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Four wires are made of the same highly resistive material, cut to the same length, and connected in series. Wire 1 has resistance R1 and cross-sectional area A. Wire 2 has resistance R2 and cross-sectional area 2A. Wire 3 has resistance R3 and cross-sectional area 3A. Wire 4 has resistance R4 and cross-sectional area 4A. A voltage V0 is applied across the series, as shown in the figure.Find the voltage V2 across wire 2.Give your answer in terms of V0, the voltage of the battery.

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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]

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