Answer:
[tex]V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}[/tex]
Explanation:
The 4 wires are connected in series: this means that the same current flow through them, and the voltage of the battery, V0, is equal to the sum of the voltages on each individual resistor:
[tex]V_0=V_1+V_2+V_3+V_4[/tex]
Also, the equivalent resistance of the series circuit is
[tex]R_{eq}=R_1+R_2+R_3+R_4[/tex]
The voltage V2 across wire 2 is given by Ohm's law:
[tex]V_2 = R_2 I[/tex] (1)
where I is the total current in the circuit, which is given by:
[tex]I=\frac{V_0}{R_{eq}}=\frac{V_0}{R_1+R_2+R_3+R_4}[/tex]
Substituting this into eq. (1), we find an expression for V2:
[tex]V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}[/tex]
