ch27-p030

# ch27-p030 - Thus by the loop rule(since the battery voltage...

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30. (a) By the loop rule, it remains the same. This question is aimed at student conceptualization of voltage; many students apparently confuse the concepts of voltage and current and speak of “voltage going through” a resistor – which would be difficult to rectify with the conclusion of this problem. (b) The loop rule still applies, of course, but (by the junction rule and Ohm’s law) the voltages across R 1 and R 3 (which were the same when the switch was open) are no longer equal. More current is now being supplied by the battery which means more current is in R 3 , implying its voltage-drop has increased (in magnitude).
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Unformatted text preview: Thus, by the loop rule (since the battery voltage has not changed) the voltage across R 1 has decreased a corresponding amount. When the switch was open, the voltage across R 1 was 6.0 V (easily seen from symmetry considerations). With the switch closed, R 1 and R 2 are equivalent (by Eq. 27-24) to 3.0 Ω , which means the total load on the battery is 9.0 Ω . The current therefore is 1.33 A which implies the voltage-drop across R 3 is 8.0 V. The loop rule then tells us that voltage-drop across R 1 is 12 V – 8.0 V = 4.0 V. This is a decrease of 2.0 volts from the value it had when the switch was open....
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## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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