Question

3. In the circuit shown in fig. 3.28, there are 5 cells
in series each of e.m.f. 2 V and internal resistance
1 ohm .Calculate the current through the external resistance
(R) and p.d. across it.


in gig r =15 ohm
Fig. 3.28
8.​

Answer 1

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In the circuit shown in fig. 3.28, there are 5 cells

in series each of e.m.f. 2 V and internal resistance

1 ohm .Calculate the current through the external resistance

(R) and p.d. across it.

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Total resistance = 1 x 5 + 15 = 20 ohms,

total potential difference = 2 x 5 = 10 V,

So , current = 10 V /20 ohms = 0.5 Ampere,

P.d. across 15 ohm resistor = 0.5 A x 15 ohm = 7.5 V.

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Answer 2

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In the circuit shown in fig. 3.28, there are 5 cells

in series each of e.m.f. 2 V and internal resistance

1 ohm .Calculate the current through the external resistance

(R) and p.d. across it.

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Total resistance = 1 x 5 + 15 = 20 ohms,

total potential difference = 2 x 5 = 10 V,

So , current = 10 V /20 ohms = 0.5 Ampere,

P.d. across 15 ohm resistor = 0.5 A x 15 ohm = 7.5 V.

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