How Does a 30.0 W Charger Affect Battery Charging Times and Efficiency?

[phys-unit]

Homework Statement

A cordless drill battery charger is rated at 30.0 W but requires only 13.5 V. It contains a transformer that converts 120 V household voltage.


(b) What is the current in the secondary coil while the battery is charging?


(c) What is the current in the primary coil while the battery is charging?


(d) The battery is rated at 3150 mA·h life and generates 12.0 V. This means, for instance,that it could supply 100 mA of current at 12.0 V for 31.5 hours. How much energy does the battery store?
4 J

(e) What is the minimum time required for the 30.0 W charger to fully charge the battery? While such a time is possible in principle, in practice charging times are much longer.

Homework Equations

Vs/Vp = Ns/Np (I think this is what I should be using)

The Attempt at a Solution

I just don't get how to set any of it up.


EDIT! ALRIGHT, I got A and B A is 2.222 and B is .25



If someone could show me how to do it I'd be really happy :\

 

[anathema]

Solution:

(a) The power rating of the charger is 30.0 W, which means it can provide a maximum power output of 30.0 W. This does not necessarily mean that it will always provide this much power, but it is the maximum amount it is capable of.

(b) To find the current in the secondary coil, we can use the formula P=IV, where P is power, I is current, and V is voltage. We know that the power output is 30.0 W and the voltage is 13.5 V. Plugging these values in, we get:

30.0 W = I * 13.5 V

Solving for I, we get:

I = 30.0 W / 13.5 V = 2.22 A

So, the current in the secondary coil is 2.22 A while the battery is charging.

(c) To find the current in the primary coil, we can use the formula Vs/Vp = Ns/Np, where Vs is the voltage in the secondary coil, Vp is the voltage in the primary coil, Ns is the number of turns in the secondary coil, and Np is the number of turns in the primary coil.

We know that Vs = 13.5 V and Vp = 120 V. We also know that the transformer is ideal, which means that the input power is equal to the output power. So, we can use the formula P=IV to find the current in the primary coil.

P = IV

30.0 W = I * 120 V

Solving for I, we get:

I = 30.0 W / 120 V = 0.25 A

So, the current in the primary coil is 0.25 A while the battery is charging.

(d) The energy stored in the battery can be calculated using the formula E=Pt, where E is energy, P is power, and t is time.

We know that the battery can supply 100 mA of current at 12.0 V for 31.5 hours. So, the power output of the battery is:

P = IV = (100 mA) * (12.0 V) = 1.2 W

Using this value for P and t = 31.5 hours, we can calculate the energy stored in the battery:

E

 

[tomcue]

Solution:

(a) To determine the current in the secondary coil while the battery is charging, we can use the formula P=IV where P is power, I is current and V is voltage. We are given that the charger is rated at 30.0 W and requires 13.5 V. Therefore, we can rearrange the formula to solve for current: I = P/V = 30.0 W / 13.5 V = 2.22 A. This is the current in the secondary coil while the battery is charging.

(b) To determine the current in the primary coil while the battery is charging, we can use the formula Vs/Vp = Ns/Np, where Vs and Vp are the voltages in the secondary and primary coils respectively, and Ns and Np are the number of turns in the secondary and primary coils respectively. We are given that the voltage in the secondary coil is 13.5 V and the voltage in the primary coil is 120 V. We can rearrange the formula to solve for the current in the primary coil: I = Ns/Np * Vs/Vp = (1/9) * (13.5 V / 120 V) = 0.125 A. This is the current in the primary coil while the battery is charging.

(c) To determine the energy stored in the battery, we can use the formula E = P*t, where E is energy, P is power, and t is time. We are given that the battery is rated at 3150 mA·h, which is equivalent to 3.15 A·h. We can convert this to A·s by multiplying by 3600 s/h, giving us 11,340 A·s. We are also given that the battery generates 12.0 V. Therefore, we can rearrange the formula to solve for energy: E = P*t = (3.15 A * 12.0 V) * (11,340 A·s / 3.15 A) = 11,340 J. This is the amount of energy stored in the battery.

(d) The minimum time required for the 30.0 W charger to fully charge the battery can be determined by using the formula t = Q/I, where t is time, Q is charge, and I is current. We are given that the battery can supply 100 mA of current at 12.0