A conductor connected to a dry cell or battery has the necessary electric potential to influence the flow of charges towards one direction, hence producing current.
Battery + - Conductor Charges Direction of charges
These positive charges are not actual particles. They are called holes , vacant spaces where there should be an electron. The charge of a hole is +1.6 x 10 -19 C .
Electrons are known as negative charge carriers . Holes are known as positive charge carriers .
Current per unit of cross-sectional area of a conductor.
A vector quantity with the same direction as the motion of positive charge carriers.
Where: I = electric current (A) J = current density (A/m 2 ) n = charge concentration v d = drift velocity (m/s) e = charge of electron A = cross-sectional area of conductor(m 2 )
A 491 gauge copper wire has a nominal diameter of 0.64 mm. This wire carries a constant current of 1.67 A to a 4,910 watts lamp. The density of free electron is 8.5 x 10 28 electrons/m 3 . Find the current density and the magnitude of drift velocity.
Measure by placing the material between two plates with constant electric field ( E ) and taking the ratio of electric field and current density ( J ) .
Varies with temperature
Where: ρ = resistivity ( Ω m) E = electric field (N/c) J = current density (A/m 2 )
Where: ρ = resistivity ( Ω m) ρ 0 = resistivity at room temperature ( Ω m) T = temperature (Kelvin,K) T 0 = room temperature (K) α = coefficient of resistivity (K -1 )
What is the resistivity of iron at 200K? Use the values of resistivity (at room temperature) and temperature coefficient of the resistivity in the handout.
The current I (Ampere, A) is directly proportional to the potential difference V (Volt,V) with resistance R (ohms, Ω ) as the proportionality constant.
Resistors R 1 = 2.00 ohms, R 2 = 3.00 ohms and R 3 = 4.00 ohms are in series connection with a voltage source of 100.0 volts. Find the equivalent resistance, electric current and electric potential difference.
Resistors R 1 = 3.00 ohms, R 2 = 5.00 ohms and R 3 = 7.00 ohms are in parallel connection with a voltage source of 110.0 volts. Find the equivalent resistance, electric current and electric potential difference.
At any point in a circuit, the sum of the currents leaving the junction point is equal to the sum of all the current entering the junction point. (Using current direction).
KIRCHHOFF’S LAW R 2 + ε 1 + - R 1 + ε 2 + - R 3 Junction point I 1 I 3 I 2 +
R + - C S 1 S 2 ε + - Where: ε = Batteries (Emf) S 1 & S 2 = Switches R = Resistor C = Capacitor Open Close
65.
Charging a capacitor R + - C S 1 S 2 ε + - I I I I I closed open Where: V R = Potential difference across the resistor. V C = Potential difference across the capacitor. I
A resistor with resistance R=1.0 x 10 6 Ω , capacitor with capacitance C=2.2 x 10 -6 F, a voltage source with ε = 100 v, and a switch are all connected in a single loop series circuit. The switch is initially open. When the switch is closed, calculate: