Question

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.

Answer 1

V = IR

I is the rate of flow in amps

R is the resistance in Ohms

V = (2.0)(50)

V = 100

The potential difference of the batteries is 1.0 E2 Volts

Answer 2

We can solve the problem by using Ohm's law, which can be written as:

`\Delta V=IR`
where

`\Delta V` is the potential difference across the resistor
I is the current flowing in the circuit
R is the resistance

In our circuit, I=2.0 A and the resistance is
`R=50 \Omega`, so the potential difference across the batteries (equal to the potential difference of the battery) is

`\Delta V = IR=(2.0 A)(50 \Omega)=100 V`