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A battery of `9 v` is connected in series with resistors of `0.2 Omega, 0.3 Omega, 0.4 Omega, 0.5 Omega and 12 Omega`. How much current would flow through the `12 Omega` resistor ? |
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Answer» Since all the resistors are in series, equivalent resistance, `R_s = 0.2 Omega + 0.3 Omega + 0.4 Omega + 0.5 Omega+ 12 Omega = 13.4 Omega` Current through the circuit, `I = (V)/(R_s) = (9 V)/(13.4 Omega) = 0.67 A` In series, same current `(I)` flows through all the resistors. Thus, current flowing through `12 Omega` resistor =`0.67 A`. |
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