The circuit shown in the figure below is connected for 2.50 min. (Assume R 1 = 7

Question

The circuit shown in the figure below is connected for 2.50 min. (Assume R1 = 7.70 ?, R2 = 2.60 ?, and V = 10.0 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.

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

Let the current flowing in the left loop in the anti-clockwise direction be i1 and the current in the right loop flowing in the anti-clockwise direction be i2

Conserving potential drop in the left loop:

4 - (i1-i2) x (5 +1) - i1 x 7.7 =0

or, 13.7 x i1 - 6 x i2 = 4 ------- Eqn 1

Conserving potential drop in the right loop:

10 - i2 x (2.6 +3) + (i1 - i2) x (5+1) - 4 = 0

or, 6 - 11.6 x i2 + 6 x i1 = 0 -----Eqn 2

Solving for i1 and i2

i1 = 0.67 A and i2= 0.86 A

Therefore, current in the left branch = 0.67 A

Current in the middle branch = 0.86 - 0.67 = 0.19 A

Current in the right branch = 0.86 A

Energy delivered by the 4 V battery = 4 x 0.19 x 2.5 x 60 J = 114 J

Energy delivered by the 10 V battery = 10 x 0.86 x 2.5 x 60 J = 1.29 kJ

Energy delivered at 7.7 ohm resistor = 7.7 x 0.672 = 3.457 J

Energy delivered at 5 ohm resistor = 5 x 0.192 = 0.1805 J

Energy delivered at 1 ohm resistor = 5 x 0.192 = 0.0361 J

Energy delivered at 3 ohm resistor = 3 x 0.862 = 2.2188 J

Energy delivered at 2.6 ohm resistor = 2.6 x 0.862 = 1.92296 J

In the circuit the chemical energy of the battery is converted into electrical energy and vice versa.

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