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The power dissipated in the circuit shown in the is 30 watts. The value of `R` is A. `10Omega`B. `30Omega`C. `20Omega`D. `15Omega` |
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Answer» `P_("Total")=(V^(2))/(R_(eq))` So `R_(eq)=(V^ (2))/( P_("total"))=(10xx10)/(30)=(10)/(3)` As `R_(eq)=(R_(1)R_(2))/(R_(1)+R_(2))` `rArr (10)/(3)=(5R)/(5+R)=10Omega` |
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