Initially a battery circuit consists of a 25 ohm load connected to a battery with 3 ohms of internal resistance and the circuit draws 1.5A of current.

What is the terminal voltage and EMF of the batter?

How much (%) power is wasted?

Please help, this doesn't make sense given only V(term) = E - Ir

emf=IRt=1.5(28) volts

terminal voltage=above-1.5*3

waster power: 1.5^2*3

percent=wasted power/EMF*1.5

total R = 28

so i = EMF/28 = 1.5

so EMF = 1.5*28 = 42 volts

voltage drop internal = 1.5*3 = 4.5 volts

so terminal V = 42 -4.5 = 37.5 volts

power waste = i^2R = (1.5)^2 * (3)
= 6.75 Watts

To find the terminal voltage and EMF (electromotive force) of the battery, as well as the percentage of power wasted, we can use the formula V(term) = E - Ir, where V(term) is the terminal voltage, E is the EMF, I is the current, and r is the internal resistance.

Given:
Load resistance, R_load = 25 ohms
Internal resistance of the battery, r = 3 ohms
Current in the circuit, I = 1.5A

1. To find the terminal voltage, substitute the given values into the formula V(term) = E - Ir:
V(term) = E - I * r
V(term) = E - (1.5A * 3 ohms)
V(term) = E - 4.5V

2. To find the EMF of the battery, isolate E in the formula E = V(term) + Ir:
E = V(term) + I * r
E = (V(term) + 4.5V) / 1.5A

3. To find the percentage of power wasted, we can calculate the power consumed by the internal resistance and divide it by the total power output.

The power consumed by the internal resistance is given by P(r) = I^2 * r.

The total power output is given by P(total) = I * V(term).

Finally, the percentage of power wasted can be calculated using the formula:

Percentage of power wasted = (P(r) / P(total)) * 100%

Substituting the given values:
P(r) = (1.5A)^2 * 3 ohms
P(r) = 6.75W

P(total) = 1.5A * V(term)

Percentage of power wasted = (6.75W / (1.5A * V(term))) * 100%

Now, by simply solving the equations, you can find the terminal voltage, EMF, and percentage of power wasted.