Question

The circuit shown in the figure below is connected for 1.70 min. (Assume R, = 7.30 2, R2 = 2.10 2, and V = 16.0 V.) 35.00 W 3.00 12 R $1.0022 + 4.00 V (a) Determine the current in each branch of the circuit. branch magnitude (A) direction left branch ---Select--- middle branch ---Select--- right branch ---Select-- (b) Find the energy delivered by each battery. 4.00 v battery 16.0 V battery (c) Find the energy delivered to each resistor. 7.30 resistor 5.00 resistor 1.00 resistor 3.00 2 resistor 2.10 2 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

Answer 1

Α- 3.00 Ω 5.00 Ω 2.10 4.00V 16V

by using Kirchoffs law

from loop 1

16 – 2.11 - 31 - 611 -4=0 = 12 – 5.11 – 611 = 0 = 61 +5.11 = 12

from loop 2

4+612 – 7.3(1-1) = 0 = 7.3 – 1 – 13.311 = 4

By solving both the equations we get
I = 1.64472A
,
11=0.601988 A
and
I - 11 = 1.042732A

so the current in left branch = 1.042732A and is directed downwards.

current in the middle branch = 0.601988A and is directed downwards

current in the right branch = 1.64472A and is directed upwards.

b)

Energy delivered by 4.0V battery

E=VI Xt= -4 x 0.601988 x (1.7minutes) = -4 x 0.601988 x 102 E= -245.611104)

negative sign is because the current is changing the battery instead of discharging from the battery.

Energy delivered by 16.0V battery

E=VI x t = 16 x 1.64472 x (1.7minutes) = 16 x 1.64472 x 102 E=2684.183043 = 2.68418304KJ

c)

Energy delivered to

7.392 = i Rt = (1-11)27.3x(1.7min) = 1.0427322 x 7.3x 102 = 809.5961J

Energy delivered to

5.092 = i Rt = (11)25 X (1.7min) = 0.6019882 x 5 x 102 = 184.8187)

Energy delivered to

1.022 = i- Rt = (11)21 x (1.7min) = 0.6019882 x 1 x 102 = 36.9637J

Energy delivered to

312 = 12 Rt = 12(3) < (1.7min) = 1.644722 x 3 x 102 = 827.7617J

Energy delivered to

2.122 = 1Rt = 14 (2.1) (1.7min) = 1.644722 x 2.1 x 102 = 579.4332J

d) In the operation of the circuit the some energy is used in charging the 4V battery which is transforming from electric to chemical energy. and remaining energy is transformed in to heat energy at the resistors.

e)The total enrgy stored in the internal energy of the resistors = 809.5961+184.8187+36.9637+827.7617+579.4332 = 2483.5734J = 2.4835734KJ