The circuit shown in the figure below is connected for 2.50 min. (Assume R1 = 8.

Question

The circuit shown in the figure below is connected for 2.50 min. (Assume R1 = 8.40 ?, R2 = 1.90 ?, and V = 14.0 V.)
http://www.webassign.net/serpse8/28-p-021-alt.gif


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

branch: magnitude (A) direction
left branch: ?
middle branch: ?
right branch: ?


(b) Find the energy delivered by each battery.
4.00 V battery: ? J
14.0 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
1.90 ? resistor: ? J



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

Explanation / Answer

applying kvl

assuming 1st loop anticlockwise

4 - 6i1 + 6i2 - 8.4i1 = 0

-> 14.4i1 - 6i2 = 4 -> 1

2nd loop anti clockwise

14 - 4.9i2 -6i2 + 6i1 - 4 = 0

-> 6i1 - 10.9i2 = -10 -> 2

solving 1 and 2 we get

i1 = 0.856A
i2 = 1.389A

a). left branch = 0.856A downwards (anticlockwise)
middle branch = 0.533A downwards
right branch = 1.389A upwards (anticlockwise)

b).
energy delivered by
4V battery = -4 * 0.533 = -2.132J
14Vbatter = 14 * 1.389 = 19.446J = 0.0194kJ

c). energy delivered
8.4 ohms= 0.856^2 * 8.4 = 6.15J
5ohms = 5*0.533^2 = 1.42J
1ohm = 1*0.533^2= 0.284J
3ohms = 3 * 1.389^2 = 5.79J
1.9ohms = 1.9 * 1.389^2 = 3.66J

d). 17.314J

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