Explanation 1) Concept: Here, we need to use the equations of Kirchhoff’s junction rule and loop rule. Using these equations, we can get the differential equation of the current through the resistor. The solution of this differential equation is the function of time, and from that, we can get the expression for the current through the inductor. The expression for time at which the current through the resistor and inductor is equal can be found by equating the equations of current and solving for time. 2) Formula: i) V R = i 1 R ii) V L = L × d i 2 d t 3) Calculations: (a) Applying junction rule to writ the expression for current in circuit: i = i 1 + i 2 Here I is the current from the constant current source. i 1 is current through the resistor R i 2 is current through the inductor L Taking derivative with respect to t of equation of junction rule, we get d i 1 d t = - d i 2 d t Now, V R is the voltage drop across resistor, and s is given by V R = i 1 R And, V L is voltage drop across inductor given as, V L = L × d i 2 d t Now applying voltage rule, we get i 1 R - L × d i 2 d t = 0 i 1 R + L d i 1 d t = 0 This is differential equation, and its solution can be written as i 1 t = i 0 e - R t L where, i 0 is the current in the circuit at time = 0 . When switch is closed, the current through the inductor is zero

10th Edition

ISBN: 9781118230725

Author: David Halliday, Robert Resnick, Jearl Walker

Publisher: Wiley, John & Sons, Incorporated

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Chapter 30, Problem 59P

To determine

To find:

a) Current through the inductor.

b) Time at which the current through the resistor is equal to the current through the inductor.

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Chapter 30, Problem 58P

Chapter 30, Problem 60P

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Chapter 30 Solutions

Fundamentals of Physics Extended

Ch. 30 - If the circular conductor in Fig. 30-21 undergoes...Ch. 30 - Prob. 2Q Ch. 30 - Prob. 3Q Ch. 30 - Prob. 4Q Ch. 30 - Prob. 5Q Ch. 30 - Prob. 6Q Ch. 30 - Prob. 7Q Ch. 30 - Prob. 8Q Ch. 30 - Prob. 9Q Ch. 30 - Prob. 10Q

Ch. 30 - Figure 30-31 shows three situations in which a...Ch. 30 - Figure 30-32 gives four situations in which we...Ch. 30 - Prob. 1P Ch. 30 - A certain elastic conducting material is stretched...Ch. 30 - Prob. 3P Ch. 30 - A wire loop of radius 12 cm and resistance 8.5 is...Ch. 30 - Prob. 5P Ch. 30 - Figure 30-37a shows a circuit consisting of an...Ch. 30 - In Fig. 30-38, the magnetic flux through the loop...Ch. 30 - Prob. 8P Ch. 30 - Prob. 9P Ch. 30 - Prob. 10P Ch. 30 - A rectangular coil of N turns and of length a and...Ch. 30 - Prob. 12P Ch. 30 - Prob. 13P Ch. 30 - GO In Fig. 30-42a, a uniform magnetic field B...Ch. 30 - GO A square wire loop with 2.00 m sides is...Ch. 30 - GO Figure 30-44a shows a wire that forms a...Ch. 30 - A small circular loop of area 2.00 cm2 is placed...Ch. 30 - Prob. 18P Ch. 30 - ILW An electric generator contains a coil of 100...Ch. 30 - At a certain place, Earths magnetic field has...Ch. 30 - Prob. 21P Ch. 30 - A rectangular loop area = 0.15 m2 turns in a...Ch. 30 - SSM Figure 30-47 shows two parallel loops of wire...Ch. 30 - Prob. 24P Ch. 30 - GO Two long, parallel copper wires of diameter 2.5...Ch. 30 - GO For the wire arrangement in Fig. 30-49, a =...Ch. 30 - ILW As seen in Fig. 30-50, a square loop of wire...Ch. 30 - Prob. 28P Ch. 30 - Prob. 29P Ch. 30 - Prob. 30P Ch. 30 - Prob. 31P Ch. 30 - A loop antenna of area 2.00 cm2 and resistance...Ch. 30 - GO Figure 30-54 shows a rod of length L = 10.0 cm...Ch. 30 - Prob. 34P Ch. 30 - Prob. 35P Ch. 30 - Prob. 36P Ch. 30 - Prob. 37P Ch. 30 - Prob. 38P Ch. 30 - Prob. 39P Ch. 30 - Prob. 40P Ch. 30 - A circular coil has a 10.0 cm radius and consists...Ch. 30 - Prob. 42P Ch. 30 - Prob. 43P Ch. 30 - Prob. 44P Ch. 30 - Prob. 45P Ch. 30 - Prob. 46P Ch. 30 - Inductors in series. Two inductors L1 and L2 are...Ch. 30 - Prob. 48P Ch. 30 - Prob. 49P Ch. 30 - Prob. 50P Ch. 30 - ILW The current in an RL circuit drops from 1.0 A...Ch. 30 - Prob. 52P Ch. 30 - Prob. 53P Ch. 30 - Prob. 54P Ch. 30 - Prob. 55P Ch. 30 - Prob. 56P Ch. 30 - In Fig. 30-65, R = 15 , L = 5.0 H, the ideal...Ch. 30 - Prob. 58P Ch. 30 - Prob. 59P Ch. 30 - Prob. 60P Ch. 30 - Prob. 61P Ch. 30 - A coil with an inductance of 2.0 H and a...Ch. 30 - Prob. 63P Ch. 30 - Prob. 64P Ch. 30 - Prob. 65P Ch. 30 - A circular loop of wire 50 mm in radius carries a...Ch. 30 - Prob. 67P Ch. 30 - Prob. 68P Ch. 30 - ILW What must be the magnitude of a uniform...Ch. 30 - Prob. 70P Ch. 30 - Prob. 71P Ch. 30 - Prob. 72P Ch. 30 - Prob. 73P Ch. 30 - Prob. 74P Ch. 30 - Prob. 75P Ch. 30 - Prob. 76P Ch. 30 - Prob. 77P Ch. 30 - Prob. 78P Ch. 30 - SSM In Fig. 30-71, the battery is ideal and = 10...Ch. 30 - Prob. 80P Ch. 30 - Prob. 81P Ch. 30 - A uniform magnetic field B is perpendicular to the...Ch. 30 - Prob. 83P Ch. 30 - Prob. 84P Ch. 30 - Prob. 85P Ch. 30 - Prob. 86P Ch. 30 - Prob. 87P Ch. 30 - Prob. 88P Ch. 30 - A coil with an inductance of 2.0 H and a...Ch. 30 - Prob. 90P Ch. 30 - Prob. 91P Ch. 30 - Prob. 92P Ch. 30 - Prob. 93P Ch. 30 - A long cylindrical solenoid with 100 turns/cm has...Ch. 30 - Prob. 95P Ch. 30 - A square loop of wire is held in a uniform 0.24 T...Ch. 30 - Prob. 97P Ch. 30 - The inductance of a closely wound coil is such...Ch. 30 - The magnetic field in the interstellar space of...Ch. 30 - Prob. 100P Ch. 30 - A toroid has a 5.00 cm square cross section, an...

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a) Current through the inductor. b) Time at which the current through the resistor is equal to the current through the inductor.

Solution Summary: The author explains how the expression for time at which the current through the resistor and inductor is equal can be found by equating the equations of current.

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To find: a) Current through the inductor. b) Time at which the current through the resistor is equal to the current through the inductor.

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Chapter 30 Solutions

To find:

a) Current through the inductor.

b) Time at which the current through the resistor is equal to the current through the inductor.