To solve this, we will approach the problem by first determining the power handling limit of each resistor and then calculating the maximum allowable voltage that can be applied across the entire circuit without exceeding the power rating of any resistor.

Given:

• Resistors:
  • $R_1 = 3.5 \, \text{k}\Omega$
  • $R_2 = 2.0 \, \text{k}\Omega$
  • $R_3 = 1.4 \, \text{k}\Omega$
• Power rating of each resistor: $P_{\text{max}} = 0.25 \, \text{W}$

Step 1: Calculate the equivalent resistance of the parallel resistors

The two resistors $R_1$ and $R_2$ are connected in parallel. The formula for the equivalent resistance $R_{\text{eq, parallel}}$ of two resistors in parallel is:

$\frac{1}{R_{\text{eq, parallel}}} = \frac{1}{R_1} + \frac{1}{R_2}$

Substitute the values of $R_1$ and $R_2$:

$\frac{1}{R_{\text{eq, parallel}}} = \frac{1}{3500 \, \Omega} + \frac{1}{2000 \, \Omega}$

Solving this will give us $R_{\text{eq, parallel}}$.

Step 2: Add the series resistor

The equivalent resistance of the parallel combination is then added in series with the $1.4 \, \text{k}\Omega$ resistor. The total equivalent resistance $R_{\text{total}}$ of the entire network is:

$R_{\text{total}} = R_{\text{eq, parallel}} + R_3$

Step 3: Calculate maximum current

For each resistor, the power is related to the current by the formula:

$P = I^2 R$

Since $P_{\text{max}} = 0.25 \, \text{W}$, the maximum current $I_{\text{max}}$ through each resistor can be calculated as:

$I_{\text{max}} = \sqrt{\frac{P_{\text{max}}}{R}}$

We need to calculate the maximum current for each resistor separately and determine which one limits the current.

Step 4: Calculate maximum voltage

Once we know the limiting current, the maximum voltage $V_{\text{max}}$ that can be applied across the entire network is given by:

$V_{\text{max}} = I_{\text{max}} \times R_{\text{total}}$

I will now calculate these values step by step.### Step-by-Step Results:

1. Equivalent resistance of the parallel combination $R_{\text{eq, parallel}}$:
   $R_{\text{eq, parallel}} = 1272.73 \, \Omega$

2. Total equivalent resistance of the whole network $R_{\text{total}}$:
   $R_{\text{total}} = 2672.73 \, \Omega$

3. Maximum current through each resistor:

   • $I_{\text{max, R1}} = 8.45 \, \text{mA}$
   • $I_{\text{max, R2}} = 11.18 \, \text{mA}$
   • $I_{\text{max, R3}} = 13.36 \, \text{mA}$

   The limiting current is determined by $R_1$, which can handle a maximum current of $8.45 \, \text{mA}$.

4. Maximum voltage across the entire network:
   $V_{\text{max}} = 22.59 \, \text{V}$

Thus, the maximum voltage that can be applied across the whole network without exceeding the power rating of any resistor is approximately 22.59 V.

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Do you want more details or have any further questions?

Here are 5 related questions to expand on:

1. How would the maximum voltage change if the power rating of one resistor increased?
2. What happens if we change one resistor value in the parallel combination?
3. How would the maximum current differ for a different resistor power rating?
4. How does the total power consumption change with increased applied voltage?
5. What if the resistors were connected in a different configuration?

Tip: Always ensure that no resistor exceeds its power rating to avoid overheating and damage.