A battery with an emf of 10.3 V and internal resistance of 1.03 Ω carries a positive current of 0.550 A from its negative terminal to its positive terminal.

(a) How long is required for 1.05 C of charge to pass through the battery?
t = Q/I = (1.05)/(.550) = 1.909sec

(b) By how much will the electrical potential energy of 1.05 C of charge increase in crossing the battery terminals?

(c) By how much will the chemical energy decrease?

(d) How much electrical power is supplied by the battery?

(e) At what rate is chemical energy used?
P = VI = (10.3)(.550) = 5.665 W

****I'm completely lost with how to do b,c, and d.

To answer parts (b), (c), and (d) of the question, we need to understand the concepts of electrical potential energy and power.

(b) The electrical potential energy gained or lost by a charge moving through an electric field can be calculated using the formula: ∆U = q∆V, where ∆U is the change in potential energy, q is the charge, and ∆V is the change in voltage.

In this case, we need to calculate the change in electrical potential energy (∆U) when 1.05 C of charge passes through the battery. Since the battery has an electromotive force (emf) of 10.3 V, ∆V is equal to 10.3 V. Therefore, ∆U = (1.05 C)(10.3 V) = 10.815 J. The electrical potential energy will increase by 10.815 J.

(c) The chemical energy of the battery decreases when it supplies electrical energy to the circuit. The decrease in chemical energy is equal to the change in electrical potential energy (∆U), which we calculated in part (b) as 10.815 J.

(d) The electrical power supplied by the battery can be calculated using the formula: P = VI, where P is the power, V is the voltage, and I is the current.

In this case, the voltage of the battery is given as 10.3 V, and the current passing through the battery is 0.550 A. Therefore, P = (10.3 V)(0.550 A) = 5.665 W. The battery is supplying electrical power at a rate of 5.665 W.

I hope this helps! Let me know if you have any further questions.

To find the increase in electrical potential energy of 1.05 C of charge crossing the battery terminals, you can use the formula:

ΔPE = Q * emf

where ΔPE is the change in electrical potential energy, Q is the charge, and emf is the electromotive force (emf) of the battery.

Substituting the given values:

ΔPE = (1.05 C) * (10.3 V)
ΔPE = 10.855 J

Therefore, the electrical potential energy of 1.05 C of charge will increase by 10.855 J.

To find the decrease in chemical energy, you can use the formula:

ΔE_chem = -Q * emf

Since the charge transfers from the negative terminal to the positive terminal, the change in chemical energy is negative.

Substituting the given values:

ΔE_chem = -(1.05 C) * (10.3 V)
ΔE_chem = -10.815 J

Therefore, the chemical energy will decrease by 10.815 J.

To find the electrical power supplied by the battery, you can use the formula:

P = I * V

where P is the power, I is the current, and V is the voltage across the battery.

Substituting the given values:

P = (0.550 A) * (10.3 V)
P = 5.665 W

Therefore, the battery is supplying electrical power at a rate of 5.665 W.

I hope this helps! Let me know if you have any further questions.