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Calculate the current drawn from the battery in the given network . |
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Answer» SOLUTION :The circuit can be redrawn as given in Fig. It is a Wheatstone bridge arrangement in which `R_1/R_5 = R_4/R_3` i.e., the bridge is balanced one. Hence resistance `R_2 = 5Omega`is superfluous and may be ignored. ` THEREFORE`VALUE of resistance` R_15`of series combination of `R_1` and `R_5 = 1 + 2 =3Omega` Value of resistance `R_43` of series combination of `R_4` and `R_3 = 2 +4=6Omega` ` therefore ` Net resistance of network R = resistance of parallel grouping of `R_15` and `R_43` `R = (R_15 xx R_43)/(R_15 xx R_43) = (3 xx 6)/(3+6) = 2 Omega` ` therefore ` Current drawn from the battery in the given network `I = epsi/R = (4V)/(2Omega) = 2A` |
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