The potential different across the Coolidge tube is `20 kV and 10 m A` current flows through the voltage supply. Only `0.5%` of the energy carried by the electrons striking the largest is converted into X-ray. The power carried by the X-ray beam is `p`. ThenA. `P = 0.1 W`B. `P = 1 W`C. `P = 2 W`D. `P = 10 W`

The potential different across the Coolidge tube is `20 kV and 10 m A`

1.	The potential different across the Coolidge tube is `20 kV and 10 m A` current flows through the voltage supply. Only `0.5%` of the energy carried by the electrons striking the largest is converted into X-ray. The power carried by the X-ray beam is `p`. ThenA. `P = 0.1 W`B. `P = 1 W`C. `P = 2 W`D. `P = 10 W`
Answer» Correct Answer - B `P = VI` Therefore, total power down by Coolidge tube `P_(T) = VI = 200 W`. As `0.5 %` of the energy is corried by X-ray is `0.5 % of P_(T) = (0.5)/(100) xx 200 = 1 W`

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