Question

Problem (6) For the circuit diagrammed below, please (a) replace the 3.00 0,5.00 , 2.00 0,6.00 Q resistors with a single equivalent resistor, and state the value of the equivalent resistance. 20.00 ww 2.00 1200V Please find 3.00 6.00 60.0 V 5.00 (b) the power supplied by the ideal 120.0 V source, 7.00 and the power dissipated by the 5.00 resistor.

Answer 1

a) Req = (2+6)*(3+5)/(2 + 6 + 3 + 5)

= 4 ohms <<<<<<---------Answer

b) let V1 = 60 V
R1 = 7 ohm
V2 = 120 V
R2 = 20 ohms

let I1 is the current through R1 (towards right)
let I2 is the current through R2 (towards left)
let I3 is the current through Req (towards down)

using Kirchoff's junction law, I2 = I1 + I3 ----(1)

Apply KVL in left loop,

60 - 7*I1 + 4*I3 = 0 ------(2)

Apply KVL in right loop,

120 - 20*I2 - 4*I3 = 0 ------(3)

in solving the above three equations we get
I1 = 7.74 A
I2 = 6.29 A
I3 = -1.45 A

b) powee supplied by the ideal 120 V battery = V2*I2

= 120*6.29

= 755 W <<<<<<<<<<<-----------------Answer

c) current through 5 ohm resistor = I3/2

= 1.45/2

= 0.725 A

power dissipated by 5 ohm resistor = 0.725^2*5

= 2.63 W <<<<<<<<<<<-----------------------Answer