# Electricity Resistance Voltage

1.

A wire 1 mm in diameter is connected to one end of a wire of the same material 2 mm in diameter of twice the length. A voltage source is connected to the wires and a current is passed through the wires. If it takes time T for the average conduction electron to traverse the 1-mm wire, how long does it take for such an electron to traverse the 2-mm wire?

Possible answers T/4 , T , 4T , 8T

2.

A wire has resistance R. A second wire has twice the length, twice the diameter, and twice the resistivity of the first wire. What is its resistance?

Possible answers 8R , R, R/4 , Resistance not given

3.

Resistor A has twice the resistance of resistor B. When individually connected across a given potential difference, which one dissipates the most power; and when connected in series across the same potential difference, which one dissipates the most power?

Possible answers A,A , A,B , B,B , B,A

4.

If a light bulb has half the resistance of a 100-W lightbulb, what would be its wattage? Assume both bulbs are attached to the same 120-V circuit.

Possible answers 200W , 50W , 25W , More information required

5.

When the voltage across a nonohmic resistor is doubled, the current through it triples. What happens to the power delivered to this resistor?

Possible answers This cannot be answered with the information given ,

Power decreases to 2/3 original amount , Power increases 1.5 times the original amount , The power increases to 6 times the original amount

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1.

A wire 1 mm in diameter is connected to one end of a wire of the same material 2 mm in diameter of twice the length. A voltage source is connected to the wires and a current is passed through the wires. If it takes time T for the average conduction electron to traverse the 1-mm wire, how long does it take for such an electron to traverse the 2-mm wire?

Possible answers T/4 , T , 4T , 8T

The speed of an electron is proportional to the ratio of the current to the resistance square:

(1.1)

The current is the ratio between the voltage V and the resistance R, therefore:

(1.2)

The resistance is proportional to the ratio between the resistor's length to it's cross section:

(1.3)

Where r is the wire's radius.

Therefore:

(1.4)

The time it takes the electron to move the length L is simply:

(1.5)

Therefore the time it will take the electron to move through the second wire with respect to the first is:

(1.6)

The answer is T/4

â€ƒ

2.

A wire has resistance R. A second wire has twice the length, twice the diameter, and twice the resistivity of the first wire. What is its resistance?

Possible answers 8R , R, R/4 , Resistance not given

As shown before the resistance is given by:

(2.1)

When we increase all these values by a factor of two we get a new resistance of:

(2.2)

The two wires resistances are equal to R.

3.

Resistor A has twice the resistance of resistor B. When individually connected across a given potential difference, which one dissipates the most power; and when connected in series across the same potential difference, which one dissipates the most power?

The power dissipated across a resistor is given by:

(3.1)

If they are connected individually across the same voltage source and then

(3.2)

When connected in series, the total resistance is

(3.3)

The current in the circuit (through both resistors) is:

(3.4)

Then:

(3.5)

In the first case more power is dissipated on resistor B, while in the latter case (series connection) more power is dissipated on resistor A (B,A).

4.

If a light bulb has half the resistance of a 100-W lightbulb, what would be its wattage? Assume both bulbs are attached to the same 120-V circuit.

Again, the power dissipated on a resistor is:

(4.1)

Then:

(4.2)

Where are the power and resistance of the 100W bulb. are the power and resistance of the unknown bulb.

We know that , therefore:

(4.3)

Then

â€ƒ

5.

When the voltage across a non-ohmic resistor is doubled, the current through it triples. What happens to the power delivered to this resistor?

Power is given by:

(5.1)

When we multiply the voltage by 2 and the current by 3 we get:

(5.2)

The power will increase by a factor of 6.

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