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Three resistors of 4 ohms and 12 ohms are connected in parallel. The combination of above resistors is connected in series to a resistance of 2 ohms and then to a battery of 6 volts. Draw a circuit diagram and calculate. (a) Current in main circuit. Current flowing through each of the resistors in parallel (c ) p.d and the power used by the 2 ohm resistor. |
Answer» Solution :Circuit diagram for the given data is  (a) Effecitive resistance when `R_(1), R_(2)` and `R_(3)` are connected in parallel is given by `(1)/(R_(P)) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3)) implies (1)/(R_(P)) = (1)/(4) + (1)/(6) + (1)/(12)` `implies (1)/(R_(P)) = (3 + 2 + 1)/(12) = (6)/(12) = (1)/(2)` `:. R_(P) = 2 Omega` Total resistance in the circuit `R = R_(P) + R_(4) = 2 + 2 = 4 Omega` `:.` Current in main circuit `I = (V)/(R) = (6)/(4) = 1.5 A` (b) Current flowing through `R_(1), I_(1) = (ir_(P))/(R_(1)) = (1.5 xx 2)/(4) = 0.74 A` current flowing through `R_(2), I_(2) = (IR_(P))/(R_(2)) = (1.5 xx 2)/(6) = 0.5 A` Current flowing through `R_(3), I_(3) = (IR_(P))/(R_(3)) = (1.5 xx 2)/(12) = 0.25 A` (c ) P.D ACROSS `2 Omega` resistor (i.e., `R_(4)`), `V_(4) = IR_(4) = 1.5 xx 2= 3` volt. POWER used by `2 Omega` resistor, `P = V_(4) I = 3 xx 1.5 = 4.5 W` |
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