Dear Student, let the current in the main line be Icurrent in 16 ohm i=I12||812||8+16=0 231Icurrent through 12 ohm i'=I16||816||8+12=0 31 Icurrent through 8 ohm i''=I16||1216||12+8=0 46 Ipower in 16 ohm=0 231I216power in 3 ohm=0 46I23ratio=0 231I2160 46I23=1 345Regards

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No links please In the circuit given in Fig. 3.19, calculate the ratio of the powers consumed in the resistances 16 and 3 n . 16 n 12 n Fi 3.19

Dear Student,

l e t t h e c u r r e n t i n t h e m a i n l i n e b e I c u r r e n t i n 16 o h m i = I (\frac{12 | | 8}{12 | | 8 + 16}) = 0.231 I c u r r e n t t h r o u g h 12 o h m i' = I (\frac{16 | | 8}{16 | | 8 + 12}) = 0.31 I c u r r e n t t h r o u g h 8 o h m i'' = I (\frac{16 | | 12}{16 | | 12 + 8}) = 0.46 I p o w e r i n 16 o h m = {(0.231 I)}^{2} 16 p o w e r i n 3 o h m = {(0.46 I)}^{2} 3 r a t i o = \frac{{(0.231 I)}^{2} 16}{{(0.46 I)}^{2} 3} = 1.345 R e g a r d s

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No links please No links please In the circuit given in Fig. 3.19, calculate the ratio of the powers consumed in the resistances 16 and 3 n . 16 n 12 n Fi 3.19

No links please

No links please In the circuit given in Fig. 3.19, calculate the ratio of the powers consumed in the resistances 16 and 3 n . 16 n 12 n Fi 3.19