Question

A flat coil of wire has an inductance of 40.0 mH and a resistance of 5.90 ?. It is connected to a 26.3-V battery at the instant t = 0. Consider the moment when the current is 2.90 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

Answer 1

The rate at which energy is being delivered by the battery is 76.07 W. The power being delivered to the resistance of the coil is 48.47 W. The rate at which energy is being stored in the magnetic field of the coil is 5.0 mJ.

Step-by-step explanation:

(a) To determine the rate at which energy is being delivered by the battery, we can use the formula:

P = VI

where P is power, V is voltage, and I is current. In this case, the voltage is 26.3 V and the current is 2.90 A. Substituting these values into the formula:

P = (26.3 V) imes (2.90 A) = 76.07 W

So the rate at which energy is being delivered by the battery is 76.07 W.

(b) The power being delivered to the resistance of the coil can be calculated using the formula:

P = I^2R

where P is power, I is current, and R is resistance. In this case, the current is 2.90 A and the resistance is 5.90 Ω. Substituting these values into the formula:

P = (2.90 A)^2 imes (5.90 Ω) = 48.47 W

So the power being delivered to the resistance of the coil is 48.47 W.

(c) The rate at which energy is being stored in the magnetic field of the coil can be calculated using the formula:

PE = ½LI^2

where PE is the energy stored in the magnetic field, L is inductance, and I is current. In this case, the inductance is 40.0 mH and the current is 2.90 A. Substituting these values into the formula:

PE = ½(40.0 mH) imes (2.90 A)^2 = 5.0 mJ

So the rate at which energy is being stored in the magnetic field of the coil is 5.0 mJ.

Answer 2

• The energy is being delivered by the battery is 76.27 W
• The resistance to the coil is 49.617W
• Energy being stored in the magnetic field of the coil is 26.653W

How to solve for the values

The rate of energy

The formula is given as:

`\[ P = V * I \]`

Where we have V = voltage = 26.3

I = current = 2.90

`P = 26.3 * 2.90`

P = 76.27

At 76.27 energy being delivered by the battery is delivered

B. The power to resistance

The formula is

`\[ P_R = I^2 * R \]`

R = 5.9

I = 2.9

Then we have

`2.9^2 * 5.9\\`

= 49.617

The resistance to the coil is 49.617

C. Rate energy is stored in magnetic field

The rate at which energy is stored in the magnetic field of the coil can be found by subtracting the power dissipated in the resistance from the total power delivered by the battery.

The formula is

`\[ P_L = P - P_R \]`

76.27 - 49.617

= 26.653

Hence we have the solutions as:

• The energy is being delivered by the battery is 76.27 W
• The resistance to the coil is 49.617W
• Energy being stored in the magnetic field of the coil is 26.653W