University:
Kentucky State University
Course:
PHY 212 | General Physics II
Academic year:

2024
Views:

171

Pages:

10
Author:

lankstoetp11

DC Currents • In DC (direct current) circuit, currents flow in one direction. • we will not encounter AC (alternating currents) • analysis of ac current is more complex! Electromotive Force  & Battery How to model a real battery (voltage source) ? • real battery consists of ideal battery + internal resistance Hmm …  is NOT a force, it is a misleading term!  has unit of volt. The real “battery” is everything inside the green box. The battery is the power supply in the circuit. Voltage of ideal battery is called electromotive force (emf)  a + - r  b internal resistance r Example 1: a battery is known to have an emf of 9 volts. If the voltage across the resistor R (the load) is measured to be 3 volts. What is the internal resistance of the battery? R=1  Recall Ohm’s law V = IR: I a ε - Ir - IR =0 Given: IR = 3 volts ≡ Vab emf + - r  b ε - Ir - IR = 0 Ir = ε - IR ε - IR r= I ε r= -R I Vab = IR Vab I= R εR r= -R Vab  ε  r = R - 1  Vab  9  r = 1  - 1  =  3- 1 = 2 3  I R=1  emf a + - b As a battery ages, its r increases until you have a dead “out of juice” battery! A rather unrealistically large value for the internal resistance of a 9V battery. Example 2 : A 12 V battery with 2  internal resistance is connected to a 4  resistor. Calculate (a) the rate at which chemical energy is converted to electrical energy in the battery, (b) the power dissipated internally in the battery, and (c) the power output of the battery. R=4 +- I r=2  = 12 V Calculate (a) the rate at which chemical energy is converted to electrical energy in the battery. I R=4 +- r=2  = 12 V  - I R2 - I R4 = 0 I =  / (R2 + R4) = 12 V / 6  = 2 A Energy is converted at the rate Pconverted=I=(2 A)(12 V)=24W. 24W is the total power supplied by the battery. Part of it will be “constructively” delivered to the load R, while the leftover will be “destructively” delivered to r. The electrical engineer is eager to optimize PR i.e. to minimize Pr . Notice the values of parts (b) & (c). Calculate (b) the power dissipated internally in the battery. R=4 +- I=2A r=2  = 12 V Pdissipated = I2r = (2 A)2 (2 ) = 8 W. Calculate (c) the power output of the battery. Poutput = Pconverted - Pdissipated = 24 W - 8 W = 16W. Calculate (c) the power output of the battery (cross-check). R=4 +- I=2A r=2  = 12 V The output power is delivered to the resistor: Poutput = Presistor = I2 R = (2 A)2 (4 ) = 16W. Example 3 : a 3 volt and 6 volt battery* are connected the “wrong” way (+ to + or - to -), along with a 6 ohm resistor. What is the power dissipated in (delivered to) the resistor? What is the power in the 3 volt battery? 6 I - + +- 3V 6V Starting at point a... +6–3–6I=0 a Next lecture: Kirchhoff’s Loop Rule: The sum of potential changes around any closed path in a circuit is zero. V =0 around any closed loop I = (6 – 3) / 6 = 0.5 A *Assume zero internal resistance unless the problem suggests otherwise. What is the power dissipated in (delivered to) the resistor? 6 I = 0.5 A - + +- 3V 6V a PR = I2R = (0.5)2 (6) = 1.5 W What is the power dissipated in the 3 volt battery? P3V = IV = (0.5) (3) = 1.5 W Note: P6V = IV = (0.5) (6) = 3 W = PR + P3V