Problem 24 - Physics for Scientists and Engineers with Modern Physics

Question

(III) A \(2.8-\mathrm{k} \Omega\) and a 3.7 \(\mathrm{k} \Omega\) resistor are connected in parallel; this combination is connected in series with a \(1.8-\mathrm{k} \Omega\) resistor. If each resistor is rated at \(\frac{1}{2} \mathrm{W}\) (maximum without overheating), what is the maximum voltage that can be applied across the whole network?

Step-by-Step Solution

Answer

The maximum voltage that can be applied across the whole network is approximately 39.34 V.

Step 1: Calculate Equivalent Resistance of Parallel Resistors

The two resistors, 2.8 kΩ and 3.7 kΩ, are connected in parallel. The equivalent resistance \( R_p \) can be calculated using the formula for resistors in parallel: \[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \].In our case:\[ \frac{1}{R_p} = \frac{1}{2800} + \frac{1}{3700} \]\[ \frac{1}{R_p} = \frac{3700 + 2800}{2800 \times 3700} \]\[ \frac{1}{R_p} = \frac{6500}{10360000} \]Therefore:\[ R_p = \frac{10360000}{6500} \approx 1593 \Omega \].

Step 2: Calculate Total Equivalent Resistance

The parallel combination \( R_p \) is in series with the 1.8 kΩ resistor. Calculate the total resistance \( R_t \) by adding these resistances:\[ R_t = R_p + 1800 = 1593 + 1800 = 3393 \Omega \].

Step 3: Determine Maximum Current per Resistor

Each resistor is rated at 0.5 W. The maximum current \( I \) through a resistor can be found using the power formula \( P = I^2 R \):- For the 2.8 kΩ resistor: \[ I^2 \times 2800 = 0.5 \Rightarrow I = \sqrt{\frac{0.5}{2800}} \approx 0.0133 \mathrm{A} \].- For the 3.7 kΩ resistor: \[ I^2 \times 3700 = 0.5 \Rightarrow I = \sqrt{\frac{0.5}{3700}} \approx 0.0116 \mathrm{A} \].- For the 1.8 kΩ resistor: \[ I^2 \times 1800 = 0.5 \Rightarrow I = \sqrt{\frac{0.5}{1800}} \approx 0.0167 \mathrm{A} \].Note the maximum current through the network will be limited by the smallest current which is 0.0116 A for the 3.7 kΩ resistor.

Step 4: Calculate Maximum Voltage

The maximum voltage \( V \) across the whole network can be calculated using Ohm's Law \( V = I R_t \), where \( I = 0.0116 \) A (maximum current identified from Step 3) and \( R_t = 3393 \Omega \):\[ V = 0.0116 \times 3393 \approx 39.34 \mathrm{V} \].Thus, the maximum voltage that can be applied across the network is approximately 39.34 V.

Key Concepts

Resistors in Parallel

When resistors are connected in parallel, the total or equivalent resistance decreases. This is because each resistor provides an additional path for the flow of electricity, making it easier for current to pass through.

• The formula used to calculate the equivalent resistance of two resistors in parallel is given by \[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \].
• The key point is that the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances.
• Once you have the reciprocal, you can invert it to find the actual equivalent resistance, \( R_p \).

In our example, for resistors of 2.8 \( k\Omega \) and 3.7 \( k\Omega \), this gives an equivalent resistance of approximately 1593 \( \Omega \). Having a lower equivalent resistance allows higher currents to flow through the parallel section compared to a single resistor configuration.

Ohm's Law

Ohm's Law is a fundamental principle used to calculate the relationship between voltage, current, and resistance in electrical circuits. It states that the voltage \( V \) across a resistor is the product of the current \( I \) flowing through it and its resistance \( R \).

• The formula is expressed as: \[ V = IR \].
• This law helps determine the required voltage to drive a current through a known resistance, or vice versa.

In the context of our exercise, Ohm's Law is crucial for calculating the maximum voltage across the network by first determining the current (limited by the smallest resistor) and then applying \( V = I R_t \) with the total equivalent resistance. This process ensures each component remains within its safe operational limits while the network performs optimally.

Power in Electrical Circuits

Power in electrical circuits relates to the rate at which energy is consumed or produced by an electrical component. Power can be calculated using several formulas, one of which incorporates Ohm's Law:

• \[ P = IV \] where \( P \) is power, \( I \) is current, and \( V \) is voltage.
• For a resistor, power can also be calculated as \[ P = I^2 R \] or \[ P = \frac{V^2}{R} \].

For this textbook exercise, we use the formula \( P = I^2 R \) to find the maximum current each resistor can handle without exceeding its power rating of 0.5 W. By identifying the smallest permissible current, we ensure that no resistor overheats, which is crucial for maintaining the functionality and safety of the circuit.

Equivalent Resistance

The concept of equivalent resistance is essential when analyzing complex circuits involving combinations of resistors. It simplifies a network of resistors into a single resistor that has the same overall effect on the circuit, allowing for easier calculations of current and voltage.

• To find the total equivalent resistance \( R_t \), combine parallel and series resistances.
• In series circuits, resistances are simply added: \[ R_t = R_1 + R_2 + \ldots \].

In our scenario, the parallel combination of 2.8 \( k\Omega \) and 3.7 \( k\Omega \) is first calculated. Then, this is added to the series resistor (1.8 \( k\Omega \)) to obtain the total resistance of the entire circuit: 3393 \( \Omega \). Knowing the equivalent resistance simplifies the task of applying Ohm’s Law to find overall voltage or current, thus streamlining problem-solving for complex resistor networks.