The current in a wire at time t is given by the expression
a. Find an expression for the total amount of charge (in coulombs) that has entered the wire at time t. The initial conditions are
b. Graph Q versus t for the interval
Want to see the full answer?
Check out a sample textbook solutionChapter 27 Solutions
Physics for Scientists and Engineers: A Strategic Approach with Modern Physics, Books a la Carte Edition; Student Workbook for Physics for Scientists ... eText -- ValuePack Access Card (4th Edition)
- The quantity of charge through a conductor is modeled asQ(t)=(4.7)t4+(−1.2)t+(6.8) C What is the current at time t= 4 s?arrow_forwardAn electric current in a conductor varies with time according to the expression I(t) = 94 sin (180?t), where I is in amperes and t is in seconds. What is the total charge passing a given point in the conductor from t = 0 to t = 1/240 s?arrow_forwardAn electric current in a conductor varies with time according to the expression I(t) = 100sin(120лt), where I is in amperes and t is in seconds. What is the total charge passing a given point in the conductor from t=0 to t=1/240 s?arrow_forward
- An electric current of I = 0.266 A runs though a wire. How many electrons flow past a particular point in the wire in a time of t = 191 seconds? Express your answer in scientific notation using the "e" format, and keep 3 significant figures.arrow_forwardResistance in metals increases with increasing temperature according to the equation, ρ(T) = ρo(1 + α(T - To)) where α is the temperature coefficient of resistivity and ρo is the resistivity at temperature To. For a particular wire α = 9.5 × 10-3 1/°C and the resistivity is ρo = 4.5 × 10-7 Ω⋅m at To = 154 °C. a). Input an expression for the temperature T2 at which the resistance of a wire will be twice as high as at T0. b). If the wire is L = 1 m long with a radius of r = 1 cm, what is its resistance R, in Ω, at 2To?arrow_forwardThe total amount of charge that has entered a wire at time t is Q(t) = (27 C) (1. – e-t/2.1), with t in seconds and t ≥ 0. a. Graph Q vs t for the interval 0 ≤ t ≤ 10 s. b. Find an expression for the current in the wire at time t. c. What is the maximum value of the current? d. Graph the current in the wire as a function of time for the same interval as in part a.arrow_forward
- The quantity of charge through a conductor is modeled as Q = (3.00 mc/s4)t4 - (4.00 mc/s)t + 7.00 mc. What is the current (in A) at time t = 2.00 s?arrow_forwardThe amount of charge "q" that has passed through a conductor varies as a function of time according to the equation q (t) = 4 sin (2 Pi t) - 11 Cos (4 Pi t) + 6, where t is in seconds and q is C What is the instantaneous current that passes through the conductor at t = 12 s? NOTE: The angle of the trigonometric functions is given in radians.arrow_forwardThe electric charge that moves through a conductor varies in accordance with the equation q(t) =950t, where Q and t are given in μC and msec respectively. Calculate the current in A.arrow_forward
- Consider the circuit shown in the accompanying figure, where R₁ =13.0 22 and R₂ = 7.5 22. iz i4 1.592 -30V i3 Determine the current i4. Express your answer in A, to at least one digit after the decimal point. www www R₁ R₂ www www R₁ R₂ is R₂2.002 wwwwarrow_forwardAnswer must be in standard form scientific notation with SI units that do not have prefixes except for kg. Provide the answer with the correct amount of significant figures. Thank you so much I greatly appreciate itarrow_forwardA high voltage transmission line of diameter 3.73 cm and length 3.76189 km carries a steady current of 1.9 x103 A. If the conductor is copper with a free charge density of 6 x 1028 electrons/m3, how long (in seconds ) does it take one electron to travel the full length of the cable? (e = 1.6 x 10−19 C).For this problem use scientific/exponential notation to represent your answer. Eg., -0.0001 can be written as 1.0e-4 or as 1.0E-4. Spaces are not allowed.arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning