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प्रश्न
A cell of e.m.f 2.0 V and internal resistance 1Ω is connected to the resistors of 3Ω and 6Ω in series. Calculate:
(i) the current drawn from the cell,
(ii) the p.d. across each resistor,
(iii) the terminal voltage of the cell and
(iv) the voltage drop.
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उत्तर
Total resistance of the circuit R = 1 + 3 + 6 = 10 Ω
(i) The current drawn from the cell = `"E"/"R" = 2.0/10 = 0.2` A
(ii) The p.d. across 3 Ω resistor V1 = IR1 = 0.2 × 3 = 0.6V
The p.d. across 6 Ω resistor V2 = IR2 = 0.2 × 6 = 1.2 V
(iii) The terminal voltage of the cell V = V1 + V2
= 0.6 + 1.2 = 1.8 V
(iv) The voltage drop = E - V = 2.0 - 1.8 = 0.2 V
संबंधित प्रश्न
Which of the following statements correctly defines a volt?
(a) a volt is a joule per ampere.
(b) a volt is a joule per coulomb.
How much work is done when one coulomb charge moves against a potential difference of 1 volt?
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What will be the nature of the graph between the current and potential difference? (Do not draw a graph.)
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Which of the following represents voltage?
An electric charge always flows from a point at a ______ to a point at a ______.
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