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(a) Three resistors 2 Omega 4 Omega and 5 Omega are combined in parallel. What is the total resistance of the combination ? (b) If the combination is connected toa bettery of emf 20 V and neglibible internal resistance, determine the current through each resistor, and the total current drawn from the battery. |
Answer» Solution : (i) Current passing through `R_(1)` is , `I_(1) = (EPSILON)/(R_(1)) = (20)/(2) = 10 `A (II) Current passing through`R_(2)` is , `I_(2) = (epsilon)/(R_(2)) = (20)/(4) = 5`A (iii) Current passing through`R_(3)` is , `I_(3) = (epsilon)/(R_(3)) = (20)/(5) = 4 ` A Applying Kirchhoff.s current law, (KCLalso known as junction rule) at junction a, `sum I = 0 ` `therefore I- I_(1) - I_(2) - I_(3) = 0` `therefore I = I_(1) + I_(2) + I_(3) ` `= 10 + 5 + 4 ` `therefore I = 19 A ` . |
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