problem26_62

problem26_62 - 20 ) A 2 ( ) 10 ( par = = = RI V Current in...

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26.62: a) First do series/parallel reduction: Now apply Kirchhoff’s laws and solve for . ε A 25 . 4 ) A 25 . 2 ( A 2 A 2 A 25 . 2 0 ) 20 ( V 5 ) A 2 )( 20 ( : 0 1 2 1 2 2 adefa = - - = = + - = = - - - = I I I I I V reversed. be should polarity V; 109 0 ) A 25 . 2 ( ) 20 ( A) 25 . 4 ( ) 15 ( : 0 abcdefa - = = - - + = ε ε V b) Parallel branch has a resistance 10 . V
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Unformatted text preview: 20 ) A 2 ( ) 10 ( par = = = RI V Current in upper part: A 3 2 30 V 20 = = = R V I s 5 . 13 J 60 ) 10 ( A 3 2 2 2 = = = = t t U Rt I U Pt...
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