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Key is the position 2 for time t. Thereafter, it is in position 1. Resistance of the bulb and inductance of inductor are marked in the figure choose the figure choose the correct alternative. |
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Answer» Bulb 2 dies as soon as key is wsitched into position 1. `i_(0)=(epsilon)/(R_(2))` Energy stored in the inductor `U=(1)/(2)Li_(0)^(2)=(1)/(2)L((epsilon)/(R_(2)))^(2)=(L epsilon^(2))/(2R_(2)^(2))` When key is in position 1, this energy will convert into heat energy through RESISTOR. |
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