1.

Find the circuit flowing through the following electric circuit:

Answer»

Solution :In circuit COMBINED resistance `R_1` of the series grouping of `1 Omega and 3 Omega` RESISTORS
`R_1=1+3=4 Omega`
Combined resistance of parallel grouping of `6 Omega and R_1=4 Omega` resistors is `R_2` , where
`1/R_2=1/6+1/4=(2+3)/12=5/12 implies R_2=2.4 Omega`
`THEREFORE` TOTAL resistance of circuit `R=3.6+R_2+3=3.6+2.4+3=9 Omega`
`therefore` Current flowing `I=V/R=(4.5 V)/(9 Omega)=0.5A`


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