Two resistors of `6 Omega " and " 9 Omega` are connected in series to a 120 V source. The power consumed by the `6 Omega` resistor is

Options

A. 384 W
B. 616 W
C. 1500 W
D. 1800 W

Answer

To solve the problem of finding the power consumed by the 6 ohm resistor when connected in series with a 9 ohm resistor to a 120 V source, we can follow these steps:

### Step 1: Calculate the Total Resistance
In a series circuit, the total resistance (R_total) is the sum of the individual resistances.

\[ R_{\text{total}} = R_1 + R_2 = 6 \, \Omega + 9 \, \Omega = 15 \, \Omega \]

**Hint:** Remember that in a series circuit, resistances simply add up.

### Step 2: Calculate the Total Current
Using Ohm's Law (V = IR), we can find the total current (I) flowing through the circuit.

\[ I = \frac{V}{R_{\text{total}}} = \frac{120 \, V}{15 \, \Omega} = 8 \, A \]

**Hint:** Ohm's Law relates voltage, current, and resistance; make sure to use the total resistance for the entire circuit.

### Step 3: Calculate the Voltage Drop across the 6 Ohm Resistor
The voltage drop (V1) across the 6 ohm resistor can be calculated using Ohm's Law again.

\[ V_1 = I \times R_1 = 8 \, A \times 6 \, \Omega = 48 \, V \]

**Hint:** The voltage drop across a resistor in a series circuit can be found by multiplying the current by the resistance of that resistor.

### Step 4: Calculate the Power Consumed by the 6 Ohm Resistor
The power (P1) consumed by the 6 ohm resistor can be calculated using the formula:

\[ P_1 = \frac{V_1^2}{R_1} = \frac{(48 \, V)^2}{6 \, \Omega} = \frac{2304}{6} = 384 \, W \]

**Hint:** Power can also be calculated using the voltage drop across the resistor and its resistance.

### Final Answer
The power consumed by the 6 ohm resistor is **384 watts**.