Circuits Rs r Ig Is I r >> Rs A + - Chapter 27 R V r e I + -

Single-Loop Circuits To produce a steady flow of charge, you need a “charge pump,” a device that—by doing work on the charge carriers—maintains a potential difference between a pair of terminals. We call such a device an emf device, and the device is said to provide an emf , which means that it does work on charge carriers. Figure shows an emf device (consider it to be a battery) that is part of a simple circuit containing a single resistance R. The emf device keeps one of its terminals (called the positive terminal and often labeled +) at a higher electric potential than the other terminal (called the negative terminal and labeled -). We can represent the emf of the device with an arrow that points from the negative terminal toward the positive terminal as in Figure. A small circle on the tail of the emf arrow distinguishes it from the arrows that indicate current direction.

Single-Loop Circuits An emf device does work on charges to maintain a potential difference between its output terminals. If dW is the work the device does to force positive charge dq from the negative to the positive terminal, then the emf (work per unit charge) of the device is An ideal emf device is one that lacks any internal resistance. The potential difference between its terminals is equal to the emf. A real emf device has internal resistance. The potential difference between its terminals is equal to the emf only if there is no current through the device.

When an emf device is connected to a circuit, the device transfers energy to the charge carriers passing through it. This energy can then be transferred from the charge carriers to other devices in the circuit, for example, to light a bulb. Figure 27-2a shows a circuit containing two ideal rechargeable (storage) batteries A and B, a resistance R, and an electric motor M that can lift an object by using energy it obtains from charge carriers in the circuit. Note that the batteries are connected so that they tend to send charges around the circuit in opposite directions. The actual direction of the current in the circuit is determined by the battery with the larger emf, which happens to be battery B, so the chemical energy within battery B is decreasing as energy is transferred to the charge carriers passing through it. However, the chemical energy within battery A is increasing because the current in it is directed from the positive terminal to the negative terminal. Thus, battery B is charging battery A. Battery B is also providing energy to motor M and energy that is being dissipated by resistance R. Figure 27-2b shows all three energy transfers from battery B; each decreases that battery's chemical energy.

Calculating Current in a Single-Loop Circuits Energy Method Equation, P=i2R, tells us that in a time interval dt an amount of energy given by i2R dt will appear in the resistor (shown in the figure) as thermal energy. This energy is said to be dissipated. (Because we assume the wires to have negligible resistance, no thermal energy will appear in them.) During the same interval, a charge dq = i dt will have moved through battery B, and the work that the battery will have done on this charge is From the principle of conservation of energy, the work done by the (ideal) battery must equal the thermal energy that appears in the resistor: Which gives us  (Unless otherwise indicated, we assume that wires in circuits have negligible resistance. Their function, then, is merely to provide pathways along which charge carriers can move.)

Calculating Current in a Single-Loop Circuits Potential Method In the figure, let us start at point a, whose potential is Va, and mentally walk clockwise around the circuit until we are back at point a, keeping track of potential changes as we move. Our starting point is at the low-potential terminal of the battery. Because the battery is ideal, the potential difference between its terminals is equal to . When we pass through the battery to the high-potential terminal, the change in potential is + . After making a complete loop, our initial potential, as modified for potential changes along the way, must be equal to our final potential; that is, The value of Va cancels from this equation, which becomes Which gives us

Calculating Current in a Single-Loop Circuits (Kirchoff’s loop rule)

Single-Loop Circuits Internal Resistance Figure (a) shows a real battery, with internal resistance r, wired to an external resistor of resistance R. The internal resistance of the battery is the electrical resistance of the conducting materials of the battery and thus is an un removable feature of the battery. Figure (b) shows graphically the changes in electric potential around the circuit. Now if we apply the loop rule clockwise beginning at point a, the changes in potential give us Solving for the current we find,

Sources and Internal Resistance V r e i + - r e + - A B – i r e + - A B – i

Single-Loop Circuits – Resistance in Series Figure (a) shows three resistances connected in series to an ideal battery with emf . The resistances are connected one after another between a and b, and a potential difference is maintained across a and b by the battery. The potential differences that then exist across the resistances in the series produce identical currents i in them. To find total resistance Req in Fig. (b), we apply the loop rule to both circuits. For Fig. (a), starting at a and going clockwise around the circuit, we find or For Fig. (b), with the three resistances replaced with a single equivalent resistance Req, we find Equating them, we get,

Resistors in Series

Single-Loop Circuits Resistance in Series Answer: (a) current is same for all resistors in series. (b) V1 , V2 , and V3

Concept Check – Resistors in Series Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? 1. 12 V 2. zero 3. 3 V 4. 4 V 5. you need to know the actual value of R 9 V

Concept Check – Resistors in Series Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? 1. 12 V 2. zero 3. 3 V 4. 4 V 5. you need to know the actual value of R Since the resistors are all equal, the voltage will drop evenly across the 3 resistors, with 1/3 of 9 V across each one. So we get a 3 V drop across each. 9 V Follow-up: What would be the potential difference if R = 1 W, 2 W, 3 W ?

Concept Check – Resistors in Series (2) In the circuit below, what is the voltage across R1? 1. 12 V 2. zero 3. 6 V 4. 8 V 5. 4 V 12 V R1= 4 W R2= 2 W

Concept Check – Resistors in Series (2) In the circuit below, what is the voltage across R1? 1. 12 V 2. zero 3. 6 V 4. 8 V 5. 4 V The voltage drop across R1 has to be twice as big as the drop across R2. This means that V1 = 8 V and V2 = 4 V. Alternately, you could find the current i = V/R = (12 V)/(6 W) = 2 A, then use Ohm’s Law to get voltages. 12 V R1= 4 W R2= 2 W

Single-Loop Circuits – Potential Difference Potential Difference across a real battery: In the Figure, points a and b are located at the terminals of the battery. Thus, the potential difference Vb - Va is the terminal-to-terminal potential difference V across the batter and is given by: Grounding a Circuit: Grounding a circuit usually means connecting one point in the circuit to a conducting path to Earth’s surface (actually to the electrically conducting moist dirt and rock below ground) Power of emf Device: The rate Pemf at which the emf device transfers energy both to the charge carriers and to internal thermal energy is

Sample 27.01

Sample Problem 27.01

Multiloop Circuits 𝑖 1 + 𝑖 3 = 𝑖 2 Figure shows a circuit containing more than one loop. If we traverse the left-hand loop in a counterclockwise direction from point b, the loop rule gives us If we traverse the right-hand loop in a counterclockwise direction from point b, the loop rule gives us If we had applied the loop rule to the big loop, we would have obtained (moving counterclockwise from b) the equation which is the sum of two small loops equations. When there are multiple currents (unknowns), you must create as many equations as there are unknowns and solve the system of equations.

Multiloop Circuits Resistances in Parallel Figure (a) shows three resistances connected in parallel to an ideal battery of emf . . The applied potential difference V is maintained by the battery. Fig. b, the three parallel resistances have been replaced with an equivalent resistance Req. To derive an expression for Req in Fig. (b), we first write the current in each actual resistance in Fig. (a) as where V is the potential difference between a and b. If we apply the junction rule at point a in Fig. (a) and then substitute these values, we find If we replaced the parallel combination with the equivalent resistance Req (Fig. b), we would have and thus substituting the value of i from above equation we get,

Resistors in Parallel

Resistance and capacitors Multiloop Circuits Resistance and capacitors Answer: (a) Potential difference across each resistor: V/2 Current through each resistor: i (b) Potential difference across each resistor: V Current through each resistor: i/2

Concept Check – Resistors in Parallel In the circuit below, what is the voltage across R1? 1. 10 V 2. zero 3. 5 V 4. 2 V 5. 7 V 10 V R1= 5 W R2= 2 W

Concept Check – Resistors in Parallel In the circuit below, what is the voltage across R1? 1. 10 V 2. zero 3. 5 V 4. 2 V 5. 7 V The voltage across circuit elements in parallel is the same. Both resistors share the same potential difference across their ends. 10 V R1= 5 W R2= 2 W

Concept Check – Resistors in Parallel (2) In the circuit below, what is the current through R1? 1. 10 A 2. zero 3. 5 A 4. 2 A 5. 7 A 10 V R1= 5 W R2= 2 W

Concept Check – Resistors in Parallel (2) In the circuit below, what is the current through R1? 1. 10 A 2. zero 3. 5 A 4. 2 A 5. 7 A The voltage is the same (10 V) across each resistor because they are in parallel. Thus, we can use Ohm’s Law, V1 = i1 R1 to find the current i1 = 2 A. 10 V R1= 5 W R2= 2 W Follow-up: What is the total current through the battery?

Concept Check – Resistors in Parallel (2) Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit? 1. increases 2. remains the same 3. decreases 4. drops to zero

Concept Check – Resistors in Parallel (2) Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit? 1. increases 2. remains the same 3. decreases 4. drops to zero As resistors are added in parallel, the overall resistance of the circuit drops. Since V = iR, and V is held constant by the battery, when resistance decreases, the current must increase.

Concept Check – Resistors in Combination What happens to the voltage across the resistor R1 when the switch is closed? The voltage will: 1. increase 2. decrease 3. stay the same V R1 R3 R2 S

Concept Check – Resistors in Combination What happens to the voltage across the resistor R1 when the switch is closed? The voltage will: 1. increase 2. decrease 3. stay the same With the switch closed, the addition of R2 to R3 decreases the equivalent resistance, so the current from the battery increases. This will cause an increase in the voltage across R1 . V R1 R3 R2 S Follow-up: What happens to the current through R3?

Concept Check – Resistors in Combination (2) What happens to the voltage across the resistor R4 when the switch is closed? 1. increase 2. decrease 3. stay the same V R1 R3 R4 R2 S

Concept Check – Resistors in Combination (2) What happens to the voltage across the resistor R4 when the switch is closed? 1. increase 2. decrease 3. stay the same We just saw that closing the switch causes an increase in the voltage across R1 (which is VAB). The voltage of the battery is constant, so if VAB increases, then VBC must decrease! V R1 R3 R4 R2 S A B C

Concept Check – Resistors in Combination (2) Which resistor has the greatest current going through it? Assume that all the resistors are equal. 1. R1 2. both R1 and R2 equally 3. R3 and R4 4. R5 5. all the same V R1 R2 R3 R5 R4

Concept Check – Resistors in Combination (2) Which resistor has the greatest current going through it? Assume that all the resistors are equal. 1. R1 2. both R1 and R2 equally 3. R3 and R4 4. R5 5. all the same The same current must flow through left and right combinations of resistors. On the LEFT, the current splits equally, so i1 = i2. On the RIGHT, more current will go through R5 than R3 + R4 since the branch containing R5 has less resistance. V R1 R2 R3 R5 R4

Concept Check – The Node Rule What is the current in branch P? 1. 2 A 2. 3 A 3. 5 A 4. 6 A 5. 10 A 5 A 8 A 2 A P

Concept Check – The Node Rule What is the current in branch P? 1. 2 A 2. 3 A 3. 5 A 4. 6 A 5. 10 A The current entering the node in red is 8 A, so the current leaving must also be 8 A. One exiting branch has 2 A, so the other branch (at P) must have 6 A. 5 A 8 A 2 A P

Concept Check – Kirchhoff’s Rules (2) Which of the equations is valid for the circuit below? 1. 2 – i1 – 2i2 = 0 2. 2 – 2i1 – 2i2 – 4i3 = 0 3. 2 – i1 – 4 – 2i2 = 0 4. i3 – 4 – 2i2 + 6 = 0 5. 2 – i1 – 3i3 – 6 = 0 2 V 2  6 V 4 V 3  1  i1 i3 i2

Concept Check – Kirchhoff’s Rules (2) Which of the equations is valid for the circuit below? 1. 2 – i1 – 2i2 = 0 2. 2 – 2i1 – 2i2 – 4i3 = 0 3. 2 – i1 – 4 – 2i2 = 0 4. i3 – 4 – 2i2 + 6 = 0 5. 2 – i1 – 3i3 – 6 = 0 Equation 3 is valid for the left loop: The left battery gives +2V, then there is a drop through a 1W resistor with current i1 flowing. Then we go through the middle battery (but from + to – !), which gives –4V. Finally, there is a drop through a 2W resistor with current I2. 2 V 2  6 V 4 V 3  1  I1 I3 I2

Sample Problem 27.02

Sample Problem 27.02

Sample Problem 27.02

Sample Problem 27.02

Sample Problem 27.03

Sample Problem 27.03 =

Sample Problem 27.04

Galvanometer r G The input that causes a maximum deflection of the needle is called the full-scale current and is usually very small (ig ≈ 1mA).

The Ammeter and The Voltmeter An instrument used to measure currents is called an ammeter. To measure the current in a wire, you usually have to break or cut the wire and insert the ammeter so that the current to be measured passes through the meter. In the figure, ammeter A is set up to measure current i. It is essential that the resistance RA of the ammeter be very much smaller than other resistances in the circuit. Otherwise, the very presence of the meter will change the current to be measured. Rs r ig is i r >> Rs + -

The Ammeter and The Voltmeter A meter used to measure potential differences is called a voltmeter. To find the potential difference between any two points in the circuit, the voltmeter terminals are connected between those points without breaking or cutting the wire. In the Figure, voltmeter V is set up to measure the voltage across R1. It is essential that the resistance RV of a voltmeter be very much larger than the resistance of any circuit element across which the voltmeter is connected. This is to insure that only a negligible current passes through the voltmeter, otherwise, the meter alters the potential difference that is to be measured. R r r << R + - ig

RC Circuits 𝑞=𝐶ℰ 1− 𝑒 − 𝑡 𝑅𝐶 charging a capacitor Charging a capacitor: The capacitor of capacitance C in the figure is initially uncharged. To charge it, we close switch S on point a. This completes an RC series circuit consisting of the capacitor, an ideal battery of emf , and a resistance R. The charge on the capacitor increases according to in which 𝐶 =𝑞0 is the equilibrium (final) charge and RC=𝜏 is the capacitive time constant of the circuit. Figure: RC circuit 𝑞=𝐶ℰ 1− 𝑒 − 𝑡 𝑅𝐶 charging a capacitor The plot shows the buildup of charge on the capacitor of the above figure.

RC Circuits 𝑖= 𝑑𝑞 𝑑𝑡 = ℰ 𝑅 𝑒 − 𝑡 𝑅𝐶 (charging a capacitor) Charging a capacitor: During the charging, the current is And the voltage is: The product RC is called the capacitive time constant of the circuit and is represented with the symbol t. 𝑖= 𝑑𝑞 𝑑𝑡 = ℰ 𝑅 𝑒 − 𝑡 𝑅𝐶 (charging a capacitor) Figure: RC circuit 𝑉 𝐶 = 𝑞 𝐶 =ℰ 1− 𝑒 − 𝑡 𝑅𝐶 (charging a capacitor) 𝜏=𝑅𝐶 time constant The plot shows the buildup of charge on the capacitor of the above figure.

RC Circuits 𝑞= 𝑞 0 𝑒 − 𝑡 𝑅𝐶 (discharging a capacitor) Discharging a capacitor: Assume now that the capacitor of the figure is fully charged to a potential V0 equal to the emf of the battery. At a new time t=0, switch S is thrown from a to b so that the capacitor can discharge through resistance R. When a capacitor discharges through a resistance R, the charge on the capacitor decays according to where 𝑞0 (=𝐶𝑉0) is the initial charge on the capacitor. Figure: RC circuit 𝑞= 𝑞 0 𝑒 − 𝑡 𝑅𝐶 (discharging a capacitor) A plot shows the decline of the charging current in the circuit of the above figure.

RC Circuits 𝑖= 𝑑𝑞 𝑑𝑡 =− 𝑞 0 𝑅𝐶 𝑒 − 𝑡 𝑅𝐶 (discharging a capacitor) During the discharging, the current is 𝑖= 𝑑𝑞 𝑑𝑡 =− 𝑞 0 𝑅𝐶 𝑒 − 𝑡 𝑅𝐶 (discharging a capacitor) Figure: RC circuit A plot shows the decline of the charging current in the circuit of the above figure.

Sample Problem 27.05

Summary Emf Series Resistance Parallel Resistance Single-Loop Circuits The emf (work per unit charge) of the device is Series Resistance When resistances are in series Eq. 27-7 Eq. 27-1 Parallel Resistance When resistances are in parallel Single-Loop Circuits Current in a single-loop circuit: Eq. 27-24 Eq. 27-4 Power The rate P of energy transfer to the charge carriers is The rate Pr at which energy is dissipated as thermal energy in the battery is The rate Pemf at which the chemical energy in the battery changes is RC Circuits The charge on the capacitor increases according to During the charging, the current is During the discharging, the current is Eq. 27-14 3 Eq. 27-33 Eq. 27-34 Eq. 27-16 Eq. 27-40 Eq. 27-17