600Hz) * (2.5 × 10⁻⁶F)) = 106Ω Answer 3 (c) In this part, the two reactance X_C and X_L are the same but at a specific frequency f. We are asked to determine the value of this frequency. X_C = X_L 1/(2πfC) = 2πfL f = 1 / (2π√(LC)) Now plug the values for L and C into equation (3) to get f f = 1 / (2π√(0.45H)(2.5×10⁻⁶F)) = 150 Hz Question 2 Find the annual percent increase or decrease that y = 0.35(2.3)^x models. Answer Given model y = 0.35(2.3)^x (1) We know the exponential growth function y = A(1 + γ)^x (2) Where A = initial amount γ = annual percent x = numbers of years From equation (1) and (2) A = 0.35 1 + γ = 2.3 ⇒ γ = 2.3 -1 ⇒ γ = 1.3 ⇒ γ = 1.3 × 100% γ = 130% Question 3 Two light bulbs have constant resistances of 400 Ω and 800 Ω. If the two light bulbs are connected in series across a 120 V line, find the power dissipated in each bulb Answer Given: The resistance across first bulb is R₁ = 400Ω The resistance across second bulb is R₂ = 800Ω The voltage across the builbs is V = 120 V. The formula to calculate the power dissipated in first bulb is, P₁ = V²/R₁ Here, P₁ is the power dissipated in first bulb, V is the voltage and R₁ is the resistance across first bulb. Substitute the khown values in the formula to calculate the power dissipated in first bulb. P₁ = (120 V)² / 400Ω = 36 W The formula to calculate the power dissipated in second bulb is, P₂ = V²/R₂ Here, P₂ is the power dissipated in first bulb and R₂ is the resistance across second bulb. Substitute the known values in the formula to calculate the power dissipated in second bulb. P₂ = (120 V)² / 800Ω = 18 W Thus, the power dissipated in first bulb is 36 W and the power dissipated in second bulb is 18 W. Question 4 Rachel has good distant vision but has a touch of presbyopia. Her near point is 0.60 m. When she wears +2.0 D reading glasses, what is her near point? Her far point? Answer Given Power = +2D Initial near point =0.6m a) power (P) = 1/f = +2D, V = -0.6 m Here, using the lens equation as following below, 1/V - 1/u = 1/f 1/-0.6 - 1/u = 2 1/u = 1/-0.6 -2 u = -0.27 m Her near point is 0.27 m b) Now, her new near point V = ∞ P = 1/f 2 = 1/f f = 1/2 m = 0.5 m Now, using the lens equation, 1/V - 1/u = 1/f 1/∞ - 1/u = 1/0.5 1/u = 1/-0.5 u = -0.5m Raches far point is 0.5 m Question 5 Suppose the price level and value of the U.S dollar in year 1 are 1 and $1 respectively. a. If the price level rises to 1.40 in year 2, what is the new value of the dollar? b. If, instead the price level falls to 0.80, what is the value of dollar? Answer Given, Price level = 1 Value of U.S dollar = $1 The relation between the price level or inflation rate and value of a currency is as follows: Value of currency = nominal currency value /price level a. If the price level rises to 1.40, the value of dollar can be computed as: Value of dollar = 1 / price level = 1 / 1.40 = 0.025 b. When price level fall to 0.80, the value of dollar is: Value of dollar = 1 / price level = 1 / 0.80 = 1.25