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A cell of emf E and internal resistance r is connected in series with an external resistance nr. The ratio of the terminal potential difference to emf is
- a)1 / n
- b)1 / n + 1
- c)n / n + 1
- d)n + 1 / n
Correct answer is option 'C'. Can you explain this answer?
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A cell of emf E and internal resistance r is connected in series with...
The terminal potential difference (V) of a cell is the potential difference across its terminals when a current is flowing through it. The emf (E) of a cell is the maximum potential difference it can provide when no current is flowing through it.
When a cell is connected in series with an external resistance (R), the current flowing through the circuit is given by Ohm's law: I = (E - V) / (r + R), where I is the current, E is the emf of the cell, V is the terminal potential difference, r is the internal resistance of the cell, and R is the external resistance.
To find the ratio of V to E, we can rearrange the equation to solve for V: V = E - I(r + R).
Now, let's consider the scenario when the external resistance is n times larger than the internal resistance, i.e., R = nr.
- Case 1: When n = 1
In this case, the external resistance is equal to the internal resistance, so R = r. Plugging this into the equation, we get V = E - I(2r). Therefore, the ratio of V to E is V/E = (E - I(2r)) / E = 1 - (2rI) / E.
- Case 2: When n > 1
In this case, the external resistance is larger than the internal resistance, so R > r. Plugging this into the equation, we get V = E - I(r + nr) = E - I(1 + n)r = E - I(r + R). Therefore, the ratio of V to E is V/E = (E - I(r + R)) / E = 1 - (r + R)I / E = 1 - (1 + n)rI / E.
From the above calculations, we can see that the ratio of V to E depends on the value of n:
- When n = 1, the ratio is 1 - (2rI) / E.
- When n > 1, the ratio is 1 - (1 + n)rI / E.
Therefore, the correct answer is option 'C': n / (n + 1).
When a cell is connected in series with an external resistance (R), the current flowing through the circuit is given by Ohm's law: I = (E - V) / (r + R), where I is the current, E is the emf of the cell, V is the terminal potential difference, r is the internal resistance of the cell, and R is the external resistance.
To find the ratio of V to E, we can rearrange the equation to solve for V: V = E - I(r + R).
Now, let's consider the scenario when the external resistance is n times larger than the internal resistance, i.e., R = nr.
- Case 1: When n = 1
In this case, the external resistance is equal to the internal resistance, so R = r. Plugging this into the equation, we get V = E - I(2r). Therefore, the ratio of V to E is V/E = (E - I(2r)) / E = 1 - (2rI) / E.
- Case 2: When n > 1
In this case, the external resistance is larger than the internal resistance, so R > r. Plugging this into the equation, we get V = E - I(r + nr) = E - I(1 + n)r = E - I(r + R). Therefore, the ratio of V to E is V/E = (E - I(r + R)) / E = 1 - (r + R)I / E = 1 - (1 + n)rI / E.
From the above calculations, we can see that the ratio of V to E depends on the value of n:
- When n = 1, the ratio is 1 - (2rI) / E.
- When n > 1, the ratio is 1 - (1 + n)rI / E.
Therefore, the correct answer is option 'C': n / (n + 1).
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A cell of emf E and internal resistance r is connected in series with...
E - i(r) = Terminal voltage
= Terminal voltageAttention NEET Students!
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A cell of emf E and internal resistance r is connected in series with an external resistance nr. The ratio of the terminal potential difference to emf isa)1 / nb)1 / n + 1c)n / n + 1d)n + 1 / nCorrect answer is option 'C'. Can you explain this answer?
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