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**Direct-Current Circuits**

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**Resistors in Series and Parallel**

As with capacitors, resistors are often in series and parallel configurations in circuits Series Parallel The question then is what is the equivalent resistance

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Resistors in Series Since these resistors are in series, we have the same current in all three resistors We also have that the sum of the potential differences across the three resistors must be the same as the potential difference between points a and b

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**Resistors in Series Then using We have that**

Now the equivalent resistor, R, will also have the same potential difference across it as Vab, and it will also have the same current I Equating these last two results, we then have that The equivalent resistance for a sequence of resistors in series is just the sum of the individual resistances

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Resistors in Parallel Here we have that the voltage across each resistor has to be the same (work done in going from a to b is independent of the path, independent of which resistor you go through)

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**Resistors in Parallel We now deal with currents through the resistors**

At point a the current splits up into three distinct currents We have that the sum of theses three currents must add to the value coming into this point We also have that The equivalent resistor, R, will have also have the current I going through it

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**Resistors in Parallel Using**

and combining with the previous equations, we then have or The inverse of the effective resistance is given by the sum of the inverses of the individual resistances

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**Solving Resistor Networks**

Make a drawing of the resistor network Determine whether the resistors are in series or parallel or some combination Determine what is being asked Equivalent resistance Potential difference across a particular resistance Current through a particular resistor

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**Solving Resistor Networks**

Solve simplest parts of the network first Then redraw network using the just calculated effective resistance Repeat calculating effective resistances until only one effective resistance is left

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**Solving Resistor Networks**

Given the following circuit What is the equivalent resistance and what is the current through each resistor We see that we have two resistors in parallel with each other and the effective resistance of these two is in series with the remaining resistor

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**Solving Resistor Networks**

Step 1: Combine the two resistors that are in parallel yielding Step 2: Combine the two resistors that are in series yielding

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**Solving Resistor Networks**

Current through this effective resistor is given by The current through the resistors in the intermediate circuit of Step 1 is also 3 Amps with the voltage drop across the individual resistors being given by

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**Solving Resistor Networks**

To find the current through the resistors of the parallel section of the initial circuit, we use the fact that both resistors have the same voltage drop – 6 Volts

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Consistency Check There is a check that can be made to see if the answers for the currents make sense: The power supplied by the battery should equal the total power being dissipated by the resistors The power being supplied by the battery is given by where I is the total current The power being dissipated by each of the resistors is given by

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Example 1 Two identical light bulbs are represented by the resistors R2 and R3 (R2 = R3 ). The switch S is initially open. If switch S is closed, what happens to the brightness of the bulb R2? a) It increases b) It decreases c) It doesn’t change The power dissipated in R2 is given by When the switch is closed neither V nor R changes So the brightness does not change

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Example 2 Two identical light bulbs are represented by the resistors R2 and R3 (R2 = R3 ). The switch S is initially open. What happens to the current I, after the switch is closed ? a) Iafter = 1/2 Ibefore b) Iafter = Ibefore c) Iafter = 2 Ibefore Initially the current is given by After the switch is closed the net resistance is given by The new current is then

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**Kirchoff’s Rules Not all circuits are reducible**

There is no way to reduce the four resistors to one effective resistance or to combine the three voltage sources to one voltage source

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**Kirchoff’s Rules First some terminology**

A junction, also called a node or branch point, is is a point where three or more conductors meet A loop is any closed conducting path

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**Kirchoff’s Rules Kirchoff’s Rules are basically two statements**

1. The algebraic sum of the currents into any junction is zero A sign convention: A current heading towards a junction, is considered to be positive, A current heading away from a junction, is considered to be negative Be aware that all the junction equations for a circuit may not be independent of each other

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Kirchoff’s Rules 2. The algebraic sum of the potential differences in any loop including those associated with emfs and those of resistive elements must equal zero Procedures to apply this rule: Pick a direction for the current in each branch If you picked the wrong direction, the current will come out negative

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Kirchoff’s Rules Pick a direction for traversing a loop – this direction must be the same for all loops Note that there is a third loop along the outside branches As with the junction equations not all the loop equations will be independent of each other.

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Kirchoff’s Rules Starting at any point on the loop add the emfs and IR terms An IR term is negative if we traverse it in the same sense as the current that is going through it, otherwise it is positive An emf is considered to be positive if we go in the direction - to +, otherwise it is negative Need to have as many independent equations as there are unknowns

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**Kirchoff’s Rules For loop I we have For loop II we have**

Junction equation at a gives us We now have three equations for the three unknown currents

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Kirchoff’s Rules Assume that the batteries are: e1 = 19 V; e2 = 6 V; e3 = 2 V and the resistors are: R1 = 6W; R2 = 4W; R3 = 4W; R4 = 1W you should end up with: I1 = 1.5 A; I2 = -0.5 A; I3 = 2.0 A The minus sign on I2 indicates that the current is in fact in the opposite direction to that shown on the diagram Complete details can be found here

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RC Circuits Up until now we have assumed that the emfs and resistances are constant in time so that all potentials, currents, and powers are constant in time However, whenever we have a capacitor that is being charged or discharged this is not the case Now consider a circuit that consists of a source of emf, a resistor and a capacitor but with an open switch With the switch open the current in the circuit is zero and zero charge accumulates on the capacitor

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**RC Circuits Now close the switch**

Initially the full potential will be across the resistor as the potential across the capacitor is zero since q is zero Initially the full potential is across the resistor The initial current in the circuit is then given by As the current flows a charge will accumulate on the capacitor At some time t, the current in the circuit will be I and the charge on the capacitor will be q

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**According to Kirchoff’s 2nd rule we have Using a counterclockwise loop**

RC Circuits According to Kirchoff’s 2nd rule we have Using a counterclockwise loop Solving for the current As time increases, the charge on the capacitor increases, therefore the current in the circuit decreases Current will flow until the capacitor has a charge on it given by

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**RC Circuits We remember that So we then have Rearranging we have**

Setting up the integration we have The resultant integration yields

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RC Circuits We exponentiate both sides of this last equation and rearrange to obtain where Qf is the final charge on the capacitor given by Ce The constant RC is known as the time constant of the circuit We see that the charge on the capacitor increases exponentially

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Example 3 I1 I3 I2 e R2 C R1 At t = 0 the switch is closed in the circuit shown. The initially uncharged capacitor then begins to charge. What will be the voltage across the capacitor a long time after the switch is closed? (a) VC = 0 (b) VC = e R2/(R1+ R2) (c) VC = e After a long time the capacitor is completely charged, so no current flows through it The circuit is then equivalent to a battery with two resistors in series The voltage across the capacitor equals the voltage across R2 (since C and R2 are in parallel)

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**RC Circuits The current in the circuit is given by and looks like**

Note that is also how the voltage across the resistor behaves

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**RC Circuits – Charging Summary**

For the simple RC circuit we have the following for the voltage drops across the capacitor and the resistor

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RC Circuits We now start from a situation where we have a charged capacitor in series with a resistor and an open switch The capacitor will now act as a source of emf, but one whose value is not constant with time

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**RC Circuits We now close the switch and a current will flow**

Kirchoff’s 2nd rule gives us Rearranging we have To find q as a function of time we integrate the above equation

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RC Circuits Exponentiation of both sides of the equation on the right yields We see that the charge on the capacitor decreases exponentially

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RC Circuits The current in the circuit is obtained by taking the derivative of the charge equation The quantity Q0 / C is just the initial voltage, Vo , across the capacitor But then V0 / R is the initial current I0 So we then have that The voltage across the resistor is given by

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Example 4 The two circuits shown below contain identical fully charged capacitors at t = 0. Circuit 2 has twice as much resistance as circuit 1. Compare the charge on the two capacitors a short time after t = 0 a) Q1 > Q2 b) Q1 = Q2 c) Q1 < Q2 Initially, the charges on the two capacitors are the same. But the two circuits have different time constants: t1 = RC and t2 = 2RC Since t2 > t1 it takes circuit 2 longer to discharge its capacitor Therefore, at any given time, the charge on capacitor 2 is larger than that on capacitor 1

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**Example 5 a) b) c) (a) t0 < ta (b) t0 = ta (c) t0 > ta R 3R C**

The capacitor in the circuit shown is initially charged to Q = Q0. At t = 0 the switch is connected to position a. At t = t0 the switch is immediately flipped from position a to position b. a) Which of the following graphs best represents the time dependence of the charge on C? a) b) c) b) Which of the following correctly relates the value of t0 to the time constant ta while the switch is at a? (a) t0 < ta (b) t0 = ta (c) t0 > ta

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Example 5 C a b R 3R The capacitor in the circuit shown is initially charged to Q = Q0. At t = 0 the switch is connected to position a. At t = t0 the switch is immediately flipped from position a to position b. a) Which of the following graphs best represents the time dependence of the charge on C? a) b) c) For 0 < t < t0, the capacitor is discharging with time constant t = RC For t > t0, the capacitor is discharging with time constant t = 3RC, i.e., much more slowly Therefore, the answer is a)

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**Example 5 a) b) c) (a) t0 < ta (b) t0 = ta (c) t0 > ta R 3R C**

The capacitor in the circuit shown is initially charged to Q = Q0. At t = 0 the switch is connected to position a. At t = t0 the switch is immediately flipped from position a to position b. a) b) c) b) Which of the following correctly relates the value of t0 to the time constant ta while the switch is at a? (a) t0 < ta (b) t0 = ta (c) t0 > ta We know that for t = ta, the value of the charge is e-1 = 0.37 of the value at t = 0 Since the curve shows Q(t0) ~ 0.6 Q0, t0 must be less than ta

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**Capacitors Circuits, Qualitative**

Basic principle: Capacitor resists rapid change in Q resists rapid changes in V Charging It takes time to put the final charge on Initially, the capacitor behaves like a wire (DV = 0, since Q = 0). As current starts to flow, charge builds up on the capacitor it then becomes more difficult to add more charge the current decreases After a long time, the capacitor behaves like an open switch. Discharging Initially, the capacitor behaves like a battery. After a long time, the capacitor behaves like a wire.

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