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a) The current through the 8-ohm resistor.
b)The total rate of dissipation of electrical energy in the 8-ohm resistor and the internal resistance of the batteries
c) In one of the batteries, chemical energy is being converted into electrical energy. In which one is this happening and at what rate?
d) In one of the batteries, electrical energy is being converted into chemical energy. In which one is this happening and at what rate?
e) show that the overall rate of production of electrical energy equals the overall rate of production equals the overall rate of production of electrical energy in the circuit.11

a The current through the 8ohm resistor bThe total rate of dissipation of electrical energy in the 8ohm resistor and the internal resistance of the batteries c class=

Respuesta :

egenriether
Take a Kirchoff voltage loop to see the effective voltage is 4V and the total resistance is 10ohms. This gives 400mA in the loop. This flows through everything since it's a series circuit.
Rate of energy is power. So the total power dissipated in the resistors is I^2R=(0.4^)2(10)=1.6W
For C and D know that the 12V battery is charging the 8V. So electric energy is produced in the 12V battery at a rate of P=IV. In the 8V, electric energy is converted to chemical energy because it's charging. The power here has the same sign as a load, not a source. Hope that helps

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