Electricity Magnetism Part II Current Electricity Table of

Electricity & Magnetism Part II – Current Electricity

Table of Contents Nature of Electrical Current AC v. DC Resistance (is not futile) Ohm’s Law Power Dissipated by a Resistor Series & Parallel Circuits

What is Current? Current is the flow of electrical charge. Current (Amperes) Charge (Coulombs) q I = /t Time (Seconds) Current does not flow spontaneously. Water does not flow through a pipe unless there is a difference in pressure at the ends of the pipe. Similarly, electrical current will not flow through a conductor unless there is a voltage difference pushing the flow.

AC/DC Current may be direct (DC), which means it continuously flows in the same direction through a wire…. . . Or current may be alternating (AC), which means that the direction of flow rapidly shifts back and forth. Batteries supply DC voltage; wall plugs AC. Either way, electron flow speed is actually incredibly slow, at about 0. 01 cm/s!

Resistance – property of a conductor which is a measure of how much the object hinders the flow of charge. Symbol: R Unit: Ohm (Ω) Resistance depends on both the material used as well as its shape.

Consider a wire of length L and cross-sectional area A: A I L In general, the longer L is and the narrower A is, the greater resistance R will be: R ~ L/A This proportionality becomes an equation when you include the constant ρ, which represents “resistivity”: R = ρ (L / A ) Note: Resistivity ρ is an intrinsic property of the material of the conductor, though it may depend on the temperature as well.

A note about current direction Note that on the previous slide, current was depicted going toward the right. What was really going on was that electrons were moving to the left. Wait, what? The convention for the direction of current was established in the 18 th century, who defined current to be the direction of the flow of positive charge. Today we know that charge carriers are electrons, whose charge is negative, yet we adhere to the century’s old convention. Guess who came up with this (as we know now, incorrect) convention? Hint: If you have his picture in your wallet, you’re probably pretty happy!

Thanks, Ben Franklin!

Ohm’s Law So electrical current increases with voltage, but decreases with resistance: I ~ V/R If current I is measured in amps, voltage V is measured in volts, and resistance R is measured in ohms, the constant of proportionality equals 1, and we can write I = V/R, or V = IR

Power Dissipated in a Resistor Recall the defining equations for power and voltage: P = W/t and V = W/q Solving the latter for W and substituting that expression into the former yields P = Vq/t But q/t = I So, the power dissipated by a resistor may be written P = IV

Example Problem What’s the resistance and current through a 60 -W light bulb plugged into a 110 -V wall socket? P = IV I = P/V = 60 W/110 V = 0. 545 A V = IR R = V/I = 110 V/0. 545 A = 2. 0 x 102 Ω

Circuits All functioning circuits are composed of one or more complete loops. Note: “Closed circuit” means a complete loop with no gaps or disconnections; “open circuit” means an incomplete loop.

“Closed Circuits” v. “Closed store” Note: the words “open” and “closed” pretty much mean the opposite thing when applied to circuits compared to when applied to stores. A “closed circuit” is one that is “open for business”!

Series Circuits If two or more components of a circuit lie along the same path, they are said to be “in series”. Consider a simple circuit with a power supply and two resistors in series: In this case, + V - R 1 Rtotal = R 1 + R 2

Example: Find the current going through this circuit as well as the voltage across each resistor. Rtotal = R 1 + R 2 = 20. 0 Ω + 120 V - 12. 0 Ω V = IR, so Itotal = V/Rtotal Itotal = 120 V/20 Ω = 6. 0 A 8. 0 Ω V 1 = Itotal. R 1 = (6 A)(12 Ω) = 72 V V 2 = Itotal. R 2 = (6 A)(8 Ω) = 48 V

+ 120 V - 12. 0 Ω 8. 0 Ω Note: We can use the same value of current I for Itotal, I 1, and I 2 since there’s only one path, and thus the current is the same everywhere.

V 1 = 72 V + 120 V - 12. 0 Ω 8. 0 Ω V 2 = 48 V Also note: the voltages across individual components in the circuit add up to the total voltage supplied.

So, for a series circuit: Total Resistance Total Voltage Total Current Rtotal = R 1 + R 2 + R 3 + … Vtotal = V 1 + V 2 + V 3 + … Itotal = I 1 = I 2 = I 3 = …

Parallel Circuits: What if more than one path is available for the current? R 1 R 2 + - V

Parallel Circuits: What if more than one path is available for the current? I 1 In this case, the current forks: ITotal splits into I 1 and I 2. R 1 I 2 R 2 ITotal + - V

Since both resistor #1 and resistor #2 have direct connection to the voltage supply, the voltage across both resistors equals the total voltage: R 1 V 1 = V 2 = VTotal R 2 + - V

Since both resistor #1 and resistor #2 have direct connection to the voltage supply, the voltage across both resistors equals the total voltage: R 1 V 1 = V 2 = VTotal R 2 + - VTotal

V 1 = VTotal R 1 Since both resistor #1 and resistor #2 have direct connection to the voltage supply, the voltage across both resistors equals the total voltage: V 1 = V 2 = VTotal R 2 + - VTotal

V 1 = VTotal R 1 V 2 = VTotal Since both resistor #1 and resistor #2 have direct connection to the voltage supply, the voltage across both resistors equals the total voltage: V 1 = V 2 = VTotal R 2 + - VTotal

So how does the total equivalent resistance RTotal relate to R 1 and R 2? R 1 R 2 + - V

So how does the total equivalent resistance RTotal relate to R 1 and R 2? V 1 = VTotal Apply Ohm’s Law to each individual resistor: I 1 R 1 V 2 = VTotal = I 1 R 1 I 2 Now solve both equations for current: I 1 = Vtotal/R 1 R 2 VTotal = I 2 R 2 I 2 = Vtotal/R 2 We know ITotal = I 1 + I 2, so ITotal Itotal = Vtotal/R 1 + Vtotal/R 2 + - V

So how does the total equivalent resistance RTotal relate to R 1 and R 2? V 1 = VTotal Apply Ohm’s Law to the whole circuit: I 1 R 1 V 2 = VTotal = ITotal. RTotal, or I 2 R 2 ITotal = Vtotal/RTotal Now set both our expressions for Itotal equal to each other: Vtotal/RTotall = Vtotal/R 1 + Vtotal/R 2 + - V

So how does the total equivalent resistance RTotal relate to R 1 and R 2? V 1 = VTotal Apply Ohm’s Law to the whole circuit: I 1 R 1 V 2 = VTotal = ITotal. RTotal, or I 2 R 2 ITotal = Vtotal/RTotal Now set both our expressions for Itotal equal to each other: Vtotal/RTotal = Vtotal/R 1 + Vtotal/R 2 + - V So, 1/RTotal = 1/R 1 + 1/R 2

Example Problem: What is the total equivalent resistance of the circuit below? 1/RTotal = 1/R 1 + 1/R 2 So, 1/RTotal = 1/12 Ω + 1/8 Ω 12. 0 Ω Solve for Rtotal: Rtotal = 4. 8 Ω 8. 0 Ω + - 120 V

Wait – How can a total resistance be less than the values of the individual resistors in the circuit? ? ? Consider a flowing river. Now imagine digging a narrow trench, parallel to the river. Even though the trench is narrow, you’ve now provided an avenue for the water to flow that it did not have before, so the total rate of water flow increases. Similarly, when you add a resistor in parallel to an existing circuit path, you’re adding a new avenue for current to flow. Since overall current goes up, total resistance goes down (even though you just added a resistor. )

Example Problem Continued: So what currents would the ammeters below measure? We’ve already calculated Rtotal = 4. 8 Ω. A 1 Ammeter A will measure ITotal: 12. 0 Ω Itotal = Vtotal/Rtotal = 120 V/4. 8 Ω = 25 A A 2 Ammeter A 1 will measure I 1: 8. 0 Ω A I 1 = Vtotal/R 1 = 120 V/12 Ω = 10 A Ammeter A 2 will measure I 2: + - 120 V I 2 = Vtotal/R 2 = 120 V/8 Ω = 15 A Note: ITotal = I 1 + I 2, as it should!

So, for a parallel circuit: Total Resistance Total Voltage 1/Rtotal = 1/R 1 + 1/R 2 + 1/R 3 + … Total Current Itotal = I 1 + I 2 + I 3 + … Vtotal = V 1 = V 2 = V 3 = …

Series v. Parallel Suppose each resistor in the circuits below is the filament of a light bulb. In which circuit will the bulbs glow brighter, or will all the bulbs glow the same? + V - Since each bulb gets the full voltage of the power supply, the bulbs in the parallel circuit will glow brighter.

Series v. Parallel What happens to the other bulbs if one bulb in each circuit goes out? + V -

Series v. Parallel What happens to the other bulbs if one bulb in each circuit goes out? + V - If one bulb goes out in the series circuit, there is no longer a closed loop, and all the bulbs go out.

Series v. Parallel What happens to the other bulbs if one bulb in each circuit goes out? + V - If one bulb goes out in the series circuit, there is no longer a closed loop, and all the bulbs go out. If one bulb goes out in the parallel circuit, the other bulbs remain on closed loops experiencing the same voltage, so they don’t change.

A combination circuit may contain some series and some parallel elements…

What is the current measured by the ammeter below? 10 Ω A 20 Ω 30 Ω 25 Ω + - 120 V

These two are in series, and can be replaced with a single resistor with value RT given by: RT = R 1 + R 2 = 10 Ω + 20 Ω = 30 Ω 10 Ω A 20 Ω 30 Ω 25 Ω + - 120 V

These two are in series, and can be replaced with a single resistor with value RT given by: RT = R 1 + R 2 = 10 Ω + 20 Ω = 30 Ω A 30 Ω 25 Ω + - 120 V

These two are in parallel, and can be replaced with a single resistor with value RT given by: 1/RT = 1/R 1 + 1/R 2: 1/RT = 1/(30 Ω) + 1/(30 Ω) Solve for RT: RT = 15 Ω A 30 Ω 25 Ω + - 120 V

These two are in parallel, and can be replaced with a single resistor with value RT given by: 1/RT = 1/R 1 + 1/R 2: 1/RT = 1/(30 Ω) + 1/(30 Ω) Solve for RT: RT = 15 Ω A 15 Ω 25 Ω + - 120 V

A 15 Ω 25 Ω + - 120 V

These two are in series, and can be replaced with a single resistor with value RT given by: RT = R 1 + R 2 = 25 Ω + 15 Ω = 40 Ω A 15 Ω 25 Ω + - 120 V

These two are in series, and can be replaced with a single resistor with value RT given by: RT = R 1 + R 2 = 25 Ω + 15 Ω = 40 Ω A 40 Ω + - 120 V

A 40 Ω V T = ITR T IT = VT /RT = (120 V)/(40 Ω) = 3 A + - 120 V

Consider the following combination circuit in which each resistor is the filament of an identical light bulb. What would happen if you removed light 1? Light 2? Light 3? 1 2 3