emf=IRt=1.5(28) volts
terminal voltage=above-1.5*3
waster power: 1.5^2*3
percent=wasted power/EMF*1.5
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
terminal voltage=above-1.5*3
waster power: 1.5^2*3
percent=wasted power/EMF*1.5
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
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.