The circuit shown in the figure below is connected for 3.50 min. (Assume R 1 = 7

Question

The circuit shown in the figure below is connected for 3.50 min. (Assume R1 = 7.10 ?, R2 = 1.40 ?, and V = 18.0 V.)

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

                                      Magnitude (A)

Left branch

Middle Branch

Right Branch

(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.

This answer has not been graded yet.

(e) Find the total amount of energy transformed into internal energy in the resistors.
kJ

Explanation / Answer

on right side loop applying KVL,

18 - (3 + R2)i - (5 +1)i1 - 4 = 0

4.40i + 6i1 = 14 .............(i)

current in R1 will be (i - i1)

R1( i - i1) - (5 +1)i1 - 4 = 0

7.10i - 13.10i1 = 4 .... (ii)

from (i) and (ii) ,

i = 2.07 A

i1 = 0.816 A

a) left branch = i - i1 = 1.25 A

middle branch = i1 = 0.816 A

right branch i = 2.07 A

b) 4V :

P = VI = 4 x 0.816 =3.26 W

E = P t = 3.26 x 3.50 x 60 = 685.44 J

18V :

P = VI = 18 x 2.07 = 37.26 W

E =Pt =37.26 x 3.50 x 60 =7824 J = 7.82 kJ

c) 7.10 ohm

P = i^R

E = i^R t = 2329.68 J

5 ohm :

E = 699.15 J

1 ohm :

E = 139.83 J

3 ohm :

E = 2599.49 J

1.40 ohm

E = 1259.76 J

e) E = 7824 - 685.44 =7165.56 = 7.1 kJ

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