A 6 ohms load is connected to a 119.6volts source through a pair of 0.25 ohms conductors. calculate a) the load current and voltage. b) the voltage drop in the wires.

Answer 1

I assume the wires are in series with the load, leading to the + and - sides of the power source.

total R = 6.5 ohms

i = V/R = 119.6 / 6.5 = 18.4 amps
V = i R = 18.4 * 6 = 110.4 volts across 6 ohm load
V = i R = 18.4 * .25 = 4.6 volts across each of the two wires.

Answer 2

Pls draw the graph

Answer 3

I think your total Resistance (R) is wrong. It would be 6.25 because 6ohms + .25 ohms is equal to 6.25.

I=V/R = 119.6/6.25 = 19.136 amps

V= 6×19.136=114.816v this across 6 ohms

V2= .25×19.136=4.784v this is across .25 ohms

For the graph pm mu ko Hahaha

Answer 4

total R = 6.5 ohms
REMINDER: a PAIR of .25 ohms so the total would be 6.5 ohms
a.
i = V/R = 119.6 / 6.5 = 18.4 amps
V = i R = 18.4 * 6 = 110.4 volts across 6 ohm load
b.
V = i R = 18.4 * .25 = 4.6V
V(total)= 4.6x2= 9.2 volts