The circuit shown in the figure below is connected for 1.70 min. (Assume R 1 = 8

Question

The circuit shown in the figure below is connected for 1.70 min. (Assume R1 = 8.30 , R2 = 1.50 , and V = 14.0 V.)

(a) Determine the current in each branch of the circuit.

(b) Find the energy delivered by each battery.

(c) Find the energy delivered to each resistor.

(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 2---Select---updown middle branch 4---Select---updown right branch 6---Select---updown   5.00 3.00 1.00 (2 4.00 V

Explanation / Answer

(a)

Apply KVL for right loop,

V - l*R2 - 3*l - 5*l1 - 1*l1 - 4 = 0

14 - 4 -1.5*l - 3*l - 5*l1 - 1*l1 = 0

4.5*l + 6*l1 = 10 .......eq1

Apply KVL for left loop,

-4 + R1(l - l1) - 5*l1 - 1*l1 = 0

-4 + 8.3*l - 8.3*l1 - 5*l1 - 1*l1 = 0

8.3*l - 14.3*l1 = 4 .........eq2

By solving equation 1 and 2,

l1 = 0.466 A (current in middle branch)

l = 1.6 A (current in right branch)

current in left branch = l - l1 = 1.13 A

(b)

Energy delivered by 4 V battery,

E = V*l*t = 4*0.466*1.7*60

E = 190.12 J

Energy delivered By 14 V battery,

E = 14*1.6*1.7*60

E = 2284.8 J = 2.28 kJ

(c)

Energy delivered to 8.3 ohm resistor,

E = l^2*R*t = (1.13)^2*8.3*1.7*60 = 1081.02 J

Energy delivered to 5 ohm resistor,

E = (0.466)^2*5*1.7*60 = 110.75 J

Energy delivered to 3 ohm resistor,

E = (1.6)^2*3*1.7*60 = 783.36 J

Energy delivered to 1.5 ohm resistor,

E = (1.6)^2*1.5*1.7*60 = 391.68 J

Energy delivered to 1 ohm resistor,

E = (0.466)^2*1*1.7*60 = 22.15 J

(e)

total amount of energy transformed into internal energy in the resistors,

Et = 2.38 kJ

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