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State Kirchhoff's rules of current distribution in an electrical network. Using these rules determine the value of the currentI_1in the electric circuit of Fig. |
Answer» Solution :The circuit diagram can be REDRAWN as shown in Fig.  Here `I_3 = I_1 +I_2` In mesh CABDA, we have `-40.I_3 - 20.I_1 +40 = 0` ` - 40 (I_1+ I_2) - 20 I_1 + 40 = 0` ` RARR 60 I_1 + 40I_2 = 40`....(II) and in mesh EABFE, we have `-20.I_1 + 20I_2 + 80 = 0` `rArr 20 I_1 - 20 I_2 = 80 `....(iii) Multiplying (iii) by 2 and then adding with (ii), we get `100 I_1 = - 120 ` ` rArr I_1 = - 1.2 A ` The -ve sign SHOWS that the direction of current is OPPOSITE to that shown in figure with an arrow mark. |
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