Physics, asked by mahilunavath, 3 months ago

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 .​

Answers

Answered by Ekaro
10

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

Answered by deepakmishradm292878
0

Answer:

gfj

kg true today etude red tube yy5 true reflection

bheja 4rt red tg7 e4 we 4r 45 656 5t5 hgg

Similar questions