In conclusion, there are two requirements which must be met in order to establish an electric circuit. The requirements are: 1.There must be an energy supply capable doing work on charge to move it from a low energy location to a high energy location and thus creating an electric potential difference across the two ends of the external circuit. 2.There must be a closed conducting loop in the external circuit which stretches from the high potential, positive terminal to the low potential negative terminal.

RESISTORS IN SERIES - In a series circuit, the current is the same at all points along the wire. I T = I 1 = I 2 = I 3 - An equivalent resistance is the resistance of a single resistor that could replace all the resistors in a circuit. The single resistor would have the same current through it as the resistors it replaced. R E = R 1 + R 2 + R 3 - In a series circuit, the sum of the voltage drops equal the voltage drop across the entire circuit. V T = V 1 + V 2 + V 3

RESISTORS IN SERIES

12.4 Two resistances of 2 Ω and 4 Ω respectively are connected in series. If the source of emf maintains a constant potential difference of 12 V. a. Draw a schematic diagram with an ammeter and a voltmeter.

12.4 Two resistances of 2 Ω and 4 Ω respectively are connected in series. If the source of emf maintains a constant potential difference of 12 V. b. What is the current delivered to the external circuit? R e = R 1 + R 2 = = 6 Ω = 2 A c. What is the potential drop across each resistor? V 1 = I R 1 = 2(2) = 4 V V 2 = I R 2 = 2(4) = 8 V

PARALLEL CIRCUITS - In a parallel circuit, each resistor provides a new path for electrons to flow. The total current is the sum of the currents through each resistor. I T = I 1 + I 2 + I 3 - The equivalent resistance of a parallel circuit decreases as each new resistor is added. - The voltage drop across each branch is equal to the voltage of the source. V T = V 1 = V 2 = V 3

RESISTORS IN PARALLEL

KIRCHHOFF’S LAWS An electrical network is a complex circuit consisting of current loops. Kirchhoff developed a method to solve this problems using two laws. Law 1. The sum of the currents entering a junction is equal to the sum of the currents leaving that junction. This law is a statement of charge conservation.

A junction (j) refers to any point in the circuit where two or three wires come together. j

KIRCHHOFF’S LAWS Law 2. The sum of the emfs around any closed current loop is equal to the sum of all the IR drops around that loop. (Ohm’s Law: V = IR) This law is a statement of energy conservation.

Gustav Robert Kirchhoff ( )

12.5 The total applied voltage to the circuit in the figure is 12 V and the resistances R 1, R 2 and R 3 are 4, 3 and 6 Ω respectively. a. Determine the equivalent resistance of the circuit. R 2 and R 3 are in parallel (R P ) R p = 2 Ω R P and R 1 are in series R e = = 6 Ω

b. What is the current through each resistor? = 2 A I 1 = 2 A (series) V 1 = I 1 R 1 = 2(4) = 8 V The voltage across the parallel combination is therefore: = 4 V each = 1.33 A = 0.67 A

12.6 Find the equivalent resistance of the circuit shown. 1 and 2 are in series: 1+ 2 = 3 Ω this combination is in parallel with 6: this combination is in series with 3: = 5 Ω this combination is in parallel with 4: R P = 2.22 Ω = R eq R P = 2 Ω

12.7 A potential difference of 12 V is applied to the circuit in the figure below. a. Find the current through the entire circuit R 1 and R 2 are in parallel: R P = 2 Ω This combination is in series with R 3 : = 6 Ω This combination is now in parallel with R 4 : = 3 Ω = R eq

= 4 A I T = I 3 + I 4 = 2 A I 3 = I T - I 4 = = 2 A b. Find the current through each resistor.

The voltage for the parallel combination is: V' = V - I 3 R 3 = 12 - (2)(4) = 4 V = 1 AI 2 = = 1 A

EMF AND TERMINAL POTENTIAL DIFFERENCE Every source of emf ( Є ) has an inherent resistance called internal resistance represented by the symbol r. This resistance is a small resistance in series with the source of emf. The actual terminal voltage V T across a source of emf with an internal resistance is given by: V T = Є - I rUnits: Volts (V)

12.8 A load resistance of 8 Ω is connected to a battery whose internal resistance is 0.2 Ω a. If the emf of the battery is 12 V, what current is delivered to the load? Є = 12 V R L = 8 Ω r = 0.2 Ω = 1.46 A b. What is the terminal voltage of the battery? V T = Є - I r = (0.2) = 11.7 V

12.9 a. Determine the total current delivered by the source of emf to the circuit in the figure. V = 24 V. The resistances are 6, 3, 1, 2 and 0.4 Ω respectively. this combination is in series with R 3 : = 3 Ω R 1 and R 2 are in parallel: this combination is now in parallel with R 4 : R P = 1.2 Ω R P = 2 Ω

finally the internal resistance r is in series giving the equivalent resistance: R eq = = 1.6 Ω = 15 A

b. What is the current through each resistor? V T = Є - I r = (0.4) = 18 V V 4 = V T = 18 V = 9 A I 3 = I T - I 4 = = 6 A V 3 = I 3 R 3 = 6(1) = 6 V V 1 = V 2 = = 12 V each = 2 A = 4 A

Ammeter Voltmeter Rheostat Source of EMF Rheostat A Ammeters and Voltmeters V Emf - +

Galvanometer NS The galvanometer uses torque created by small currents as a means to indicate electric current. A current I g causes the needle to deflect left or right. Its resistance is R g.. The sensitivity is determined by the current required for deflection.

Operation of an Ammeter The galvanometer is often the working element of both ammeters and voltmeters. A shunt resistance in parallel with the galvanometer allows most of the current I to bypass the meter. The whole device must be connected in series with the main circuit. I = I s + I g RgRgRgRgI RsRsRsRs IsIsIsIs IgIgIgIg The current I g is negligible and only enough to operate the galvanometer. [ I s >> I g ]

Operation of an Voltmeter The voltmeter must be connected in parallel and must have high resistance so as not to disturb the main circuit. A multiplier resistance R m is added in series with the galvanometer so that very little current is drawn from the main circuit. V B = I g R g + I g R m RgRgRgRgI VBVBVBVB IgIgIgIg The voltage rule gives: RmRmRmRm

CAPACITORS IN SERIES AND PARALLEL These are the symbols used in different arrangements of capacitors:

CAPACITORS IN SERIES

 Series capacitors always have the same charge.  The voltage across the equivalent capacitor C eq is the sum of the voltage across both capacitors.

CAPACITORS IN PARALLEL

 Parallel Capacitors always have the same voltage drop across each of them.  The charge on the equivalent capacitor C eq is the sum of the charges on both capacitors.

19.9 a. Find the equivalent capacitance of the circuit. C 2 and C 4 are in series = 1.33 μF C 2 = 2 μF, C 3 = 3 μF, C 4 = 4 μF V = 120 V C 3 is now in parallel with C 2,4 C eq = C 3 + C 2,4 = = 4.33 μF

b. Determine the charge on each capacitor. The total charge of the system Q T = C eq V = 4.33 (120) = 520 μC Q 3 = C 3 V = 3(120) = 360 μC Q 2 and Q 4 have the same charge since they are in series: Q 2 = Q 4 = Q T - Q 3 = = 160 μC Q 3 = 360 μC, Q 2 = Q 4 =160 μC

c. What is the voltage across the 4 μF capacitor? = 40 V The remaining voltage ( = 80 V) goes through the C 2 capacitor.

RC CIRCUITS A resistance-capacitance (RC) circuit is simply a circuit containing a battery, a resistor, and a capacitor in series with one another. An RC circuit can store charge, and release it at a later time. A couple of rules dealing with capacitors in an RC circuit: 1. An empty capacitor does not resist the flow of current, and thus acts like a wire. 2. A capacitor that is full of charge will not allow current to flow, and thus acts like a broken wire.

When the switch is closed, the capacitor will begin to charge. RC Circuits

If an isolated charged capacitor is connected across a resistor, it discharges: RC Circuits

19.10 Three identical resistors, each with resistance R, and a capacitor of 1.0 x 10 ‑ 9 F are connected to a 30 V battery with negligible internal resistance, as shown in the circuit diagram above. Switches S I and S 2 are initially closed, and switch S 3 is initially open. A voltmeter is connected as shown.

a. Determine the reading on the voltmeter.

b. Switches S l and S 2 are now opened, and then switch S 3 is closed. Determine the charge Q on the capacitor after S 3 has been closed for a very long time.

After the capacitor is fully charged, switches S 1 and S 2 remain open, switch S 3 remains closed, the plates are held fixed, and a conducting copper block is inserted midway between the plates, as shown below. The plates of the capacitor are separated by a distance of 1.0 mm, and the copper block has a thickness of 0.5 mm.

c. i. What is the potential difference between the plates?

ii. What is the electric field inside the copper block?

iii. On the diagram, draw arrows to clearly indicate the direction of the electric field between the plates.

iv. Determine the magnitude of the electric field in each of the spaces between the plates and the copper block.