Curent, Resistance ,Direct-current Circuits

• Electric current• The current is the rate at which charge

flows through this surface• Suppose ΔQ is the amount of charge that

flows through an area A in a time interval Δt and that the direction of flow is perpendicular to the area. Then the current I is equal to the amount of charge divided by the time

• I= Δ Q/Δt• SI C/s=A (Ampere)

• Resistance and Ohm’s Law

• Resistance is the ration of the voltage across the conductor to the current:

• R = ΔV/ I

• SI units :1V/A=1Ω (Ohms)

• Ohm’s Law: the resistance remains constant over a wide range of applied voltage or currents:

• ΔV =IR

• A resistor : is a conductor that provides a specified resistance in an electric circuit

• Materials that obey Ohm’s law, are called ohmic

• Materials having resistance that changes with voltage or current are nonohmic (Ex: dioded = a circuit element that acts like one way valve for current)

• The resistance is proportional to the conductor’s length l and inversely proportional to its cross-sectional area A:

• R =ρ l/A• Ρ –resistivity of the material (ct. of

proportionality) SI unit: Ω m

• Electrical energy and power• The rate at which the system loses potential

energy as the charge passes through resistor is equal to the rate at which the system gains internal energy in the resistor.

• (ΔQ/ Δt) ΔV =I ΔV• Power P- the rate at which energy is

delivered to the resistor:• P =I ΔV• P =I2R = ΔV2/R• SI: kilowatt-hour • 1kWh =(103W)(3600s) =3.6x106J

• Sources of EMF (electromotive force)- a “ charge pump” that forces electrons to move in a direction opposite the electrostatic field inside the source

• The emf Σ of a source is the work done per unit charge (SI unit: V)

• ΔV =Σ –Ir• Σ = the terminal voltage when the current is zero

(open circuit voltage)• ΔV= IR (R- the external resistance)• Σ =IR +Ir; I =Σ/ (R+r)• IΣ =I2R + I2r if r<<R (the power delivered by the

battery is transferred to the load resistance)

• Resistors in series• ΔV = IR1 +IR2 =I(R1+R2)• ΔV = I Req

• IReq = I(R1+R2)• Req =R1+R2

• Req =R1+R2+….• The equivalent resistane of a series

combination of resistors is the algebraic sum of the individual resistances and is always greatter than any individual resistance

• Resistors in parallel• The potential differences across resistors are the

same because each is connected directly across the battery terminals

• Because charge is conserved, the current I that enters point a must equal the total current I1+I2 leaving that point

• I =I1+I2

• I1=ΔV /R1 ; I2=ΔV /R2 ;• I =ΔV /Req ; 1/Req = 1/R1 +1/R2 • The inverse of the equivalent resistance of two

or more resistors connected in parallel is the sum of the inverses of the individual resistances and is always less than the smallest resistance in the group

• Kirchhoff’s Rules:

• 1. The sum of the urrents entering any jonction must equal the sum of the currents leaving that junction (junction rule)

• 2. The sum of the potential differences across all the elments around any closed circuit loop must be zero (loop rule)

• a) if the resistor is traversed in the direction of the current, the charge in electric potential across the resistor is -IR

• b) If a resistor is traversed in the direction opposite the current, the charge in electric potential across resistors is +IR

• c) if a source of emf is traversed in the direction of the emf (from – to +) the charge in electric potential is +Σ

• d) if a source of emf is traversed in the direction opposite the emf, the charge in electric potential is -Σ

• Problem solving strategy:• 1. Assign labels and symbols to all the known

and unknown quantities• 2. Assign directions to the currents in each part

of the circuit• 3. Apply the junction rule to any junction in the

circuit• 4. Apply Kirchhoff’s loop rule, to as many loops

in the circuit as are needed to solve for the unknowns

• 5. solve the equations• 6. Check your answers by substituting them into

the original equations