Question

A battery made of 5 cells each of 2 V and have internal resistance 0.1 ,0.2,0.3,0.4,0.5 is connected across 10ohm resistance. Calculate the current drawing through 10 ohm resistance

Answer 1

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Answer 2

Answer:

The current drawing through 10 ohm resistance is 0.86 A.

Explanation:

Given that,

Voltage V = 2 volt

Internal resistance

r₁= 0.1

r₂=0.2

r₃ =0.3

r₄ =0.4

r₅ = 0.5

Resistance R = 10 ohm

Number of cell = 5

The battery is connected in series with the resistance.

So, the total internal resistance is

$r = r_{1}+r_{2}+r_{3}+r_{4}+r_{5}$.....(I)

Put the all value in the equation (I)

$r = 0.1+0.2+0.3+0.4+0.5$

$r=1.5\ \omega$

Now, The total voltage is

$V= V_{1}+V_{2}+V_{3}+V_{4}+V_{5}$.....(II)

Put the all value in the equation (II)

$V = 2+2+2+2+2$

$V = 10\ volt$

Using ohm's law

The voltage of the battery is the product of the current and resistance.

$V = I\ R$

$I =\dfrac{V}{r+R}$

Where, r = total internal resistance

R = resistance

V = total voltage

$I = \dfrac{10}{11.5}$

$I = 0.86\ A$

Hence, The current drawing through 10 ohm resistance is 0.86 A.