Problem Analysis:

This circuit involves two batteries ($\mathcal{E}_0$ and $\mathcal{E}$) and resistors $R_0$, $R_1$, $R_2$, and $R_3$. The problem states that the current through resistor $R_2$ is zero. This condition implies there is no potential difference across $R_2$, meaning the potentials at its two terminals are equal.

We need to calculate $\mathcal{E}$, given the following values:

• $\mathcal{E}_0 = 19 \, \text{V}$
• $R_0 = 150 \, \Omega$
• $R_1 = 145 \, \Omega$
• $R_2 = 280 \, \Omega$
• $R_3 = 50 \, \Omega$

Solution:

Step 1: Potential Across $R_2$ is Zero

If the current through $R_2$ is zero, the potential at both ends of $R_2$ must be the same. Let us denote the potentials of the top and bottom nodes connected to $R_2$ by $V_A$ and $V_B$, respectively.

This gives: $V_A = V_B$

Step 2: Voltage at $V_A$

The voltage at $V_A$ is determined by the battery $\mathcal{E}_0$ and the current through $R_0$. Let the current through $R_0$ be $I_0$. Using Ohm's Law: $I_0 = \frac{\mathcal{E}_0}{R_0}$

Substitute $\mathcal{E}_0 = 19 \, \text{V}$ and $R_0 = 150 \, \Omega$: $I_0 = \frac{19}{150} = 0.127 \, \text{A}$

The voltage at $V_A$ is: $V_A = \mathcal{E}_0 - I_0 R_0$

Substitute: $V_A = 19 - (0.127 \times 150) = 19 - 19 = 0 \, \text{V}$

Step 3: Voltage at $V_B$

The voltage at $V_B$ is determined by the battery $\mathcal{E}$, and the current through $R_1$ and $R_3$. Let the current through these resistors be $I_1$. Since $R_2$ has no current, $R_1$ and $R_3$ form a single series branch.

The total resistance in this branch is: $R_{\text{total}} = R_1 + R_3 = 145 + 50 = 195 \, \Omega$

The current in this branch is: $I_1 = \frac{\mathcal{E}}{R_{\text{total}}}$

The voltage at $V_B$ is: $V_B = \mathcal{E} - I_1 R_1$

Substitute $I_1 = \frac{\mathcal{E}}{195}$: $V_B = \mathcal{E} - \left(\frac{\mathcal{E}}{195}\right) R_1$

Substitute $R_1 = 145 \, \Omega$: $V_B = \mathcal{E} - \left(\frac{\mathcal{E}}{195}\right)(145)$

Simplify: $V_B = \mathcal{E} \left(1 - \frac{145}{195}\right)$

$V_B = \mathcal{E} \left(\frac{50}{195}\right)$

Step 4: Equating $V_A$ and $V_B$

Since $V_A = V_B$, we have: $0 = \mathcal{E} \left(\frac{50}{195}\right)$

Solve for $\mathcal{E}$: $\mathcal{E} = 0 \, \text{V}$

Final Answer:

The value of $\mathcal{E}$ is: $\boxed{\mathcal{E} = 0 \, \text{V}}$

Let me know if you'd like further clarifications or assistance! 😊

Follow-up Questions:

1. How does the condition of zero current through $R_2$ simplify the circuit analysis?
2. What would happen if $\mathcal{E}_0$ was different (e.g., $25 \, \text{V}$)?
3. Why does the potential difference across $R_2$ being zero imply $V_A = V_B$?
4. How would the problem change if $R_2$ had a current flowing through it?
5. Can we calculate the power dissipated in the resistors for this circuit?

Tip:

Always simplify a circuit based on given conditions (e.g., zero current through a branch) to reduce unnecessary calculations.