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For a circuit shown in Fig. find the value of resistance R_2 and current I_2 flowing through R_2 |
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Answer» SOLUTION :If equivalent resistance of parallel combination of `R_(1) and R_(2)` is R,then `R=(R_(1)R_(2))/(R_(1)+R_(2))=(10R_(2))/(10+R_(2))` According to Ohm.s law, `R=(V)/(I)` `R=(50)/(10)=5Omega RARR (10R_(2))/(10+R_(2))=5 rArr R_(2)=10Omega`. The current is EQUALLY DIVIDED into `R_(1) and R_(2)`. Hence `I_(2)=5A`. |
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