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?

Question

Charles Aeschliman · Accepted Answer

#4 Omega#

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 #V_A - V_B#

Now, it is given that the potential drop across the cell is #0.96V#; that means this drop is due to some potential drop across the #0.5 Omega# internal resistance in the cell, we can say the rest i.e #(1.08-0.96)=0.12V# has dropped across #0.5 Omega#.

So, if the current flowing through the circuit is #I#, then we can say, #I×0.5=0.12#

Or, #I=0.24 A#

Now, the voltage drop across resistor #R# is #(1.08-0.12)=0.96V# (note it is the same as #V_A-V_B#)

So,if voltage drop across #R# is #0.96 V# and Current flowing is #0.24A#

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

#R=0.96/0.24=4 Omega#

Layla Ahmed · Answer

undefined 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 $.