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Question
A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in following figure.
Find:
- The reading of the ammeter,
- The potential difference across the terminals of the cells, and
- The potential difference across the 4.5 Ω resistor.
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Solution
The total resistance of the circuit is
`"R"_"eq" = "R"_"cell" + "R"_"ammeter" + "R"_1 || "R"_2`
`therefore "R"_"eq" = 1.2 + 0.8 + ("R"_1"R"_2)/("R"_1 + "R"_2)`
`therefore "R"_"eq" = 2 + (4.5 xx 9)/(4.5 + 9)`
= `2 + 40.5/13.5`
`therefore "R"_"eq" = 5 Omega`
(a) Therefore, the current through the ammeter is
I = `"E"_"cell"/"R"_"eq" = 2/5 = 0.4 "A"`
(b) The potential difference across the ends of the cells is
`"V"_"cell" = "E"_"cell" - "IR"_"cell"`
`therefore "V"_"cell" = 2 - 0.4 xx 1.2`
`therefore "V"_"cell" = 2 - 0.48 = 1.52 "V"`
(c) The potential difference across the 4.5 Ω resistor is
`"V"_4.5 = "V"_"cell" - "V"_"ammeter"`
`therefore "V"_4.5 = 1.52 - 0.4 xx 0.8`
`therefore "V"_4.5 = 1.52 - 0.32`
`therefore "V"_4.5 = 1.2 "V"`
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