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By a cell a current of 0.9 A flows through 2 ohm resistor and 0.3 A through 7 ohm resistor. The internal resistance of the cell is [KCET 2003]
- a)0.5Ω
- b)1.0Ω
- c)1.2Ω
- d)2.0Ω
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
By a cell a current of 0.9 A flows through 2 ohm resistor and 0.3 A th...
In Istcas E-V*R+r=i.
E-V*2+r=.9
in 2nd case
E-V*7+r=0.3
on solving both equation
you will get ans.0.5A
E-V*2+r=.9
in 2nd case
E-V*7+r=0.3
on solving both equation
you will get ans.0.5A
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By a cell a current of 0.9 A flows through 2 ohm resistor and 0.3 A th...
To find the internal resistance of the cell, we can use Ohm's Law and Kirchhoff's Laws.
Let's assume that the internal resistance of the cell is R.
According to Ohm's Law, the voltage across a resistor is equal to the current through it multiplied by its resistance.
For the 2 ohm resistor, the voltage is V1 = 0.9 A * 2 ohm = 1.8 V.
For the 7 ohm resistor, the voltage is V2 = 0.3 A * 7 ohm = 2.1 V.
According to Kirchhoff's Law, the total voltage across the resistors in series is equal to the sum of the individual voltages.
So, V1 + V2 = E, where E is the voltage of the cell.
1.8 V + 2.1 V = E
3.9 V = E
According to Ohm's Law, the voltage of the cell is E = I * (R + r), where I is the total current and r is the internal resistance.
Substituting the given values:
3.9 V = (0.9 A + 0.3 A) * (2 ohm + R)
3.9 V = 1.2 A * (2 ohm + R)
3.9 V = 2.4 ohm + 1.2 A * R
1.5 V = 1.2 A * R
R = 1.5 V / 1.2 A
R = 1.25 ohm
Therefore, the internal resistance of the cell is approximately 1.25 ohm.
Let's assume that the internal resistance of the cell is R.
According to Ohm's Law, the voltage across a resistor is equal to the current through it multiplied by its resistance.
For the 2 ohm resistor, the voltage is V1 = 0.9 A * 2 ohm = 1.8 V.
For the 7 ohm resistor, the voltage is V2 = 0.3 A * 7 ohm = 2.1 V.
According to Kirchhoff's Law, the total voltage across the resistors in series is equal to the sum of the individual voltages.
So, V1 + V2 = E, where E is the voltage of the cell.
1.8 V + 2.1 V = E
3.9 V = E
According to Ohm's Law, the voltage of the cell is E = I * (R + r), where I is the total current and r is the internal resistance.
Substituting the given values:
3.9 V = (0.9 A + 0.3 A) * (2 ohm + R)
3.9 V = 1.2 A * (2 ohm + R)
3.9 V = 2.4 ohm + 1.2 A * R
1.5 V = 1.2 A * R
R = 1.5 V / 1.2 A
R = 1.25 ohm
Therefore, the internal resistance of the cell is approximately 1.25 ohm.
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By a cell a current of 0.9 A flows through 2 ohm resistor and 0.3 A through 7 ohm resistor. The internal resistance of the cell is [KCET 2003]a)0.5Ωb)1.0Ωc)1.2Ωd)2.0ΩCorrect answer is option 'A'. Can you explain this answer?
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