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Power P. Do Now: What quantity determines the brightness of a bulb on a circuit? What can be increased to increase bulb brightness?. IB Circuit Diagram Symbols. Internal Resistance, r & emf , e . Work done on Charges.
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Power P Do Now: What quantity determines the brightness of a bulb on a circuit? What can be increased to increase bulb brightness?
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 work done per C of charge to push the e- across is emf– e. e also measured in Volts W/q. • Term –voltageor pd used for resistors, emf or voltage for source. • Emf can be thought of as totalp.d. of cell.
e = Wq Voltagepd across resistor-amount of E per C to heat. Emf – total energy available to be supplied by source (battery) to charges per C. Conservation of E – chem E converted by cell = E turned to heat/light other in resistors.
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? • 2 C/s x 10 s = 20 C. • V = E/q 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 currentflows, is often less than e- some E heats cell.
Sketching Cell with internal resistance Represent battery with r by placing box (resistor) next to cell symbol. Measure earound both symbols.Sketch box around the 2 showing they are the same unit (cell).
If e = 1.5 V with 1-A & cell has 0 internal r, the bulb has 1.5V & 1-A (current on). 1 A 1 A e = 1.5 V r = 0 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 0.5 A e = 1.5 V r = 1.5 W. 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 + Vlost. • IR + Ir • See table • e = I( R +r ).
Internal Resistance Ex 2: 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? Vterm = e - Ir. Vterm = 12 V – (3.0 A)(1.5 W ) Vterm = 7.5 V.
Ex 3. A battery of e = 9-V & r=1W, is connected to a 2 W resistor. Sketch the circuit. Calculate the current flow. 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 2 W
Ex 4: 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.
I = V/R 1.2 V = 0.10 A12 W IR = e - Ire = IR + Ir = 0.10 A (12.0 +1.50) W = 1.35 V. Find the current in the circuit:
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.