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ioda
3 years 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:
Paraphin [41]3 years ago
8 0

Answer:

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

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:

V_0=V_1+V_2+V_3+V_4

Also, the equivalent resistance of the series circuit is

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

The voltage V2 across wire 2 is given by Ohm's law:

V_2 = R_2 I (1)

where I is the total current in the circuit, which is given by:

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

Substituting this into eq. (1), we find an expression for V2:

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

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