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7 -- Part 2: . Direct Current Circuits

- If the resistors are dissipating a total of 24 J every second, then they need a lot of power.
- It's easy to check: P is IV and A is V.

- The current will flow counterclockwise if the battery's emf is V 2.

- We use the overall emf, not the individual battery emfs.

- All real batteries have internal resistance.

- There is no conducting pathway from the positive terminal of the battery to the negative terminal before the switch is closed.
- When the switch is closed, the resistance of the circuit is 2 + 3 + 5 + 10, so the current in the circuit is 2 A.

- There is a distinction between the emf of the battery and the actual voltage it provides once the current has begun.
- The battery's effective voltage is lower than the ideal one because it drops across the internal resistance.
- It is V B - Ir - 20 V - 4 V - 16 V.

- A student has an ideal 90 V battery.
- Compare the current drawn from and the power supplied by the battery when the resistors are arranged in parallel.

- The equivalent resistance greater than any of the individual resistances is always provided by the resistors in series.

- The power will be supplied by the battery.

- If the resistors are in series, the equivalent resistance is 30 + 30 + 90, and the current drawn is only I + V. The power supplied by the battery is just P, IV, and 90 W.

- A voltmeter is a device used to measure the voltage between points in a circuit.
- An ammeter is used to measure.
- Determine the readings on the ammeter in the circuit below.

- Ammeters measure current and current stays the same in series.

- The voltage is measured in parallel because it stays the same in parallel.

- The ammeter is ideal because it doesn't alter the current that it's trying to measure.
- The voltmeter draws negligible current away from the circuit because it has an extremely high resistance.

- We want to find the equivalent resistance in the circuit.
- The 600 and 300 resistors are in close proximity.
- The batteries internal resistance is r, and R 3.
- The equivalent resistance is 50 + 200 + 150 + 400, so the current supplied by the battery is V/ R (2400 V)/(400 ) The current splits at the junction.
- Half as much current will flow through R 1 as through R 2 since R 1 is twice R 2.
- Matching voltages verify the values of currents I 1 and I 2 for each of these resistors.
- The ammeter is in the branch that contains R 2.

- The potential at point b is 900 V lower than at point R.

- The diagram below shows a point at potential V of 20 V connected by a combination of resistors to a point that is grounded.
- The ground is considered to be zero.

- We need another method for analyzing the circuit when the resistors are not in series or parallel.

- Any closed loop in a circuit must have zero potential differences.

- The total current that leaves the junction must be equal to the total current that enters the junction.

- By the time we get back to the same point by following any closed loop, we have to be back to the same potential.
- The total rise in potential must be equal to the total drop in potential.
- The Loop Rule says that all the decreases in electrical potential energy must be balanced by all the increases in electrical potential energy.
- The loop rule is a re-statement of the law of energy.

- It's a good idea to know that the total drop must equal the total rise in potential.
- No more and no less must be used if 60 V came out of the battery.

- The charge that goes into a junction must be equal to the charge that comes out.
- This is a statement about the law of charge.
- The Junction Rule is easy to apply.

- It doesn't matter if you choose clockwise or counterclockwise for the loop.
- The potential drops by IR when the loop goes across a Resistor in the same direction as the current.
- The potential increases when the loop goes across a Resistor in the opposite direction.
- The potential increases when the loop goes from the negative to the positive terminal.

- Let's start with the points in the circuit.

- junctions are the points c and f. We have three branches, two of which are fabc and cdef.
- Each branch has a current.
- The current in branch cdef I 2 must be I 1 - I 2 if we label the current in Fabc I 1 and the current in branch cdef I 2 with the directions shown in the diagram below.

- Pick a loop in the counterclockwise direction.
- The potential drops by I 1 R 1 when we go to b and then across R 1 in the direction of the current.
- The potential drops when we go up through R 2 in the direction of the current.
- The potential increases when we travel through V 1 from the negative to the positive terminal.

- Since we have two unknowns, we need two equations, so we need to pick another loop and choose cdefc in the counterclockwise direction.
- The potential drops by I 2 R 3 when we travel from c to d. The potential drops when we travel through V 2 from the positive to the negative terminal.
- We travel across R 2 but in the opposite direction to the current, so the potential increases.

- I 1 is the current through R 2.

- The direction of the currents at the beginning of the solution was arbitrary.
- Don't worry about trying to guess the direction of the current in a branch.
- Pick a direction and follow it.
- When you solve for the values of the branch current, a negative value will alert you that the direction of the current is different to the direction you chose in the diagram.

- You learned about the electric field they store in Chapter 6.
- Capacitors are often used in circuits.
- When a potential difference is applied to them, they are devices that store charges.

- Capacitors can be arranged in electric circuits.
- The parallel combination and the series combination are the same type of arrangement.

If they all share the same potential 888-609- 888-609- 888-609- 888-609- 888-609-

- The top and bottom plates are connected by a wire and form a single equipotential.
- The difference in potential between one and the other is the same.

- We want to find 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 to 888-353-1299 888-353-1299 to 888-353-1299 would 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 to 888-353-1299 would 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 888-353-1299 The charge on the firstCapacitor is Q 1 V and the charge on the secondCapacitor is C 2 V.

- Adding the individual capacitances to the collection ofCapacitors is the equivalent of a collection ofCapacitors in parallel.

- A collection of capacitors are 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 888-282-0476 This is similar to a resistors in a series.

- If a potential difference is applied, the negative charge will be deposited on the bottom plate of the bottomCapacitor and the negative charge will be pushed away from the top plate of the bottomCapacitor.

- The potential difference between the top and bottom capacitors is V 1 + Q + C 1, and the potential difference between them is V 2 + Q + C 2.
- The total potential difference is V 1 + V 2.

- Adding the reciprocals of the individual capacitances to the collection ofCapacitors in Series is how it is found.

- C 2 and C 3 are in the same series.
- The series combination of C 2 and C 3 is equivalent to the series combination of C 1 and C 2.

- Let's take a look at the case in which a Capacitor is first charged up and then has a Dielectric inserted between its plates.
- The voltage between the plates must match the battery's.
- V will not change.
- Q will increase by a factor of k if V doesn't change and C increases by a factor of k. Since V doesn't change and Q increases by a factor of k, the stored electric potential energy increases by a factor of k. The electric field strength is brought back to its original value by the increase in Q and the fact that the molecules of the dielectric are polarized.
- More electric potential energy is stored as more charge is transferred to the plates.

- The following figures show the effects on the properties of a capacitor.

- Capacitors can be charged by batteries.
- The electrons are attracted to the positive terminal of the battery when the switch is closed on the diagram on the left.
- The electrons accumulate on the bottom plate of theCapacitor until the voltage across theCapacitor plates matches the emf of the battery The current stops when this condition is reached.

- The circuit behaves like a parallel circuit.
- There is 12 V across the 6 Resistor so there is 2 Amp of current flow through it.
- There is 12 V across the 3 Resistor and 4 Amps flowing through it.
- The equivalent resistance of the circuit is 2 and the currents add 6A through the battery.

- There is still 12V across the 6resistor and 2amps of current still flow through it.
- Current cannot flow through that branch if theCapacitor behaves like an infinite Resistor.
- There won't be a current through the 3 Resistors.
- TheCapacitor has 12 V across it and there is no voltage across the 3 Resistors.
- The equivalent resistance of the circuit is 6 and the current will be 2 Amps through the battery.

- The 6 and 12 resistors are in close proximity and can be thought of as 4.
- The 4 and 8 resistors are in close proximity and can be thought of as 2.67 (from 1/ Req to R1 + 1/ R2 ).
- The equivalent resistance of the two branches is 6.67.

- The current through the battery is 3 Amps.

- The 6 and 12 have an effective resistance of 4.
- The top branch has a 12 V drop.
- 2 and 1 Amp are the currents through the 6 and 12 resistors.

- The 4 and 8 have an effective resistance of 2.67 and a current of 3 Amps.
- The 8 V drop across the branch is from the fact that the voltage across each series section must sum to the total voltage across the battery.
- 2 and 1 Amp are the currents through the 4 and 8 resistors.

- The 6 and 4 resistors are in a series and can be thought of as a 10 resistors.
- The 12 and 8 resistors can be thought of as 20 from Req.
- Since the branches are in parallel, the current can be calculated by the amount of power that goes through the left and right branches.

- The current through the battery is 3 Amps.

- Ohm's Law can be used to get the voltage across the 6, 4, 12, and 8 resistors.
- The numbers are 12 V, 8 V, 12 V, and 8 V.

- Chapter 12 contains solutions.

- The diameter of the brass wire is four times that of the silver wire.
- The resistivity of brass is 5 times greater than that of silver.

- There are three components in a circuit.

- A student wants to know the resistivity of copper.
- She has a wire that is known to be long.

- There is a constant voltage between points a and b.
- For a long time, the voltage has been flowing.

- Three identical light bulbs are connected to a source of emf.

- A simple DC circuit is set up.
- The graph shows the difference between the current and the voltage drop.
- The graph is proportional for an ideal battery.

- The resistance of an object can be determined by the material's resistivity, length, and cross-sectional area.
- The current is the rate at which the charge is transferred.
- Many objects obey the law.
- The electrical power in a circuit is given by P IV or P I2R, which is the same power we've encountered in our discussion of energy P. V1 + V2 are in a parallel circuit.
- The total current that enters a junction must be equal to the total current that leaves it.
- Simply add their capacitances to find the combined one in parallel.
- When aCapacitor is in a circuit and is completely discharged, it will act like a closed switch.
- It acts like an open switch when the capacitor is fully charged.

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