the a TV of rating 100w is operated for 6 hours per day. find the number of units consumed in any leap year.
(solve).
Answers
Answer:
219.6 is right answer
Answer
It is given that the total power of the TV is 100 W and it is operated 6 hr/day.
The total number of days in a leap year is 366 days. Hence, the total time taken in a leap year is given as 6hours×366days=2196hours.
The electric power in watts associated with a complete electric circuit or a circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other form, e.g., heat, mechanical energy, or energy stored in electric fields or magnetic fields. The power is given by the product of applied voltage and the electric current. That is, P=VI.
The heat produced or the energy dissipated by an appliance is given by the product of power and time taken in seconds. That is, Q=Pt.
So, Q=P×t=100W×2196hrs=219600Wh=219.6kWh.
1 kWh = 1 unit.
Hence, the number of units consumed by the TV in any leap year is 219.6 units.
Answer:
ʜᴇʀᴇ ɪs ʏᴏᴜʀ ᴀɴsᴡᴇʀ ⬇️
Explanation:
➡️ ɪᴛ ɪs ɢɪᴠᴇɴ ᴛʜᴀᴛ ᴛʜᴇ ᴛᴏᴛᴀʟ ᴘᴏᴡᴇʀ ᴏғ ᴛʜᴇ ᴛᴠ ɪs 100 ᴡ ᴀɴᴅ ɪᴛ ɪs ᴏᴘᴇʀᴀᴛᴇᴅ 6 ʜʀ/ᴅᴀʏ.
➡️ᴛʜᴇ ᴛᴏᴛᴀʟ ɴᴜᴍʙᴇʀ ᴏғ ᴅᴀʏs ɪɴ ᴀ ʟᴇᴀᴘ ʏᴇᴀʀ ɪs 366 ᴅᴀʏs. ʜᴇɴᴄᴇ, ᴛʜᴇ ᴛᴏᴛᴀʟ ᴛɪᴍᴇ ᴛᴀᴋᴇɴ ɪɴ ᴀ ʟᴇᴀᴘ ʏᴇᴀʀ ɪs ɢɪᴠᴇɴ ᴀs 6ʜᴏᴜʀs×366ᴅᴀʏs=2196ʜᴏᴜʀs.
➡️ ᴛʜᴇ ᴇʟᴇᴄᴛʀɪᴄ ᴘᴏᴡᴇʀ ɪɴ ᴡᴀᴛᴛs ᴀssᴏᴄɪᴀᴛᴇᴅ ᴡɪᴛʜ ᴀ ᴄᴏᴍᴘʟᴇᴛᴇ ᴇʟᴇᴄᴛʀɪᴄ ᴄɪʀᴄᴜɪᴛ ᴏʀ ᴀ ᴄɪʀᴄᴜɪᴛ ᴄᴏᴍᴘᴏɴᴇɴᴛ ʀᴇᴘʀᴇsᴇɴᴛs ᴛʜᴇ ʀᴀᴛᴇ ᴀᴛ ᴡʜɪᴄʜ ᴇɴᴇʀɢʏ ɪs ᴄᴏɴᴠᴇʀᴛᴇᴅ ғʀᴏᴍ ᴛʜᴇ ᴇʟᴇᴄᴛʀɪᴄᴀʟ ᴇɴᴇʀɢʏ ᴏғ ᴛʜᴇ ᴍᴏᴠɪɴɢ ᴄʜᴀʀɢᴇs ᴛᴏ sᴏᴍᴇ ᴏᴛʜᴇʀ ғᴏʀᴍ.
➡️ ᴇ.ɢ., ʜᴇᴀᴛ, ᴍᴇᴄʜᴀɴɪᴄᴀʟ ᴇɴᴇʀɢʏ, ᴏʀ ᴇɴᴇʀɢʏ sᴛᴏʀᴇᴅ ɪɴ ᴇʟᴇᴄᴛʀɪᴄ ғɪᴇʟᴅs ᴏʀ ᴍᴀɢɴᴇᴛɪᴄ ғɪᴇʟᴅs. ᴛʜᴇ ᴘᴏᴡᴇʀ ɪs ɢɪᴠᴇɴ ʙʏ ᴛʜᴇ ᴘʀᴏᴅᴜᴄᴛ ᴏғ ᴀᴘᴘʟɪᴇᴅ ᴠᴏʟᴛᴀɢᴇ ᴀɴᴅ ᴛʜᴇ ᴇʟᴇᴄᴛʀɪᴄ ᴄᴜʀʀᴇɴᴛ. ➡️ᴛʜᴀᴛ ɪs, ᴘ=ᴠɪ.
➡️ᴛʜᴇ ʜᴇᴀᴛ ᴘʀᴏᴅᴜᴄᴇᴅ ᴏʀ ᴛʜᴇ ᴇɴᴇʀɢʏ ᴅɪssɪᴘᴀᴛᴇᴅ ʙʏ ᴀɴ ᴀᴘᴘʟɪᴀɴᴄᴇ ɪs ɢɪᴠᴇɴ ʙʏ ᴛʜᴇ ᴘʀᴏᴅᴜᴄᴛ ᴏғ ᴘᴏᴡᴇʀ ᴀɴᴅ ᴛɪᴍᴇ ᴛᴀᴋᴇɴ ɪɴ sᴇᴄᴏɴᴅs.
➡️ ᴛʜᴀᴛ ɪs, ǫ=ᴘᴛ.
sᴏ, ǫ=ᴘ×ᴛ=100ᴡ×2196ʜʀs=219600ᴡʜ=219.6ᴋᴡʜ.
1 ᴋᴡʜ = 1 ᴜɴɪᴛ.
➡️ʜᴇɴᴄᴇ, ᴛʜᴇ ɴᴜᴍʙᴇʀ ᴏғ ᴜɴɪᴛs ᴄᴏɴsᴜᴍᴇᴅ ʙʏ ᴛʜᴇ ᴛᴠ ɪɴ ᴀɴʏ ʟᴇᴀᴘ ʏᴇᴀʀ ɪs 219.6 ᴜɴɪᴛs.