Question

Five '2V' cells, each having internal resistance of 0.2 ohm are Connected in series to a load of resistance 14ohm. Find the Current flowing in circuit​

Answer 1

$\large \sf \underline{ \underline{Given:-}}$

Five '2V' cells, each having internal resistance of 0.2Ω are connected in series to a load of resistance 14Ω.

$\large \sf \underline{ \underline{to \: find:-}}$

Current flowing in circuit.

$\large \sf \underline{ \underline{Solution:-}}$

❖ In this case five cells of equal pd are connected in series.

Terminal voltage of each cell is 2V.

Hence net terminal voltage will be 10V.

All five resistances are connected in series. Equivalent resistance of series connection is given by

• R = R₁ + R₂ + ... + Rₙ

➠ R = 5(0.2) + 14

➠ R = 1 + 14

➠ R = 15 Ω

♦ Net current flow in circuit :

As per ohm's law current flow in circuit is directly proportional to the applied potential difference..

Mathematically, V = I R

• V denotes applied pd
• I denotes current
• R denotes resistance

By substituting the given values;

➛ V = I R

➛ 10 = I × 15

➛ I = 10/15

➛ I = 2/3

➛ I = 0.67 A

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Answer 2

Given:-

Five '2V' cells, each having internal resistance of 0.2Ω are connected in series to a load of resistance 14Ω.

To find:-

Current flowing in circuit.

Solution:-

❖ In this case five cells of equal pd are connected in series.

Terminal voltage of each cell is 2V.

Hence net terminal voltage will be 10V.

All five resistances are connected in series. Equivalent resistance of series connection is given by

R = R₁ + R₂ + ... + Rₙ

➠ R = 5(0.2) + 14

➠ R = 1 + 14

➠ R = 15 Ω

♦ Net current flow in circuit :

As per ohm's law current flow in circuit is directly proportional to the applied potential difference..

Mathematically, V = I R

V denotes applied pd

I denotes current

R denotes resistance

By substituting the given values;

➛ V = I R

➛ 10 = I × 15

➛ I = 10/15

➛ I = 2/3

➛ I = 0.67 A

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