Question

A flat coil of wire has an inductance of 40.0mH and a resistance of 5.70Ω. It is connected to a 22.5−V battery at the instant t=0. Consider the moment when the current is 3.60 A. (a) At what rate is energy being delivered by the battery? W (b) What is the power being delivered to the resistance of the coil? W (c) At what rate is energy being stored in the magnetic field of the coil? W (A) What ie the ralatinnehin amnne thaea throa nnwar waluae? This answer has not been graded yet. (e) Is the relationship described in part (d) true at other instants as well? Yes No (f) Explain the relationship at the moment immediately after t=0 and at a moment several seconds later.

Answer 1

The rate at which energy is being delivered by the battery is 81 W. The power being delivered to the resistance of the coil is 73.15 W. The rate at which energy is being stored in the magnetic field of the coil is 0.2592 W.

Step-by-step explanation:

To calculate the rate at which energy is being delivered by the battery, we can use the equation:

Power = IV

Where I is the current flowing through the circuit and V is the voltage across the circuit. In this case, the current is 3.60 A and the voltage is 22.5 V, so the power being delivered by the battery is:

Power = (3.60 A) * (22.5 V) = 81 W

The power being delivered to the resistance of the coil can be calculated using the equation:

Power = I^2R

Where I is the current flowing through the circuit and R is the resistance. In this case, the current is 3.60 A and the resistance is 5.70 Ω, so the power being delivered to the resistance of the coil is:

Power = (3.60 A)^2 * (5.70 Ω) = 73.15 W

To calculate the rate at which energy is being stored in the magnetic field of the coil, we can use the equation:

Power = 0.5LI^2

Where L is the inductance of the coil and I is the current flowing through the circuit. In this case, the inductance is 40.0 mH (or 0.040 H) and the current is 3.60 A, so the power being stored in the magnetic field of the coil is:

Power = 0.5 * (0.040 H) * (3.60 A)^2 = 0.2592 W

The relationship described in part (d) is true at any instant when the current is 3.60 A. The power delivered by the battery is equal to the power delivered to the resistance plus the power stored in the magnetic field. This relationship holds true regardless of the specific values of current, voltage, resistance, and inductance.

At the moment immediately after t=0, the current will be 0 A, so the power delivered by the battery and the power stored in the magnetic field will both be 0 W. The power delivered to the resistance will be the voltage squared divided by the resistance. Several seconds later, the current may have changed, but the relationship will still hold true: the power delivered by the battery will be equal to the power delivered to the resistance plus the power stored in the magnetic field.