In the circuit shown in the figure below, R = 5.0 ?, ?1 = 11.0 V and ?2 = 7.0 V.
ID: 2141192 • Letter: I
Question
In the circuit shown in the figure below, R = 5.0 ?, ?1 = 11.0 V and ?2 = 7.0 V.
(a) Find the current through the 5.0 ? resistor.
(b) Find the total rate of dissipation of electrical energy in the 5.0 ? resistor and in 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?
?1
?2
At what rate?
(d) In one of the batteries, electrical energy is being converted into chemical energy. In which one is this happening?
?1
?2
At what rate?
(e) Show that the overall rate of production of electrical energy equals the overall rate of consumption of electrical energy in the circuit.
Please explain showing the formulas you used. THANKS!
In the circuit shown in the figure below, R = 5.0 ?, ?1 = 11.0 V and ?2 = 7.0 V. Find the current through the 5.0 ? resistor. Find the total rate of dissipation of electrical energy in the 5.0 ? resistor and in the internal resistance of the batteries. In one of the batteries, chemical energy is being converted into electrical energy. In which one is this happening? ?1 ?2 At what rate? In one of the batteries, electrical energy is being converted into chemical energy. In which one is this happening? ?1 ?2 At what rate? Show that the overall rate of production of electrical energy equals the overall rate of consumption of electrical energy in the circuit.Explanation / Answer
a)
net current = Vnet/Rnet = (11-7)/(5+1+1) = 0.571 A
b)
rate of dissipation = I^2*R = 0.571^2*5 = 1.63 W
c)
E1 <-----answer
rate = 11*0.571 = 6.28 W
d)
E2
rate = 7*0.571 = -4 W
e)
overall rate = 6.28-4 = 2.28
rate of consuption of energy = I^2*Rnet = 0.571^2*7 = 2.28 <------same as overall rate