Current Electricity and Electrical InstrumentMedium
Question
In a discharge tube when 200 volt potential difference is applied 6.25 × 1018 electrons move from cathode to anode and 3.125 × 1018 singly charged positive ions move from anode to cathode in one second. Then the power of tube is:
Options
A.100 watt
B.200 watt
C.300 watt
D.400 watt
Solution
P = P− + P+
= 200 × (6.25 × 1018 + 3.125 × 1018) × 1.6 × 10-19 W
= 300 W.
= 200 × (6.25 × 1018 + 3.125 × 1018) × 1.6 × 10-19 W
= 300 W.
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