# For the circuit, V1 = 12.3 V and V2 = 6.4 V. Determine the current in the 4.0- \Omega resistor....

## Question:

For the circuit, {eq}V1 = 12.3{/eq} {eq}V{/eq} and {eq}V2 = 6.4{/eq} {eq}V{/eq}. Determine the current in the {eq}4.0- \Omega{/eq} resistor. (Give your answer to the nearest {eq}0.001{/eq} {eq}A{/eq}).

## The relation between the current and voltage:

First we need to understand the relation between the current and voltage in electric circuit.

Suppose {eq}R {/eq} is the resistance of the circuit and {eq}V = V_1 - V_2 {/eq} is the voltage drop between two points that is applied in the circuit, then we can write the relation between the current and voltage that is given by: {eq}\displaystyle I = \frac{V}{R} {/eq} where {eq}I {/eq} is the current which is flowed in the circuit.

Given:

• The Voltage drop between two points is: {eq}V = 12.3 - 6.4 = 5.9 \ \rm V {/eq}.
• The resistance in the circuit is: {eq}R = 4 \ \rm Ohm {/eq}.

Now we will compute the current in the circuit.

As we know the formula for the current is:

\begin{align*} \displaystyle I &= \frac{V}{R} &\text{(Where } V = 5.9 \ \rm V \text{ and } R = 4 \ \rm Ohm \text{)}\\ &= \frac{5.9}{4} \\ I \ &\boxed{= 1.475 \ \rm I } \end{align*}