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Home , Capacitance

Capacitance A different kind of :

• Find the expression for the capacitance of a • The ratio C = Q/V is a conductor’s “self cylindrical capacitor consisting of two conductors,

capacitance” each of length L. One, a cylinder of radius R1 and • Units of capacitance: / = the other a shell of larger radius R2. • A capacitor is made of two conductors with Strategy for a capacitor problem: equal but opposite charge 1. Draw the capacitor shape • Capacitance depends on shape and distance 2. Pretend there is charge + and – Q on the plates 3. Determine the E field between the plates apart between the conductors 4. Use E to find the potenal difference across the plates • Special case: plate capacitor 5. Compute C=Q/V (derivation) 6. Important: Q should cancel out!

Work must be done to charge in circuits a capacitor What’s a circuit? 1. Wires are represented by lines • stored in a capacitor is related to the potential difference across the 2. The battery has its own symbol capacitor 3. The capacitor has its own symbol 4. Think of charge as water flowing through the wires from the positive to negative terminals (current) – The battery shares its charge with the capacitor plates

Capacitor connected to a battery Parallel and series • E is zero in a conductor (wires in circuits are conductors) • Parallel… there’s a fork – So there is equal potential all along the wires of a in the road circuit, but capacitor plates don’t touch, so the potential value can be different across a capacitor. • Series… No splitting in • A battery connected to a capacitor will move the current’s path charge from one plate to the other • Sometimes it’s not – The battery retains a across its terminals and will impose the same voltage across the capacitor obvious what is is • The amount of charge moved is Q=CV, where parallel and what is in C= the “equivalent capacitance” of the whole series. Example 24-7… circuit

1 If the area of the plates of a parallel- Which of the following statements plate capacitor is doubled, the about a parallel plate capacitor is false? capacitance is A. The two plates have equal charges of the same sign. A. not changed. B. The capacitor stores charges on the plates. B. doubled. C. The capacitance is proportional to the area C. halved. of the plates. D. The capacitance is inversely proportional to D. increased by a factor of 4. the separation between the plates. E. decreased by a factor of 1/4. E. A charged capacitor stores energy.

Beginning Chapter 25 • vs. – Alternating current is what comes out of the wall outlet, but direct current is easier to If C1 < C2 < C3 < C4 for the combination analyze so here we focus on “DC of capacitors shown, the equivalent circuits” (direct current circuits) capacitance is • Definition of current

A. less than C1. • SI Units of current

B. more than C4.

C. between C1 and C4.

Introducing Resistance Example 25-3

• Ohms law • Calculate the resistance per unit length – If there is a potential drop (change in of a copper wire with a diameter of potential) in a wire, that wire has some 1.628 mm (the resistivity is found in “resistance” table 25-1, for copper, it is 1.7X10-8 • Resistance of a conducting wire Ohm-meters). • Define “resistivity” (not to be confused with density, same symbol) • A Light bulb is an example of a “”

2 16 If 4.7 × 10 electrons pass a particular Energy in Circuits point in a wire every second, what is the current in the wire? • Change in depends on the charge moved from one potential to A. 4.7 mA another B. 7.5 A • Power = rate of potential energy loss C. 2.9 A • Heating Law D. 7.5 mA • Power delivered to a resistor can be E. 0.29 A found from Ohm’s Law

Emf () Not a force!! Real battery • Energy is required to maintain a current in a wire • A real battery is not ideal, meaning it • This energy supply is called “source of emf” has internal resistance r • Emf has units of (like potential) • An example of an ideal source of emf is a battery which maintains a constant potential difference across the terminals (converts chemical energy into electrical energy) • In an ideal battery, potential and emf are interchangeable

Example 25-6 (work in groups, Example 25-7 students will present) An 11 ohm resistor is connected across a real For a battery that has an emf equal to E battery of emf 6.00V and internal resistance and internal resistance equal to r, what 1.00 ohms. Find value of external resistance R should be a) the current placed across the terminals to obtain b) the terminal voltage of the battery maximum power to the resister? c) power supplied by the battery (total power) d) power delivered to the external resistor e) and the power delivered to the battery’s internal resistance. Hint: d and e should add up to c

3 in circuits Example 25-9

• Resistors in series An ideal battery supplies 12 V across the • Resistors in parallel parallel combination of resistors (4.0 Ω and 6.0 Ω) a) Find the equivalent resistance, • Example 25-8 b) the total current (coming out of the battery), c) the current through each resistor, d) the power delivered to each resistor and e) the power supplied by the battery.

What if the circuit is just too Example 25-10 complicated? An ideal battery supplies 12 V across the series • How do you analyze a circuit that can’t be replaced by equivalent resistances? combination of resistors (4.0 Ω and 6.0 Ω) KIRCHHOFF’s RULES a) Find the equivalent resistance, • You need to be able to solve systems of b) the total current in the circuit, equations (brush up on your algebra!) c) the potential drop across each resistor, – Organize your equations, be clear and specific d) the power delivered to each resistor and about what you are solving for and what the “knowns” are e) the power supplied by the battery. – PLEASE don’t plug in numbers until you have solved completely for the variables in question

How to use these rules to Kirchhoff’s rules (2): generate useful equations: 1. “junction rule” – At any junction in a circuit (when wires • Is the junction rule useful in single loop circuits? meet) when current can divide, the sum of the currents in must equal the sum of the Steps: currents out of the junction 1. Draw the circuit 2. “loop rule” 2. Label current arbitrarily in each branch of circuit – When any closed loop is traversed in a 3. Choose the (positive) direction around which you will traverse the loop circuit, the algebraic sum of the changes in potential around the loop must be zero 4. Generate as many equations as you have unknowns 5. Solve equations with algebra

4 Example 25-16

a) Find the current in each branch of the circuit. (to be drawn) Don’t plug in numbers until (I’ll just set up this problem and leave the the end!!! algebra for you to practice on your own)

Another example for you to try…

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