InterviewSolution
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InterviewSolution
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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 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» (a) There are the resistiors of resistances, `R_(1)=2 Omega,R_(2) = 4 Omega, and R_(3) = 5 Omega` They are connected in parallel. Hence, total resistance (R) of the combination is given by, `(1)/(R)=(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))` `:. R = (20)/(19) Omega` Therefore, total resistance of the combination is `(20)/(19) Omega` (b) Emf of the battery, `V = 20 V` Current `(I_(1))` flowing through resistor `R_(1)` is given by, `I_(1)=(V)/(R_(1))` `=(20)/(2)=10 A` Current `(I_(2))` flowing through resistor `R_(2)` is given by, `I_(2)=(V)/(R_(2))` `= (20)/(4)=5A` Current `(I_(2))` flowing through resistor `R_(3)` is given by, `I_(3)=(V)/(R_(3))` `=(20)/(5) = 4A` The current, `I = I_(1)+I_(2)+I_(3)=10+5+4 = 19 A` Therefore, the current through each resistor is 10 A, 5 A and 4 A respectively and the total current is 19 A. |
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