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A `24` volt battery is connected to the arrangement of resistances shown in (Fig.) Calculate (i) the total effective resistance of the circuit, (ii) the total current flowing in the circuit. . |
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Answer» (i) If `R_1` is the equivalent resistance of `5 Omega and 20 Omega` which are in series, `R_1 = 5 Omega + 20 Omega = 25 Omega` Further, if `R_2` is the equivalent resistance of `15 Omega and 10 Omega` which are in series, `R_2 = 15 Omega + 10 Omega = 25 Omega` Since `R_1 and R_2` are in parallel, their effective resistance is given by `R_p = (R_1 R_2)/(R_1 + R_2) = (25 xx 25)/(25 + 25) Omega = (625)/(50) Omega = 12.5 Omega` (ii) Total pd across the arrangement of all the resistances, i.e., `V = 24 V` Total current in the circuit, `I = (V)/(R_p) = (24 V)/(12.5 Omega) = 1.92 A`. |
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