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Question
A cell of e.m.f. 1.8V and internal resistance 2Ω is connected in series with an ammeter of resistance 0.7Ω and a resistor of 4.5Ω as shown in Fig.
- What would be the reading of the ammeter?
- What is the potential difference across the terminals of the cell?
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Solution
The resistances are series connected
∴ Total resistance of circuit = (r + R)
(r + R) = (2 + 0.7 + 4.5)
= 7.2 Ω
e.m.f. of cell = 1.8 V
- Reading of Ammeter `I = "V"/("R" + r)`
= `("e.m.f")/("R" + r)`
`I = 1.8/7.2`
= 0.25 A - Potential difference across the terminals of the cell = IR
`"V" = 1/4 xx 5.2`
V = 1.3 V
R = 4.5 + 0.7
R = 5.2 Ω
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