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