Generate an image representing a physics problem. Picture a clean workspace with a blank white paper. On top of the paper, there is a simplistic 2D circuit diagram drawn with a black pencil. The diagram features a detailed sketch of a 119.6 volts source that's connected to an indicated 6 ohms load through a pair of accurately depicted 0.25 ohms conductors. No numbers or text should appear on this image; only the drawn equipment and lines signify the values.

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.

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.

Pls draw the graph

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

a) Well, let's start with the load current. According to Ohm's Law, we can calculate it using the formula: I = V/R, where V is the voltage and R is the resistance. In this case, the voltage is 119.6 volts and the resistance of the load is 6 ohms. So, the load current would be 119.6V/6Ω. I hope it doesn't overload your brain, but the answer is... 19.93 amps (approximately)!

Now, let's move on to the load voltage. Since the resistance of the load is 6 ohms and the load current is 19.93 amps, we can use Ohm's Law once again to find the voltage across the load. Using the same formula, V = I x R, we have V = 19.93A x 6Ω. And voilà, the load voltage is approximately 119.6 volts!

b) Ah, the voltage drop in the wires! To calculate this, we need to consider the resistance of the conductors. In this case, each conductor has a resistance of 0.25 ohms. Since we already know the load current is 19.93 amps, we can calculate the voltage drop using Ohm's Law (yes, that superhero formula strikes again!). V = I x R, where V is the voltage drop, I is the current, and R is the resistance. For one wire, the voltage drop is approximately 19.93A x 0.25Ω. But since there are two wires, we need to double that amount! So, the voltage drop in the wires is approximately 9.97 volts (2 times 4.98 volts). It's not a shocking amount, just a little spark of voltage drop, if you will.

To calculate the load current and voltage, as well as the voltage drop in the wires, we can use Ohm's Law and apply it to the given values.

a) The load current can be calculated using Ohm's Law (I = V/R). Here, V represents the source voltage and R represents the load resistance. Let's substitute the values:

V = 119.6 volts
R = 6 ohms

Using Ohm's Law: I = V/R = 119.6 volts / 6 ohms = 19.9333... amps

Therefore, the load current is approximately 19.9333 amps.

To calculate the load voltage, we can use the formula V = I * R, where I is the load current and R is the load resistance:

I = 19.9333 amps
R = 6 ohms

V = I * R = 19.9333 amps * 6 ohms = 119.6 volts

Therefore, the load voltage is equal to the source voltage, which is 119.6 volts.

b) To calculate the voltage drop in the wires, we need to consider the resistance of the conductors. Given that the resistance of the conductors is 0.25 ohms, we can use Ohm's Law again to find the voltage drop.

Using Ohm's Law: V = I * R, where V is the voltage drop, I is the load current, and R is the resistance of the conductors:

I = 19.9333 amps
R = 0.25 ohms

V = I * R = 19.9333 amps * 0.25 ohms = 4.9833... volts

Therefore, the voltage drop in the wires is approximately 4.9833 volts.

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