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(a) Three resistors 2Omega ,4Omega and 5Omegaare combined in parallel. What is the total resistance of thecombination ?(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance,determine the current through each resistor, and the total current drawn from the battery. |
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Answer» SOLUTION :Here `R_1 = 2Omega , R_2 = 4Omega` and `R_3 = 5Omega` In parallel combination RESULTANT RESISTANCE R is given by `1/R = (1)/(R_1) + (1)/(R_2) + (1)/(R_3) = 1/2 + 1/4 + 1/5= (10 + 5 + 4)/(20) = 19/20` ` rArr R = 20/19 Omega` (b) Here `epsi = 20V` and r = 0 ` therefore ` current through the THREE resistors are respectively `I_1 = (epsi)/(R_1) = 20/2 = 10 A, I_2 = (epsi)/(R_2) = 20/4 = 5 A, and I_3 = (epsi)/(R_3) = 20/5 = 4A` Total current drawn `I = I_1 + I_2 + I_3 = 10 + 5 + 4 = 19A` |
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