Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 28 - Fundamentals of Circuits - Stop to Think 28.4 - Page 775: 1

Answer

The current is the same at each point, but the potential differences are different: $I_A = 2$ A; $V_A = 20$ Volts $I_B = 2$ A; $V_B= 16$ Volts $I_C = 2$ A; $V_C= 10$ Volts $I_D = 2$ A; $V_D= 8$ Volts $I_E = 2$ A; $V_E= 0$ Volts

Work Step by Step

Since the resistors are in series, we can find the total resistance by: $R_{total} = \Sigma_i^n R_i= R_1+R_2+...+R_n$ $R_{total} = 2+5+4+1 = 10$ $\Omega$ and thus, $\Delta V_{total} = IR = 2(10) = 20$ Volts Since the current into each resistor must be the same as the amount of current flowing out of the resistor (Kirchoff's junction law) as well as having no breaks in the wire (again, the wire is in series), the current at each point is the same. We can find the potential difference at each point by subtracting the potential difference of each individual resistor from the total potential difference. $\Delta V_{AB} = -IR = -2(2) = -4$ Volts $\Delta V_{BC} = -IR = -2(3) = -6$ Volts $\Delta V_{CD} = -IR = -2(1) = -2$ Volts $\Delta V_{DE} = -IR = -2(4) = 8$ Volts Now we can find the potential difference at each point $V_A = V_{total} = 20 $ Volts $V_B = V_{total} + V_{AB} = 20 - 4 = 16 $ Volts $V_C = V_{total} + V_{AB} + V_{BC} = 20 - 4 -6 = 10 $ Volts $V_D = V_{total} + V_{AB} +V_{BC}+V_{CD}= 20 - 4 -6-2= 8 $ Volts $V_D = V_{total} + V_{AB} +V_{BC}+V_{CD}-V_{DE}= 20 - 4 -6-2-8= 0 $ Volts
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