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A uniform wound solenoid coil of self inductance `1.8 xx 10^(-4) H` and resistance ohm is broken into two identical coils. These indentical coils are connected in parallel across a 12 V battery of negligible resistance. Calculate the steady state current through the battery and time constant of the circuit. |
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Answer» Here, when the soleniod is broken ino two identical coils, inductance of each of coil becomes half and so is the resistance `:. L_(1) = L_(2) = (1.8 xx 10^(-4))/(2) = 0.9 xx 10^(-4)` henry `R_(1) = R_(2) = 6//2 = ohm`. When these two coils are connected in parallel to a battery, then `(1)/(L) = (1)/(L_(1)) + (1)/(L_(2)) = (L_(2) + L_(1))/(L_(1) L_(2))` `= 0.45 xx 10^(-4) H` and `R = (R_(1) R_(2))/(R_(1) + R_(2)) = (3 xx 3)/(3 + 3) = 1.5 phm` Time constant, `tau = (L)/(R ) = (0.45 xx 10^(-4))/(1.5) = 3 xx 10^(-5) s` The steady state current does not depend upon L. `I_(0) = (E)/(R ) = (12)/(1.5) = 8 A` |
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