c- Using the figure above, which battery converts chemical energy into electrica
ID: 581897 • Letter: C
Question
c- Using the figure above, which battery converts chemical energy into electrical energy and at what rate? DESCRIBE HOW YOU KNOW which battery does conversion of chemical to electrical.
d- Using figure above, which batttery converts Electrical energy to chemical energy and at what rate? DESCRIBE how you know which battery does which conversion!
e- show that the overall rate of production of electrical energy equals the overall rate of consumption of electrical energy in the circuit.
Other ?'s - How do you know which direction charges flow based on the figure above? I know delta V = 4 but do you have to do 12 volts - 8 volts and go that way or can you do the absolute value of 8 volts minus 12 volts to get the answer?
E! = 12.0 V = 12.0 v = 1.011 =100 R:8.012 E, = 8.0 V = 1.0Explanation / Answer
The trick is, we solve for value of current in the given circuit, its direction will then tell us about the direction of flow of charge.
Using the direction of current as given in figure, we can say that
total voltage drop across all the resistances is, Vdrop = I*(8+1+1) volts
This voltage is supplied by the two batteries, which are actually opposing each other
hence Vdrop = 12-8 = 4 volts = I*10 => I = 0.4 ampere
This shows the current flows in same direction as we assumed.
For a battery if current enters its negative terminal then it is supplying energy and hence converting chemical energy into electrical energy.
If current enters its positive terminal then it is getting charged and hence converting electrical energy into chemical energy.
Using this we can see that E2 is converting electrical energy into chemical energy and E1 is converting chemical energy into electrical energy.
Rate of chemical to electrical energy conversion = 12*0.4 = 4.8 Volt-Amp
Electrical energy is used in second battery as well as the rest of resistors(I^2*R heat dissipation)
Total energy consumed = 8*0.4+ 0.4^2*(1+8+1) = 3.2+1.6 = 4.8 Volt-Amp
Hence, Proved.
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