The circuit shown in the figure below is connected for 2.90 min. (Assume R1 = 7.
ID: 2290601 • Letter: T
Question
The circuit shown in the figure below is connected for 2.90 min. (Assume R1 = 7.60 ?, R2 = 1.60 ?, and V = 13.0 V.)
(a) Determine the current in each branch of the circuit.
(b) Find the energy delivered by each battery.
4.00 V battery: _____ J
13.0 V battery: _____ kJ
(c) Find the energy delivered to each resistor.
7.60 ? resistor: _____ J
5.00 ? resistor: _____ J
1.00 ? resistor: _____ J
3.00 ? resistor: _____ J
1.60 ? resistor: _____ J
(d) Identify the type of energy storage transformation that occurs in the operation of the circuit.
(e) Find the total amount of energy transformed into internal energy in the resistors.
_____ kJ
branch magnitude (A) direction left branch middle branch right branchExplanation / Answer
consider current in left branch as i1 (upward direction.) and that in midle branch as i2 (upward direction) so current in right branch is (i1 + i2) (downward direction)
applying kirchoff's law:
1'st loop (left one)
7.6 * i1 - 6 * i2 + 4 = 0
2'nd loop (right one):
6 * i2 + 4.6 * (i1 + i2) +13 =0
on sovling for i1 and i2 we get
i1 = -1.1132 A
i2 = 0.4811 A
(a):
current in left branch = -1.1132 A
current in middle branch = 0.4811 A
current in right branch = -0.6321 A ............. that is (i1 + i2)
(b):
energy delivered by battery 4 V = V* i1 = 4.4528 jule
energy delivered by battery = 13*0.6321 = 0.0082173 kJ
(c):
energy delivered to 7.6 ohm = i1^2 * 7.6 = 9.4180 J
energy delivered to 5 ohm = i2^2 * 5= 1.1573 J
energy delivered to 1 ohm = i2^2 * 1 = 0.2315 J
energy delivered to 3 ohm = (i1+I2)^2 * 3 = 1.1987J
energy delivered to 1.6 ohm =(i1+I2)^2 * 1.6 = 0.6393 J
(D):
transformation of electrical energy into east energy through resistor
and in battery one can say, chem. energy into electrical energy
(e):
0.0082173 kJ
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