7 Ohm\'s Law Ch. 27 Begin Date: 2/20/2018 12:01:00 AM--D Due Date: 3/4/2018 11:5

Question

7 Ohm's Law Ch. 27 Begin Date: 2/20/2018 12:01:00 AM--D Due Date: 3/4/2018 11:59:00 PM End Date: 3/4/2018 11:59:00 PNM (796) Problem 10: A car battery has an internal resistance of r-.013 It is connected to a starter requiring 1-88 A and it has an internal EMF of -12V 33% Part (a) Input an expression for the voltage across the terminal when the starter is operating. 100% Potential Submissions (1% per attempt) 4 5 6 detailed view 0 I give up Feedback: deduction per feedback. 33% Part (b) what is the voltage in volts? 33% Part (c) on an especially cold day the starter requires more current to spin (1-150A), but the metallic battery's internal resistance decreases to r 9r. What is the new terminal voltage in V? AII content © 2018 Expert TA, LLC

Explanation / Answer

(a)Since we know the terminal voltage across the battery having certain internal resistance is given by

V = E - Ir

Where V is the terminal voltage, E is the internal EMF , I is the current flowing and r is the internal resistance

(b) here E = 12 volts , I = 88 amperes and r = 0.013 ohms

So terminal voltage V is equal to by the expression

V = 12 - (88 x 0.013) = 10.856 volts

(C) if I changes to I' = 150 amperes and internal resistance changes to r' = 0.9r = 0.9 x 0.013 = 0.0117 ohms

Then the new terminal voltage would be

V' = E - I'r' = 12 - (150 x 0.0117) = 10.245 volts

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