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An overhead transmission cable for electrical power is 2000 m long and consists of two parallel copper wires, each encased in insulating material. A short circuit has developed somewhere along the length of the cable where the insulation has worn thin and the two wires are in contact. As a power company employee, you must locate the short so that repair crews can be sent to that location. Both ends of the cable have been disconnected from the power grid. At one end of the cable (point A), you connect the ends of the two wires to a9.00-V battery that has negligible internal resistance and measure that 2.86 A of current flows through the battery. At the other end of the cable (point B ), you attach those two wires to the battery and measure that 1.65 A of current flows through the battery. How far is the short from point A ?

Short Answer

Expert verified

The distance between the short circuit from the point A is 732.5 m.

Step by step solution

01

Define the ohm’s law and resistance ( R ).

According to Ohm’s law, the current flowing through the conductor is directly proportional to the voltage across the two points.

V=IR

Where, Iis current in ampere A, R is resistance in ohmsΩ and V is the potential difference voltV .

Also,V=EL

Where, Eis the electric field across conductor and Lis length of the conductor.

The ratio of V to I for a particular conductor is called its resistance R :

R=VIorρLA

Where,ρ is resistivity role="math" localid="1664170807826" Ω·m, Lis length in m and A is area in m2.

02

Determine the distance between the point .

Given that, the length Lof two wires is 2000 m, the current IAat point is A, current IBat point B is 1.65 A and the voltage V at both points is 9.0 V . Consider short circuit at point C , length between the points A and C is LCAand distance between the point B and C is LCB.

So,LCB=2000-LCA

AndρCA=ρCB

RCAALCA=RCBALCBRCBLCA=RCB2000-LCARCA2000-LCA=LCARCBLCA=2000RCARCA+RCB.......1

Substitute the values in equation

LCA=2000RCARCA+RCB=2000VIAVIA+VIB=20009.0/2.869.0/2.86+9.0/1.65=732.5m

Hence, the distance between the short circuit from the point A is 732.5m.

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Most popular questions from this chapter

The definition of resistivity (ρ=EJ) implies that an electrical field exist inside a conductor. Yet we saw that in chapter 21 there can be no electrostatic electric field inside a conductor. Is there can be contradiction here? Explain.

When switch Sin Fig. E25.29 is open, the voltmeter V reads 3.08 V. When the switch is closed, the voltmeter reading drops to 2.97 V, and the ammeter A reads 1.65 A. Find the emf, the internal resistance of the battery, and the circuit resistance R. Assume that the two meters are ideal, so they don’t affect the circuit.

Fig. E25.29.

An 18-gauge copper wire (diameter 1.02 mm) carries a current

with a current density of 3.2×106Am2. The density of free electrons for

copper is8.5×1028electrons per cubic meter. Calculate (a) the current in

the wire and (b) the drift velocity of electrons in the wire.

A typical small flashlight contains two batteries, each having an emf of1.5V, connected in series with a bulb having resistance17Ω. (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for1.5hwhat is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)

(a) What is the potential difference Vadin the circuit of Fig. P25.62? (b) What is the terminal voltage of the 4.00-Vbattery? (c) A battery with emf and internal resistance 0.50Ωis inserted in the circuit at d, with its negative terminal connected to the negative terminal of the 8.00-Vbattery. What is the difference of potential Vbcbetween the terminals of the 4.00-Vbattery now?

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