# A cell has an e.m.f of 1.08 v and an internal resistor of 0.5 ohm. when it is connected in series with resistor of R , the potential difference between the terminals fell to 0.96 v . what was the value of R?

The circuit diagram in the given case appears as follows,

Here, you can place the internal resistance as a separate resistor in the circuit and in that case potential drop across the cell will be

Now, it is given that the potential drop across the cell is

So, if the current flowing through the circuit is

Or,

Now, the voltage drop across resistor

So,if voltage drop across

Then, using Ohm's Law, we can write,

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To find the value of resistor R, we can use the formula for the potential difference across a circuit:

[ V = \text{emf} - I \times R_{\text{internal}} - I \times R ]

Where:

- ( V ) is the potential difference across the circuit (0.96 V)
- (\text{emf}) is the electromotive force of the cell (1.08 V)
- (R_{\text{internal}}) is the internal resistance of the cell (0.5 ohm)
- (R) is the resistance of the external resistor (unknown)
- (I) is the current flowing through the circuit

We can rearrange the formula to solve for (R):

[ R = \frac{\text{emf} - V}{I} - R_{\text{internal}} ]

We need to find the current ((I)) flowing through the circuit. Using Ohm's law:

[ I = \frac{V}{R_{\text{internal}} + R} ]

Substitute the given values into the equation and solve for (I):

[ I = \frac{0.96 , \text{V}}{0.5 , \Omega + R} ]

Once we find (I), we can substitute it back into the equation for (R) to find the value of resistor (R).

[ R = \frac{1.08 , \text{V} - 0.96 , \text{V}}{I} - 0.5 , \Omega ]

[ R = \frac{0.12 , \text{V}}{I} - 0.5 , \Omega ]

Now, we substitute (I) back into the equation:

[ R = \frac{0.12 , \text{V}}{\frac{0.96 , \text{V}}{0.5 , \Omega + R}} - 0.5 , \Omega ]

[ R = \frac{0.12 , \text{V} \times (0.5 , \Omega + R)}{0.96 , \text{V}} - 0.5 , \Omega ]

[ R = \frac{0.06 , \text{V} + 0.12 , R}{0.96 , \text{V}} - 0.5 , \Omega ]

[ R = \frac{0.06 , \text{V}}{0.96 , \text{V}} + \frac{0.12 , R}{0.96 , \text{V}} - 0.5 , \Omega ]

[ R = \frac{1}{16} + \frac{1}{8}R - 0.5 , \Omega ]

[ R - \frac{1}{8}R = \frac{1}{16} + 0.5 , \Omega ]

[ \frac{7}{8}R = \frac{9}{16} ]

[ R = \frac{9}{16} \times \frac{8}{7} ]

[ R = \frac{9}{14} , \Omega ]

So, the value of resistor (R) is ( \frac{9}{14} , \Omega ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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