The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is
Show Hint
Ohm’s law will be used to calculate the internal resistance of the cell through the values.
1.0 Ω
0.2 Ω
0.5 Ω
0.8 Ω
The Correct Option is C
Approach Solution - 1
In the question, we have the values for external resistance, current, and voltage of the cell. As per ohm's law, these quantities are related to each other. Hence, ohm’s law will be used to calculate the internal resistance of the cell through the values.
Formula used:
As per ohm’s law, V = I(R+r)
- Where V is voltage,
- I is current,
- R is external resistance
- r is internal resistance.
Complete step-by-step solution:
Given: Internal resistance = 2.1V, Current = 0.2A, external resistance = 10Ω.
To calculate internal resistance, we will use ohm’s law.
V = I(R+r)
2.1 = (0.2)(10+r)
\(⇒10+r=\frac{21}{2}\)
\(⇒r=\frac{21}{2}−10\)
r = 0.5 Ω
Thus, the internal resistance of the given cell is 0.5Ω.
Hence, the correct option is (C).
Note: The sum of internal and external resistance is the total resistance.
Approach Solution -2
\(I = \frac {E}{r + R } \Rightarrow 0.2 = \frac {2.1 }{ r + 10} \Rightarrow r = 0.5 \Omega\)
Approach Solution -3
The current produced by a cell is equal to the voltage of the cell divided by the total resistance of the circuit. Using ohm's law formula and values, we can get the required answer.
Formula used:
The electric current produced by a cell is given as internal resistance and the voltage by the following expression.
\(I=\frac{E}{R+r}\)
Complete step-by-step answer:
We are given an electric cell whose voltage, amount of current and external resistance are given below respectively:
- E = 2.1 V
- I = 0.2 A
- R = 10Ω
Let r be the internal resistance of the cell.
The current produced by a cell is given as internal resistance and the voltage by the following expression.
\(I=\frac{E}{R+r}\)
\(r=\frac{E}{I}−R\)
Putting all these values in the above expression for the internal resistance of the cell, we get
\(r=\frac{2.1}{0.2}−10=10.5−10=0.5Ω\)
Now, \(r=0.5Ω\) is the required value of internal resistance.
So, the correct answer is “Option C”.
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Concepts Used:
Resistance
Resistance is the measure of opposition applied by any object to the flow of electric current. A resistor is an electronic constituent that is used in the circuit with the purpose of offering that specific amount of resistance.
R=V/I
In this case,
v = Voltage across its ends
I = Current flowing through it
All materials resist current flow to some degree. They fall into one of two broad categories:
- Conductors: Materials that offer very little resistance where electrons can move easily. Examples: silver, copper, gold and aluminum.
- Insulators: Materials that present high resistance and restrict the flow of electrons. Examples: Rubber, paper, glass, wood and plastic.
Resistance measurements are normally taken to indicate the condition of a component or a circuit.
- The higher the resistance, the lower the current flow. If abnormally high, one possible cause (among many) could be damaged conductors due to burning or corrosion. All conductors give off some degree of heat, so overheating is an issue often associated with resistance.
- The lower the resistance, the higher the current flow. Possible causes: insulators damaged by moisture or overheating.