FCC Exam Question: 3-10B4
In Figure 3B1, what is the voltage drop across R1?
Explanation: Figure 3B1 typically depicts a series circuit with a 12-volt DC power source connected to two resistors: R1 and R2. 1. **Identify Component Values:** From Figure 3B1, the voltage source is 12 volts, R1 is 5 ohms, and R2 is 7 ohms. 2. **Calculate Total Resistance (Rt):** In a series circuit, the total resistance is the sum of individual resistances. Rt = R1 + R2 = 5 ohms + 7 ohms = 12 ohms. 3. **Calculate Total Circuit Current (I):** Using Ohm's Law (I = V/R), the total current flowing through the circuit is: I = 12 volts / 12 ohms = 1 ampere. In a series circuit, the current is the same through every component. 4. **Calculate Voltage Drop Across R1 (V_R1):** Using Ohm's Law (V = I * R), the voltage drop across R1 is: V_R1 = 1 ampere * 5 ohms = 5 volts. Therefore, the voltage drop across R1 is 5 volts. Options A, B, and D are incorrect because they do not reflect the correct application of Ohm's Law based on the component values in Figure 3B1. For example, 7 volts would be the voltage drop across R2 (1A * 7 ohms).
3-55G3
3-40E2
3-18B2
3-77L2
3-24C6
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Includes Elements 1, 3, 6, 7R, 8, and 9.