DC Electrical Circuits Chapter 28 Electromotive Force Potential

DC Electrical Circuits Chapter 28 Electromotive Force Potential Differences Resistors in Parallel and Series Circuits with Capacitors

Resistors in Series a R 1 V 1 I c R 2 b V 2 The pair of resistors, R 1 and R 2, can be replaced by a single equivalent resistor R; one which, given I, has the same total voltage drop as the original pair. Note: the current I is the same, anywhere between a and b, but there is a voltage drop V 1 across R 1, and a voltage drop V 2 across R 2.

Resistors in Series a R 1 V 1 I R 2 b V 2 The pair of resistors can be replaced by a single equivalent resistor Req; one which, given I, has the same total voltage drop as the original pair. V = V 1 +V 2 = I R 1 +I R 2

Resistors in Series a R 1 V 1 I R 2 b V 2 The pair of resistors can be replaced by a single equivalent resistor Req; one which, given I, has the same total voltage drop as the original pair. V = V 1 +V 2 = I R 1 +I R 2 We want to write this as V= I Req

Resistors in Series a R 1 V 1 I R 2 b V 2 The pair of resistors can be replaced by a single equivalent resistor Req; one which, given I, has the same total voltage drop as the original pair. V = V 1 +V 2 = I R 1 +I R 2 We want to write this as V= I Req hence Req = R 1 + R 2

Resistors in Series a R 1 V 1 I R 2 b V 2 What is the voltage drop across each resistor ? V 1= I R 1 but I = V / (R 1 + R 2) V 1 = V R 1 / (R 1 + R 2) V 2= I R 2 but I = V / (R 1 + R 2) V 2 = V R 2 / (R 1 + R 2)

Resistors in Parallel R 1 a I I 1 I 2 R 2 b V Again find the equivalent single resistor which has the same V if I is given.

Resistors in Parallel R 1 a I I 1 I 2 R 2 b V Again find the equivalent single resistor which has the same V if I is given. Here the total I splits: I = I 1+I 2 = V / R 1 + V / R 2 = V(1/ R 1 +1/ R 2)

Resistors in Parallel R 1 a I I 1 I 2 R 2 b V Again find the equivalent single resistor which has the same V if I is given. Here the total I splits: I = I 1+I 2 = V / R 1 + V / R 2 = V(1/ R 1 +1/ R 2) We want to write this as: I = V / Req

Resistors in Parallel R 1 a I I 1 I 2 R 2 b V Again find the equivalent single resistor which has the same V if I is given. Here the total I splits: I = I 1+I 2 = V / R 1 + V / R 2 = V(1/ R 1 +1/ R 2) We want to write this as: I = V / Req Hence 1 / Req =1/ R 1 +1/ R 2

Parallel and Series Resistors Capacitors 1/R=1/R 1+1/R 2 C=C 1+C 2 R=R 1+R 2 1/C=1/C 1+1/C 2 Parallel Series

Batteries and Generators • Current is produced by applying a potential difference across a conductor (I=V/R) [This is not equilibrium so there is an electric field inside the conductor]. • This potential difference is set up by some source, such as a battery or generator [that generates charges, from some other type of energy, i. e. chemical, solar, mechanical]. • Conventionally an “applied voltage” is given the symbol E (units: volts). • For historical reasons, this applied voltage is often called the “electromotive force” (emf) [even though it’s not a force].

The Voltaic Pile Volta’s original battery Carbon Ag Zn wet cloth - + Zinc case - + electrical converter. . . . converts chemical energy to electrical energy Mixture of Ammonium Chloride& Manganese Dioxide

Electrical Description of a Battery I symbol E for battery + - R symbol for resistance • A battery does work on positive charges in moving them to higher potential (inside the battery). • The EMF E, most precisely, is the work per unit charge exerted to move the charges “uphill” (to the + terminal, inside) • . . . but you can just think of it as an “applied voltage. ” • Current will flow, in the external circuit, from the + terminal, to the – terminal, of the battery.

Resistors in Series R 1 + - I R 2 E Given Req = R 1 + R 2, the current is I=E/(R 1 + R 2) One can then work backwards to get the voltage across each resistor:

Resistors in Series R 1 + - I R 2 E Given Req = R 1 + R 2, the current is I=E/(R 1 + R 2) One can then work backwards to get the voltage across each resistor:

The Loop Method R 1 + - I R 2 E Go around the circuit in one direction. If you pass a voltage source from – to +, the voltage increases by E (or V). As you pass a resistor the voltage decreases by V = I R. The total change in voltage after a complete loop is zero.

Analyzing Resistor Networks 1. Replace resistors step by step. E + - 6 W R 1 6 W - I 3 W E = I (R 1 + R 2) 5 W I I=E/R E E – IR 1 – IR 2 = 0 2 W + E - R 2 + - 2 W E + 2. The loop method I = E / (R 1 + R 2)

Analyzing Resistor Networks Often you can replace sets of resistors step by step. 6 W E+ - 6 W 2 W

Analyzing Resistor Networks Often you can replace sets of resistors step by step. 6 W E+ - 6 W 2 W 1/6+1/6=1/(3) E+ 3 W - 2 W step 1

Analyzing Resistor Networks Often you can replace sets of resistors step by step. 6 W E+ - 6 W 2 W 1/6+1/6=1/(3) E+ - 3 W 3+2=5 E+ 5 W - 2 W step 1 step 2

Internal Resistance of a Battery r battery E + R - • One important point: batteries actually have an internal resistance r • Often we neglect this, but sometimes it is significant.

Effect of Internal Resistance Analyzed by the Loop Method I r E + R - Start at any point in the circuit. Go around the circuit in a loop. Add up (subtract) the potential differences across each element (keep the signs straight!). E - Ir - IR = 0 (using V=IR) I = E / (R + r) VR = I R = E R / (R + r) = E / [1 + (r/R)] if r << R VR = E