Understanding Resistors in Parallel and Series

The problem describes a circuit with two resistors in parallel and asks how to add a third resistor to achieve a specific total line current. We need to determine if the third resistor is added in series or parallel and what its value is.

Initial Circuit Analysis

We have a $40 \, \Omega$ resistor in parallel with an $80 \, \Omega$ resistor. The current through the $40 \, \Omega$ resistor is given as $6 \, \text{A}$.

Since the resistors are in parallel, the voltage across them is the same. We can calculate this voltage using Ohm's Law ($V = I \times R$) for the $40 \, \Omega$ resistor:

Voltage $V = I_{40\Omega} \times R_{40\Omega}$

$V = 6 \, \text{A} \times 40 \, \Omega = 240 \, \text{V}$

This voltage of $240 \, \text{V}$ is also across the $80 \, \Omega$ resistor.

Now we can calculate the current through the $80 \, \Omega$ resistor:

$I_{80\Omega} = \frac{V}{R_{80\Omega}} = \frac{240 \, \text{V}}{80 \, \Omega} = 3 \, \text{A}$

The initial total line current before adding the third resistor is the sum of the currents in the parallel branches:

$I_{\text{total\_initial}} = I_{40\Omega} + I_{80\Omega} = 6 \, \text{A} + 3 \, \text{A} = 9 \, \text{A}$

Adding the Third Resistor

The problem states that the line current is to be $10 \, \text{A}$ after adding a third resistor. The initial line current was $9 \, \text{A}$. Since the new line current ($10 \, \text{A}$) is greater than the initial line current ($9 \, \text{A}$), we must be adding a path for additional current. Adding a resistor in series would increase the total resistance (for a constant voltage source) and decrease the total current. Therefore, the third resistor must be added in parallel to the existing combination.

Adding the third resistor in parallel means the voltage across this third resistor is also $240 \, \text{V}$ (assuming the voltage source supplying the parallel branches remains at $240 \, \text{V}$).

With the third resistor ($R_3$) added in parallel, the total line current is the sum of the currents through all three parallel branches:

$I_{\text{total\_new}} = I_{40\Omega} + I_{80\Omega} + I_{R_3}$

We are given that the new total line current is $10 \, \text{A}$. The currents through the $40 \, \Omega$ and $80 \, \Omega$ resistors remain $6 \, \text{A}$ and $3 \, \text{A}$ respectively (because the voltage across them remains $240 \, \text{V}$).

$10 \, \text{A} = 6 \, \text{A} + 3 \, \text{A} + I_{R_3}$

$10 \, \text{A} = 9 \, \text{A} + I_{R_3}$

The current through the third resistor is:

$I_{R_3} = 10 \, \text{A} - 9 \, \text{A} = 1 \, \text{A}$

Now we can find the value of the third resistor using Ohm's Law ($R = \frac{V}{I}$), knowing the voltage across it is $240 \, \text{V}$ and the current through it is $1 \, \text{A}$:

$R_3 = \frac{V}{I_{R_3}} = \frac{240 \, \text{V}}{1 \, \text{A}} = 240 \, \Omega$

Conclusion

The third resistor must be added in parallel, and its value is $240 \, \Omega$.

Let's verify the options provided:

• Option 1: Parallel, $240 \, \Omega$ - This matches our calculation.
• Option 2: Parallel, $20 \, \Omega$ - Adding a $20 \, \Omega$ resistor in parallel would draw a much larger current ($240V / 20\Omega = 12A$), resulting in a total current far exceeding 10A.
• Option 3: Series, $24 \, \Omega$ - Adding in series would decrease the total current.
• Option 4: Series, $200 \, \Omega$ - Adding in series would decrease the total current.

Therefore, the third resistor is added in parallel and has a value of $240 \, \Omega$.