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What is the magnitude of the current in the resistor? (V1=6.3 V, V2=11.8 V, R=30 Ω.)


In the given circuit, we have two voltage sources, \( V_1 \) and \( V_2 \), and a resistor \( R \). To find the current flowing through the resistor, we need to apply Kirchhoff's Voltage Law (KVL), which states that the sum of the potential differences around a closed loop is zero.

### Step-by-Step Solution:

1. **Identify the voltages and resistor:**
   - \( V_1 = 6.3 \, V \)
   - \( V_2 = 11.8 \, V \)
   - \( R = 30 \, \Omega \)

2. **Apply Kirchhoff's Voltage Law (KVL):**
   In the circuit, the voltages \( V_1 \) and \( V_2 \) are opposing each other. The total effective voltage across the resistor \( R \) is:
   \[
   V_{\text{net}} = V_2 - V_1 = 11.8 \, V - 6.3 \, V = 5.5 \, V
   \]

3. **Use Ohm's Law to find the current:**
   Ohm's Law states that \( I = \frac{V}{R} \), where \( I \) is the current, \( V \) is the net voltage, and \( R \) is the resistance.
   \[
   I = \frac{V_{\text{net}}}{R} = \frac{5.5 \, V}{30 \, \Omega} = 0.1833 \, A
   \]

### Final Answer:
The magnitude of the current in the resistor is approximately \( 0.183 \, A \).

Do you need further details or have any other questions?

### Follow-up Questions:
1. How does Kirchhoff’s Voltage Law work for more complex circuits?
2. What happens if the polarities of the batteries were reversed?
3. How would you calculate power dissipation in the resistor?
4. Can you solve the same problem using Thevenin's theorem?
5. How do you calculate the total resistance if there were multiple resistors in the circuit?

### Tip:
When dealing with opposing voltage sources, always subtract the smaller voltage from the larger one to find the effective voltage across the components.

What is the magnitude of the current in the resistor? (V1 = 6.3 V, V2 = 11.8 V, R = 30 Ω.)

Learn how to calculate the current through a resistor in a simple circuit with opposing voltage sources (V1 = 6.3 V, V2 = 11.8 V) and resistance (R = 30 Ω). Using Kirchhoff's Voltage Law and Ohm’s Law, this step-by-step guide simplifies circuit analysis. Perfect for Grades 10-12.

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