Page 45 - Schaum's Outline of Theory and Problems of Electric Circuits
P. 45

CIRCUIT LAWS
                                                                                                [CHAP. 3
               34
                     Note: This is similar to the case of current division in a two-branch parallel circuit where the other resistor
                                       0
                     has been replaced by R .
               3.17  A power transmission line carries current from a 6000-V generator to three loads, A, B, and C. The loads
                     are located at 4, 7, and 10 km from the generator and draw 50, 20, and 100 A, respectively. The resistance of
                     the line is 0.1 
/km; see Fig. 3-21. (a) Find the voltage at loads A, B, C. (b) Find the maximum percentage
                     voltage drop from the generator to a load.











                                                        Fig. 3-21

                     Ans.(a) v A ¼ 5928 V; v B ¼ 5889 V; v C ¼ 5859 V;  (b) 2.35 percent

               3.18  In the circuit of Fig. 3-22, R ¼ 0 and i 1 and i 2 are unknown. Find i and v AC .
                     Ans. i ¼ 4A; v AC ¼ 24 V















                                                        Fig. 3-22


               3.19  In the circuit of Fig. 3-22, R ¼ 1 
 and i 1 ¼ 2 A.  Find, i, i 2 , and v AC .
                     Ans. i ¼ 5A; i 2 ¼ 16 A; v AC ¼ 27 V

               3.20  In the circuit of Fig. 3-23, i s1 ¼ v s2 ¼ 0, v s1 ¼ 9V, i s2 ¼ 12 A.  For the four cases of (a) R ¼ 0,
                     (b) R ¼ 6 
,  (c) R ¼ 9 
, and (d) R ¼ 10 000 
, draw the simplified circuit and find i BA and v AC .
                     Hint: A zero voltage source corresponds to a short-circuited element and a zero current source corresponds
                     to an open-circuited element.
                          8
                          > ðaÞ  i BA ¼ 7; v AC ¼ 30
                          >
                          >
                          <
                     Ans:   ðbÞ  i BA ¼ 4:2; v AC ¼ 21:6    ðAll in A and V)
                          >  ðcÞ  i BA ¼ 3:5; v AC ¼ 19:5
                          >
                          >
                          :
                            ðdÞ  i BA ¼ 0:006   0; v AC ¼ 9:02   9
               3.21  In  the  circuit  of  Fig.  3-23,  v s1 ¼ v s2 ¼ 0; i s1 ¼ 6A; i s2 ¼ 12 A:  For  the  four  cases  of  (a)
                     R ¼ 0; ðbÞ R ¼ 6 
; ðcÞ R ¼ 9 
; and ðdÞ R ¼ 10 000 
; draw the simplified circuit and find i BA and v AC .
                          8
                          > ðaÞ  i BA ¼ 6; v AC ¼ 36
                          >
                          >
                          <
                     Ans:   ðbÞ  i BA ¼ 3:6; v AC ¼ 28:8  ðAll in A and V)
                          >  ðcÞ  i BA ¼ 3; v AC ¼ 27
                          >
                          >
                          :
                            ðdÞ  i BA ¼ 0:005   0; v AC   18
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