To solve for the unknown resistor in this circuit, we can follow these steps:
Step 1: Determine Total Resistance of the Load
First, we can use Ohm's Law to determine the voltage across the load. The power drawn by the load is given as \( P = 27 , \text{W} \) with a current of \( I = 3 , \text{A} \). The voltage across the load can be calculated as follows:
\[ V = I \cdot R \]
We also know that:
\[ P = V \cdot I \]
Thus,
\[ V = \frac{P}{I} = \frac{27 , \text{W}}{3 , \text{A}} = 9 , \text{V} \]
Step 2: Determine the Voltage Across the Parallel Resistors
The total voltage across the resistors and the load is given as \( V_{total} = 48 , \text{V} \). Since the load uses 9V, the voltage across the parallel combination of the resistors is:
\[ V_{parallel} = V_{total} - V_{load} = 48 , \text{V} - 9 , \text{V} = 39 , \text{V} \]
Step 3: Calculate the Total Current through the Parallel Resistors
Next, we can calculate the total current flowing through the resistors in parallel. The total current drawn by the load is \( I = 3 , \text{A} \), so the current flowing through the parallel resistors is also 3A.
However, we need to find how this current splits through the parallel resistors.
Step 4: Calculate Equivalent Resistance of the Parallel Resistors
The equivalent resistance \( R_{eq} \) of two resistors \( R_1 \) (unknown) and \( R_2 = 40 , \Omega \) in parallel can be calculated using the formula:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
From Ohm's Law, we can also express the currents through these resistors as follows:
\[ I_{total} = I_1 + I_2 \]
where:
- \( I_1 = \frac{V_{parallel}}{R_1} \)
- \( I_2 = \frac{V_{parallel}}{40} \)
Given that \( I_{total} = 3A \):
\[ I_1 + I_2 = 3A \]
Substituting for \( I_1 \) and \( I_2 \):
\[ \frac{39}{R_1} + \frac{39}{40} = 3 \]
Step 5: Solve for \( R_1 \)
Now, let's rearrange the equation:
\[ \frac{39}{R_1} + \frac{39}{40} = 3 \]
Multiply through by \( R_1 \cdot 40 \):
\[ 39 \cdot 40 + 39 R_1 = 3 \cdot R_1 \cdot 40 \]
Simplifying this gives:
\[ 1560 + 39 R_1 = 120 R_1 \]
Now, rearranging gives:
\[ 1560 = 120 R_1 - 39 R_1 \]
\[ 1560 = 81 R_1 \]
Now, divide by 81:
\[ R_1 = \frac{1560}{81} \approx 19.26 , \Omega \]
Conclusion
The value of the unknown resistor is approximately 19.26 Ohms.