
(a)
The initial rate of increase of current.
(a)

Answer to Problem 63P
The initial rate of increase of current is
Explanation of Solution
Given:
The inductance of a coil is
The resistance of a resistor is
The voltage of ideal battery is
Formula used:
The expression for the current in the circuitis given by,
Calculation:
The initial rate of increase of current is calculated by differentiating the current with time.
Solve further as,
Conclusion:
Therefore, the initial rate of increase of current is
(b)
The rate of increase of current when the current is half of its steady-state value
(b)

Answer to Problem 63P
The rate of increase of current when the current is half of its steady-state value is
Explanation of Solution
Formula used:
The expression for the current in the circuit is given by,
Calculation:
According to the given condition,
The rate of increase of current at
Solve further as,
Conclusion:
Therefore, the rate of increase of current when the current is half of its steady-state value is
(c)
The steady state value of current.
(c)

Answer to Problem 63P
The steady state value of current is
Explanation of Solution
Formula used:
The expression for the steady state value of current is given by,
Calculation:
The steady state value of current is calculated as,
Conclusion:
Therefore, the steady state value of current is
(d)
The time taken of the current to reach
(d)

Answer to Problem 63P
The time taken of the current to reach
Explanation of Solution
Formula used:
The expression for the current in the circuit is given by,
Calculation:
According to the given condition,
Solve further as,
Conclusion:
Therefore, the time taken of the current to reach
Want to see more full solutions like this?
Chapter 28 Solutions
Physics for Scientists and Engineers
- please help me solve this questions. show all calculations and a good graph too :)arrow_forwardWhat is the force (in N) on the 2.0 μC charge placed at the center of the square shown below? (Express your answer in vector form.) 5.0 με 4.0 με 2.0 με + 1.0 m 1.0 m -40 με 2.0 μCarrow_forwardWhat is the force (in N) on the 5.4 µC charge shown below? (Express your answer in vector form.) −3.1 µC5.4 µC9.2 µC6.4 µCarrow_forward
- An ideal gas in a sealed container starts out at a pressure of 8900 N/m2 and a volume of 5.7 m3. If the gas expands to a volume of 6.3 m3 while the pressure is held constant (still at 8900 N/m2), how much work is done by the gas? Give your answer as the number of Joules.arrow_forwardThe outside temperature is 25 °C. A heat engine operates in the environment (Tc = 25 °C) at 50% efficiency. How hot does it need to get the high temperature up to in Celsius?arrow_forwardGas is compressed in a cylinder creating 31 Joules of work on the gas during the isothermal process. How much heat flows from the gas into the cylinder in Joules?arrow_forward
- The heat engine gives 1100 Joules of energy of high temperature from the burning gasoline by exhausting 750 Joules to low-temperature . What is the efficiency of this heat engine in a percentage?arrow_forwardL₁ D₁ L₂ D2 Aluminum has a resistivity of p = 2.65 × 10 8 2. m. An aluminum wire is L = 2.00 m long and has a circular cross section that is not constant. The diameter of the wire is D₁ = 0.17 mm for a length of L₁ = 0.500 m and a diameter of D2 = 0.24 mm for the rest of the length. a) What is the resistance of this wire? R = Hint A potential difference of AV = 1.40 V is applied across the wire. b) What is the magnitude of the current density in the thin part of the wire? Hint J1 = c) What is the magnitude of the current density in the thick part of the wire? J₂ = d) What is the magnitude of the electric field in the thin part of the wire? E1 = Hint e) What is the magnitude of the electric field in the thick part of the wire? E2 =arrow_forwardplease helparrow_forward
- Glencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-Hill
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax College





