A cell supplies a current of 2a when it is connected to a 59 resistance and supplies a current 1.2a. If its i connected to a resistance of 9 2. Find the emf and internal resistance of the cell.
Answers
ᴄᴏʀʀᴇᴄᴛᴇᴅ qᴜᴇꜱᴛɪᴏɴ:
- A cell supplies a current of 2a when it is connected to a 59 resistance and supplies a current 1.2a. If it is connected to a resistance of 9 Ω. Find the emf and internal resistance of the cell
ɪɴꜰᴏʀᴍᴀᴛɪᴏɴ ᴘʀᴏᴠɪᴅᴇᴅ ᴡɪᴛʜ ᴜꜱ:
Condition 1,
- Current is of 2A (I₁)
- Resistance is of 5Ω (R₁)
Condition 2,
- Current is of 1.2A (I₂)
- Resistance is of 9Ω (R₂)
ᴡʜᴀᴛ ᴡᴇ ʜᴀᴠᴇ ᴛᴏ ᴄᴀʟᴄᴜʟᴀᴛᴇ:
- emf and internal resistance of the cell
ᴡᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,
e.m.f of a cell is calculated by,
- ε = I (R + r)
Here,
- r is the internal resistance
ꜱᴜʙꜱᴛɪᴛᴜᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇꜱ ɪɴ ᴄᴏɴᴅɪᴛɪᴏɴ 1,
=> ε = 2 (5 + r) _____________[Equation no 1]
ꜱᴜʙꜱᴛɪᴛᴜᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇꜱ ɪɴ ᴄᴏɴᴅɪᴛɪᴏɴ 2,
=> ε = 1.2 (9 + r) _____________[Equation no 2]
ᴋᴇᴇᴘɪɴɢ ʙᴏᴛʜ ᴛʜᴇ ᴇqᴜᴀᴛɪᴏɴꜱ ᴇqᴜᴀʟ,
=> 2 (5 + r) = 1.2 (9 + r)
=> 2 × (5 + r) = 1.2 × (9 + r)
=> 10 + 2r = 10.8 + 1.2r
Transposing the sides,
=> 2r - 1.2r = 10.8 - 10
=> 0.8r = 0.8
=> r = 0.8 / 0.8
=> r = 08 × 10 / 08 × 10
=> r = 8 × 10 / 8 × 10
=> r = 4×10 / 4×10
=> r = 2 × 10 / 2 × 10
=> r = 1 × 10 / 1 × 10
=> r = 10 / 10
=> r = 1 / 1
=> r = 1
ꜱᴜʙꜱᴛɪᴛᴜᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇ of r in equation 1.
=> ε = 2 (5 + 1)
=> ε = 2 × (5 + 1)
=> ε = 10 + 2
=> ε = 12
ʜᴇɴᴄᴇꜰᴏʀᴛʜ, ᴇ.ᴍ.ꜰ. ɪꜱ ᴏꜰ 12V ᴀɴᴅ ɪɴᴛᴇʀɴᴀʟ ʀᴇꜱɪꜱᴛᴀɴᴄᴇ ɪꜱ ᴏꜰ 1Ω ʀᴇꜱᴘᴇᴄᴛɪᴠᴇʟʏ!
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