To find the current in the resistor $R$, we first need to calculate the net voltage across the circuit. In the given circuit, we have two voltage sources: $V_1$ and $V_2$. The polarity of the batteries suggests that they are connected in opposition to each other. Therefore, the net voltage $V_{\text{net}}$ is the difference between $V_2$ and $V_1$:

$V_{\text{net}} = V_2 - V_1$

Given:

• $V_1 = 7.8 \, \text{V}$
• $V_2 = 9.3 \, \text{V}$
• $R = 33 \, \Omega$

Now, we can calculate the current $I$ using Ohm's law:

$I = \frac{V_{\text{net}}}{R}$

Let me calculate the value for you.The magnitude of the current through the resistor is approximately $0.0455 \, \text{A}$ or 45.5 mA.

Would you like further details or have any questions?

Here are five related questions to deepen your understanding:

1. What happens if the values of $V_1$ and $V_2$ are switched?
2. How would the current change if the resistance $R$ were doubled?
3. How does the direction of current change when $V_1$ becomes greater than $V_2$?
4. What is the power dissipated by the resistor in this circuit?
5. How would adding another resistor in series affect the current?

Tip: In circuits with opposing voltage sources, always subtract the smaller voltage from the larger one to find the net voltage.