Question

Two non-ideal cells $E_1$ and $E_2$ of emf 6V and 3V respectively are connected with a variable load resistance $R_L$ as shown. Upon variation of $R_L$, it is observed that the maximum power dissipation in $R_L$ takes place for value $2.0\Omega$. The internal resistance of cell $E_1$ is known to be $1.2\Omega$. Then, the potential difference (in Volts) across cell $E_1$ for $R_L=1.0\Omega$ is ______.

Answer 1

4.8

Answer 2

The problem involves two non-ideal cells connected in series with a variable load resistance. We need to determine the potential difference across cell $E_1$ for a specific load resistance.

1. Analyze the Circuit Configuration: From the diagram, cell $E_1$ has its positive terminal at the top and negative terminal at the bottom. Cell $E_2$ has its positive terminal at the bottom and negative terminal at the top. The negative terminal of $E_1$ is connected to the negative terminal of $E_2$. This configuration means the cells are connected in series opposition. The net EMF ($E_{eq}$) of the combination is the difference between their EMFs. Since $E_1 = 6V$ and $E_2 = 3V$, the net EMF is: $E_{eq} = E_1 - E_2 = 6V - 3V = 3V$.

The direction of the current will be determined by the stronger cell, $E_1$. Thus, the current will flow clockwise in the circuit. The total internal resistance ($r_{eq}$) of the combination is the sum of their individual internal resistances: $r_{eq} = r_1 + r_2$.

2. Apply the Maximum Power Transfer Theorem: The problem states that maximum power dissipation in the load resistance ($R_L$) occurs when $R_L = 2.0\Omega$. According to the Maximum Power Transfer Theorem, maximum power is delivered to the load when the load resistance is equal to the total internal resistance of the source. Therefore, $R_{L,max\_power} = r_{eq}$. Given $R_{L,max\_power} = 2.0\Omega$ and $r_1 = 1.2\Omega$: $2.0\Omega = r_1 + r_2$ $2.0\Omega = 1.2\Omega + r_2$ $r_2 = 2.0\Omega - 1.2\Omega = 0.8\Omega$.

So, the internal resistance of cell $E_2$ is $0.8\Omega$. The total internal resistance of the equivalent battery is $r_{eq} = 1.2\Omega + 0.8\Omega = 2.0\Omega$.

3. Calculate the Current for $R_L = 1.0\Omega$: Now, we need to find the potential difference across cell $E_1$ when the load resistance $R_L = 1.0\Omega$. The total resistance in the circuit for this case is: $R_{total} = R_L + r_{eq} = 1.0\Omega + 2.0\Omega = 3.0\Omega$. The current ($I$) flowing through the circuit is given by Ohm's Law for the entire circuit: $I = \frac{E_{eq}}{R_{total}} = \frac{3V}{3.0\Omega} = 1.0A$.

4. Calculate the Potential Difference Across Cell $E_1$: Since the current flows clockwise (driven by $E_1$), cell $E_1$ is discharging (current flows out of its positive terminal). The potential difference ($V_1$) across cell $E_1$ is given by: $V_1 = E_1 - I \cdot r_1$ $V_1 = 6V - (1.0A \cdot 1.2\Omega)$ $V_1 = 6V - 1.2V$ $V_1 = 4.8V$.

The potential difference across cell $E_1$ for $R_L=1.0\Omega$ is $4.8V$.