Complete question:
Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.
Answer:
The electric field inside this metal resistor is 3125 V/m
Explanation:
Given;
length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m
diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m
the potential difference due to electric field between the two ends of the resistor, V = 10 V
The electric field inside this metal resistor is given by;
ΔV = EL
where;
ΔV is change in electric potential
E = ΔV / L
E = 10 / (3.2 x 10⁻³ )
E = 3125 V/m
Therefore, the electric field inside this metal resistor is 3125 V/m
Copper, because it has the lowest specific electrical resistance.
specific electrical resistance aka volume resistivity is a fundamental property of a material that quantifies how strongly that material opposes the flow of electric current. A low resistivity indicates a material that readily allows the flow of electric current.
1) Current in each bulb: 0.1 A
The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:
And so, the current through the circuit is (using Ohm's law):
And since the two bulbs are connected in series, the current through each bulb is the same.
2) 4 W and 8 W
The power dissipated by each bulb is given by the formula:
where I is the current and R is the resistance.
For the first bulb:
For the second bulb:
3) 12 W
The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:
Diagram B .... light shines through at an angle
A bridge supported by vertical cables which then leads to more support from larger cables.
