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The reading of a high resistance voltmeter when a cell is connected across it is 2.2 V. When the terminals of the cell are also connected to a resistance of 5 Ω the voltmeter reading drops to 1.8 V. Find the internal resistance of the cell
- a)1.2 Ω
- b)1.3 Ω
- c)1.1 Ω
- d)1.4 Ω
Correct answer is option 'C'. Can you explain this answer?
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The reading of a high resistance voltmeter when a cell is connected ac...
Given:
Voltage across high resistance voltmeter when connected to a cell, V1 = 2.2 V
Voltage across high resistance voltmeter when connected to a cell and a resistance of 5 Ω, V2 = 1.8 V
To find: Internal resistance of the cell, r
Solution:
We know that the voltage across the terminals of a cell is given by:
V = E - Ir
where V is the voltage across the terminals, E is the emf of the cell, I is the current flowing through the circuit, and r is the internal resistance of the cell.
When the high resistance voltmeter is connected across the cell, the current through the circuit is very small, so we can assume that the voltage across the cell is equal to the emf of the cell:
V1 = E
When the cell is connected in series with a resistance of 5 Ω, the current through the circuit will be:
I = (E / (r + 5))
The voltage across the cell will now be:
V2 = E - I * r
Substituting the value of I from above, we get:
V2 = E - (E / (r + 5)) * r
Simplifying the above equation, we get:
V2 = E - (Er / (r + 5))
V2 = (E * 5) / (r + 5)
Now we can use the given values of V1 and V2 to find the emf of the cell and then solve for r:
From the first equation, we know that V1 = E, so:
E = 2.2 V
Substituting this value in the second equation, we get:
1.8 V = (2.2 V * 5) / (r + 5)
Solving for r, we get:
r = 1.1 Ω
Therefore, the internal resistance of the cell is 1.1 Ω.
Answer: c) 1.1
Voltage across high resistance voltmeter when connected to a cell, V1 = 2.2 V
Voltage across high resistance voltmeter when connected to a cell and a resistance of 5 Ω, V2 = 1.8 V
To find: Internal resistance of the cell, r
Solution:
We know that the voltage across the terminals of a cell is given by:
V = E - Ir
where V is the voltage across the terminals, E is the emf of the cell, I is the current flowing through the circuit, and r is the internal resistance of the cell.
When the high resistance voltmeter is connected across the cell, the current through the circuit is very small, so we can assume that the voltage across the cell is equal to the emf of the cell:
V1 = E
When the cell is connected in series with a resistance of 5 Ω, the current through the circuit will be:
I = (E / (r + 5))
The voltage across the cell will now be:
V2 = E - I * r
Substituting the value of I from above, we get:
V2 = E - (E / (r + 5)) * r
Simplifying the above equation, we get:
V2 = E - (Er / (r + 5))
V2 = (E * 5) / (r + 5)
Now we can use the given values of V1 and V2 to find the emf of the cell and then solve for r:
From the first equation, we know that V1 = E, so:
E = 2.2 V
Substituting this value in the second equation, we get:
1.8 V = (2.2 V * 5) / (r + 5)
Solving for r, we get:
r = 1.1 Ω
Therefore, the internal resistance of the cell is 1.1 Ω.
Answer: c) 1.1
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Community Answer
The reading of a high resistance voltmeter when a cell is connected ac...
Initially the voltmeter reading is equal to the e.m.f of the cell.later on the reading is equal to the voltage across the external resistance..current is equal to 1.8 divided by 5..then (2.2-1.8) divided by the current gives the internal resistance..
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The reading of a high resistance voltmeter when a cell is connected across it is 2.2 V. When the terminals of the cell are also connected to a resistance of 5 Ω the voltmeter reading drops to 1.8 V. Find the internal resistance of the cella)1.2 Ωb)1.3 Ωc)1.1 Ωd)1.4 ΩCorrect answer is option 'C'. Can you explain this answer?
Question Description
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