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Four Cells Each of E.M.F. 2v and Internal Resistance 0.1 ω Are Connected in Series to an Ammeter of Negligible Resistance, a 1.6 ω Resistor and an Unknown Resistor R1. - Physics

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प्रश्न

Four cells each of e.m.f. 2V and internal resistance 0.1 Ω are connected in series to an ammeter of negligible resistance, a 1.6 Ω resistor and an unknown resistor R1. The current in the circuit is 2A. Draw a labelled diagram and calculate:

(i) Total resistance of the circuit,
(ii) Total e.m.f.
(iii) The value of R1 and
(iv) The p.d. across R1.

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उत्तर

(i) The resistance in the circuit

= (1.6 + 4 × 0.1 + R1) Ω

= (2 + R1) Ω

(ii) Total e.m.f. = No. of cell × e.m.f. of each cell = 4 × 2 = 8 V

(iii) Current in main circuit i = `"Total e.m.f."/"Total resistance"`

or  2 = `8/(2 + "R"_1)`

∴ R= `4/2 = 2` Ω

(iv) The p.d. across R= V = i R = 2 × 2 = 4V

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अध्याय 7: Electricity - Figure Based Long Answers

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आईसीएसई Physics [English] Class 10
अध्याय 7 Electricity
Figure Based Long Answers | Q 3

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