Suppose you have a simple circuit that includes a resistance device of 50 ohms, and the current flowing through it is 2.0 amps. Solve for the potential difference of the batteries.
1.0 E2 V
25 V
0.040 V
2.0 E2 V
E=IR=2*50
To solve for the potential difference of the batteries in the given circuit, you can use Ohm's Law, which states that the potential difference (V) across a resistor is equal to the current (I) flowing through it multiplied by the resistance (R) of the resistor (V = I * R).
In this case, the resistance is given as 50 ohms and the current is given as 2.0 amps. Plugging these values into the formula, we get:
V = I * R
V = 2.0 A * 50 Ω
Multiplying the numbers together:
V = 100 V
So the potential difference across the resistance device in the circuit is 100 volts.
Therefore, the correct answer is: 1.0 E2 V.