A 10 m long wire of uniform cross-section and 20 Ω resistance is used in a potentiometer. The wire is connected in series with a battery of 5 V along with an external resistance of 480 balanced at 6.0 m length of the wire, then the value of unknown emf is-
(i) 1.2 V
(ii) 1.02 V
(iii) 0.2 V
(iv) 0.12 V
(ii) 1.02 V
(iii) 0.2 V
(iv) 0.12 V
(iv) 0.12 V
Explanation:
(i) Total Resistance of circuit `R_{r}`
= Resistance of Potentionmeter wire + Resistance R connected in series
= 20 + 480 = 500 Ω
Current in Circuit
I = `\frac{"ε"^{'}}{"R"_{"T"}}`
= `5/500`A
= 0.01 A
Potential differences across wire AB,
= `"V"_{"AB"}` = `"I"\times "R"_{"AB"}`
= 0.01 `\times` 20 = 0.2 V
∴ Potential gradient k
= `\frac{"V"_{"AB"}}{"L"_{"AB"}}`
= `"0.2V/"10m"`
= 0.02 V/m
(ii) Value of known emf = k`\times`(balancing length AC)
= 0.02 `\times` 6 = 0.12 V
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