Battery with Voltmeter Internal Resistance, r & emf, e. Work done on Charges • Batteries & other devices do work on charges to raise PE e-. EChm is used to give q PEelcin battery. • Inside E source, the total E supplied per charge is emf – e. e also measured in Volts W/q. • Term –voltage or pd used for resistors – E released per charge. • Emf can be thought of as total E of cell. When e- travel in circuit they gain PE in cell, lose PE in resistors. Emf – total energy available to be supplied by source (battery) to charges per C. e = W q Voltage pd across resistor-amount of E per C to heat. Called Terminal Voltage. Ex 1: What is the p.d. across a resistor if 24 J of heat are produced when a current of 2 A flows through it for 10 seconds? • • • • V = DE/q q = 2 C/s x 10 s = 20 C. 24 J/20 C V = 1.2 V. • • • • • • • Measure e of battery. Connect bulb. Measure voltage on battery terminals. What happens? Speculate as to why. Measure the pd around bulb. Compare bulb pd to terminal voltage. Internal Resistance Inside cell or battery, charges collide with components as they pass thru. The cell resists current flow. Some E, converts to heat. This small resistance r, inside cell called internal resistance. All E supplies have internal resistance. The voltage between the terminals when current flows, is usually less than e , some E heats cell. 2. Visualize the Cell with internal resistance It is as if there are 2 resisters in series on the circuit. Battery with r represented by placing box (resistor) next to cell symbol. An external resistance R. Both resistors have a pd around them. The emf is the sum of the 2 pd’s. e = VT + Vr. e = IR + Ir V If e = 1.5 V with 1-A & cell has 0 internal r, the bulb has 1.5V & 1-A (current on). 1A e = 1.5 V r=0 1A 1.5 V What if r = 1.5 W? Current drops & bulb voltage drops. It’s the same as if another resistor were placed in series with the bulb. Rtot on circuit goes up, I goes down. Bulb gets dimmer. 0.5 A e = 1.5 V r = 1.5 W. 0.5 A 0.75 V Measure V across battery terminals you get >1.5-V. This is called terminal voltage VT. e = total E supplied by cell • • • • • • • With current off: e = measured V. With current on, The Vt is the same as pd around resistor e = Vt + Vbat. IR + Ir See table e = I( R +r ). Internal Resistance 3: A battery of e= 12 V and internal resistance, r = 1.5 W produces a current of 3.0 A. What is the p.d. across the terminals when current is flowing? Vt = e - Ir. Vt = 12 V – (3.0 A)(1.5 W ) Vt = 7.5 V. 4. A battery of e = 9-V & r=1W, is connected to a 2 W resistor. a. Sketch the circuit. b. Calculate the current flow. c. What is the voltage on the 2W resistor? • e = I( R +r ). • I= e R+r • • • • • I = 9V/3 W 3 A. By ohm’s law V = IR = (3A)(2W) V=6V 2W 5: A dry cell has an internal resistance of 1.50 W. A resistor of 12 W is connected in series with the cell. If the p.d. across the 12 W resistor is 1.2 V, calculate the emf of the cell. Find the current in the circuit: I = V/R 1.2 V = 0.10 A 12 W IR = e - Ir e = IR + Ir = 0.10 A (12.0 +1.50) W = 1.35 V. Internal Resistance battery “death” If the internal resistance is small, it might not be reported. If it is not small you cannot ignore it. As the oxidation-reduction rx in cells progresses, the buildup of products increase the internal resistance until the available emf cannot overcome r, there is not enough E to push charges across gap; the battery is “dead”. Let’s say you built a simple circuit with one variable resistor, R and a cell with a small constant internal resistance, r. If you were able to measure the current and voltage of the outside resistor R, how would you construct a graph to determine the emf and r? Use the equation for internal resistance and emf. • Hwk Sheet emf. M05p2 B1.