Final answer:
The Thevenin equivalent voltage of the circuit is 1.00 x 10^2 V and the Thevenin equivalent resistance is 0.027 ohm.
Step-by-step explanation:
To find the Thevenin equivalent voltage and resistance of the circuit, we need to determine the open-circuit voltage and the internal resistance. The open-circuit voltage is the voltage across the terminals of the loop when there is no load attached. In this case, the load is a 50 ohm resistor.
To find the open-circuit voltage, we can use Ohm's Law. The resistance of the loop is not mentioned in the question, so we'll assume it to be negligible. The voltage source provides a voltage of 1.00 x 10^2 V. The current flowing through the loop can be calculated by dividing the voltage by the resistance of the load: I = V/R = (1.00 x 10^2 V) / (50 ohm) = 2.00 A.
Since we have an open circuit, there is no current flowing through the loop. Therefore, the open-circuit voltage is equal to the voltage source voltage: V_oc = 1.00 x 10^2 V.
The Thevenin equivalent resistance can be found by short-circuiting the load and calculating the resistance seen by the load. In this case, the load is a 50 ohm resistor, so when it is short-circuited, the entire resistance of the loop is seen by the load.
Since the resistance of the loop is said to be one mile of copper wire, we can use the resistivity of copper to calculate the resistance. The resistivity of copper is approximately 1.68 x 10^-8 ohm/m. One mile is equal to 1609 meters, so the resistance of the loop would be: R_loop = (1.68 x 10^-8 ohm/m) * (1609 m) = 0.027 ohm. Therefore, the Thevenin equivalent resistance is 0.027 ohm.