Question

A resistor of a unknown value and a resistor of 40ohms are connected in parallel. A load of 27W, which draws a current of 3A, is connected in series with the two resistors in parallel.
Calculate the value of the unknown resistor if the circuit is connected across a 48V DC supply.( 7 marks)

Answer 1

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**.