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netineya
4 days ago
8

Four wires are made of the same highly resistive material, cut to the same length, and connected in series. Wire 1 has resistanc

e 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.
Physics
1 answer:
ValentinkaMS [3.3K]4 days ago
8 0

Answer:

V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}

Explanation:

The four wires are arranged in series: this setup indicates that the same current passes through them, while the total voltage from the battery, V0, equals the combined voltages across each resistor:

V_0=V_1+V_2+V_3+V_4

Additionally, the total resistance for this series configuration is

R_{eq}=R_1+R_2+R_3+R_4

Using Ohm's law, we can determine the voltage V2 across wire 2:

V_2 = R_2 I (1)

Here, I represents the entire current flowing through the circuit, which is calculated as:

I=\frac{V_0}{R_{eq}}=\frac{V_0}{R_1+R_2+R_3+R_4}

By substituting into equation (1), we derive V2:

V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}

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For our case, the area for a wire with 0.33 mm diameter, is the circular area of a 0.165 mm radius (0.000165 m), calculated as:

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