The circuit shown in the figure below is connected for 2.30 min. (Assume R 1 = 8

Question

The circuit shown in the figure below is connected for 2.30 min. (Assume R1 = 8.40 ?, R2 = 1.40 ?, and V = 13.0 V.)

(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.

8.40 ? resistor____J

5.00 ? resistor____J

1.00 ? resistor_____J

3.00 ? resistor_____J

1.40 ? resistor_____J

4.00 V battery____J

13.0 V battery____KJ

(c) Find the energy delivered to each resistor.

8.40 ? resistor____J

5.00 ? resistor____J

1.00 ? resistor_____J

3.00 ? resistor_____J

1.40 ? resistor_____J

Explanation / Answer

suppose loop current in left side loop is i1 (counterclock wise ) and in right side loop is i2.
(counterclock wise )

Applying KVL in left side loop:

4 - 1(i1 - i2) - 5(i1 -i2) - 8.40i1 = 0

14.4i1 - 6i2 = 4 .........(i)

in right side loop:

13 - 1.4i2 - 3i2 - 5 (i2-i1) - 1(i2-i1) - 4 = 0

-6i1 + 10.4i2 = 9 ......(ii)

solving (i) and (ii),

i1 = 0.84 A

i2 = 1.35 A

current through battery 4 V = i1 - i2 = - 0.51 A

Power = VI = 4 x - 0.51 = - 2.04 W

energy = power x time = -2.04 x (2.30 x 60 s) = - 281.64 J .....Ans

current through 13 V = i2 = 1.35

energy = VIt = 13 x 1.35 x 2.30 x 60 =2421 J = 2.42 kJ ....Ans

I(through R1) = i1 = 0.84 A

energy = i^2 R t = 0.84^2 x 8.40 x (2.30 x 60) = 817.93 J

through 5 ohm, I = i2-i1 = 0.51 A

energy = 0.51^2 x 5 x (2.30 x 60) = 179.47 J

through 1 ohm

energy = 0.51^2 x 1 x (2.30 x 60) = 35.90 J

Through 3 ohm

energy = 1.35^2 x 3 x (2.30 x 60) = 754.51 J

through R2,

energy = 1.35^2 x 1.40 x (2.30 x 60) = 352.11 J

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