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A metal wire of resistance Ris cut into two pieces of equal length. The two pieces are connected together side by side. What is the resistance of the two connected wires?

Short Answer

Expert verified

The resistance of the two connected wires is Req=R4.

Step by step solution

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01

Given Information

We have to find the resistance of the two connected wires.

02

Simplify

The resistance through the wire depends on the area and the resistivity ρof the wire and the formula is given as

localid="1648650084630" R=ρLA...(1)

Where, Lis the length of the filament, Ais the area and ρis the resistivity of the material. As shown by equation (1), the resistance is directly proportional to the length

RαL

If the wire has length Land it is divided into two pieces, the length of each wire is L/2. The resistance will be halved as the length is halved, so it will be R/2for each wire.

As when both new wires are connected side by side, so they will be in parallel and this Equation shows the equivalent resistance for the parallel connection in the form

localid="1648650104413" 1Req=1R1+1R2+...+1RN...(2)

The two wires and each wire has resistance R/2, so we can get the equivalent resistance for the combination by using equation (2)in the form

1Req=1R1+1R21Req=1R/2+1R/2Req=4R-1

The equivalent resistance is Req=R4

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Most popular questions from this chapter

The circuit of FIGURE Q28.4 has two resistors, with R1>R2. Which of the two resistors dissipates the larger amount of power? Explain.

An oscillator circuit is important to many applications. A simple oscillator circuit can be built by adding a neon gas tube to an RC circuit, as shown in figureCP28.83. Gas is normally a good insulator, and the resistance of the gas tube is essentially infinite when the light is off. This allows the capacitor to charge. When the capacitor voltage reaches a value Von, the electric field inside the tube becomes strong enough to ionize the neon gas. Visually, the tube lights with an orange glow. Electrically, the ionization of the gas provides a very-low-resistance path through the tube. The capacitor very rapidly (we can think of it as instantaneously) discharges through the tube and the capacitor voltage drops. When the capacitor voltage has dropped to a value Voff, the electric field inside the tube becomes too weak to sustain the ionization and the neon light turns off. The capacitor then starts to charge again. The capacitor voltage oscillates between Voff, when it starts charging, and Von, when the light comes on to discharge it.

a. Show that the oscillation period is

T=RCinε-Voffε-Von

b. A neon gas tube has Von=80VandVoff=20V. What resistor value should you choose to go with a 10μfcapacitor and a 90Vbattery to make a 10Hzoscillator?

he switch in figure28.38a closes at t=0sand, after a very long time, the capacitor is fully charged. Find expressions for

(a) the total energy supplied by the battery as the capacitor is being charged,

(b) total energy dissipated by the resistor as the capacitor is being charged, and

(c) the energy stored in the capacitor when it is fully charged. Your expressions will be in terms of E, R, and C.

(d) Do your results for parts a to c show that energy is conserved? Explain.

A variable resistor Ris connected across the terminals of a battery. FIGURE EX28.21 shows the current in the circuit as Ris varied. What are the emf and internal resistance of the battery?

Show that the product RC has units of s.

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