CH-25 WEB - Chapter 25 Energy and Current in DC Circuits:...

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Chapter 25 Energy and Current in DC Circuits: (Additional Questions) 1) In figure, if the potential at point P is 100 V, what is the potential at point Q? Solution: The current in the circuit is So from we get 2) (a) In figure, what is the equivalent resistance of the network shown? (b) What is the current in each resistor? Put 1 100 R = , 2 3 50 R R = = , 4 75 R = , and =6 V; assume the battery is ideal. Solution: (a) R 2 , R 3 and R 4 are in parallel. By finding a common denominator and simplifying, the equation 1 /R = 1 /R 2 + 1 /R 3 + 1 /R 4 gives an equivalent resistance of 2 3 4 2 3 2 4 3 4 (50 )(50 )(75 ) 18.75 . (50 )(50 ) (50 )(75 ) (50 )(75 )                     R R R R R R R R R R = = = + + Ω + Ω +
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Thus, considering the series contribution of resistor R 1 , the equivalent resistance for the network is 1 100 18.75 118.75 .       eq R R R = + = Ω + Ω = (b) 1 eq i R = E 6 V 0.0505 A 118.75 = = ; ( 29 ( 29
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CH-25 WEB - Chapter 25 Energy and Current in DC Circuits:...

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