CH28 - CHAPTER 28ELECTRIC CIRCUITS ActivPhysics can help...

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Unformatted text preview: CHAPTER 28ELECTRIC CIRCUITS ActivPhysics can help with these problems: Section 12, DC Circuits Section 28-1:Circuits and Symbols Problem 1. Sketch a circuit diagram for a circuit that includes a resistor R 1 connected to the positive terminal of a battery, a pair of parallel resistors R R 2 3 and connected to the lower-voltage end of R 1 , then returned to the batterys negative terminal, and a capacitor across R 2 . Solution A literal reading of the circuit specifications results in connections like those in sketch (a). Because the connecting wires are assumed to have no resistance (a real wire is represented by a separate resistor), a topologically equivalent circuit diagram is shown in sketch (b). Problem 1 Solution (a). Problem 1 Solution (b). Problem 2. A circuit consists of two batteries, a resistor, and a capacitor, all in series. Sketch this circuit. Does the description allow any flexibility in how you draw the circuit? Solution In a series circuit, the same current must flow through all elements. One possibility is shown. The order of elements and the polarity of the battery connections are not specified. Problem 2 Solution. Problem 3. Resistors R 1 and R 2 are connected in series, and this series combination is in parallel with R 3 . This parallel 208 CHAPTER 28 combination is connected across a battery whose internal resistance is R int . Draw a diagram representing this circuit. Solution The circuit has three parallel branches: one with R 1 and R 2 in series; one with just R 3 ; and one with the battery (an ideal emf in series with the internal resistance). Problem 3 Solution. Section 28-2:Electromotive Force Problem 4. What is the emf of a battery that delivers 27 J of energy as it moves 3.0 C between its terminals? Solution From the definition of emf (as work per unit charge), E = = = W q = = 27 3 9 J C V. Problem 5. A 1.5-V battery stores 4.5 kJ of energy. How long can it light a flashlight bulb that draws 0.60 A? Solution The average power, supplied by the battery to the bulb, multiplied by the time equals the energy capacity of the battery. For an ideal battery, P E = I , therefore E It = 4 5 . , kJ or t = = = 4 5 15 0 60 5 10 139 3 . ( . )( . ) . kJ V A s h. = Problem 6. If you accidentally leave your car headlights (current drain 5 A) on for an hour, how much of the 12-V batterys chemical energy is used up? Solution The power delivered by an emf is E I , so (if the voltage and current remain constant) the energy converted is E It = = ( )( )( ) . 12 5 3600 216 V A s kJ Problem 7. A battery stores 50 W h of chemical energy. If it uses up this energy moving 3 0 10 4 . C through a circuit, what is its voltage?...
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CH28 - CHAPTER 28ELECTRIC CIRCUITS ActivPhysics can help...

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