Let's solve **Question 1** in **two steps**, with detailed explanations. --- ### **Question 1 Restated** If the original resistor \( R_a = 1452.9\, \Omega \), what is the value of the shunting resistor \( R_s \) needed to **reduce the combined resistor value by 1%**? --- ### **Step 1: Find the Target Combined Resistance** When two resistors \( R_a \) and \( R_s \) are in parallel, the equivalent resistance is: \[ R_{eq} = \frac{R_a R_s}{R_a + R_s} \] We want \( R_{eq} \) to be **1% less** than \( R_a \): \[ R_{eq} = R_a - (1\% \text{ of } R_a) = R_a \times (1 - 0.01) = 0.99 R_a \] Plug in the value: \[ R_{eq} = 0.99 \times 1452.9 = 1438.371\, \Omega \] --- ### **Step 2: Solve for the Shunt Resistor \( R_s \)** Set up the parallel equation: \[ \frac{R_a R_s}{R_a + R_s} = 1438.371 \] Plug in \( R_a = 1452.9 \): \[ \frac{1452.9 \times R_s}{1452.9 + R_s} = 1438.371 \] Multiply both sides by \( (1452.9 + R_s) \): \[ 1452.9 \times R_s = 1438.371 \times (1452.9 + R_s) \] Expand: \[ 1452.9 R_s = 1438.371 \times 1452.9 + 1438.371 R_s \] Move all terms involving \( R_s \) to one side: \[ 1452.9 R_s - 1438.371 R_s = 1438.371 \times 1452.9 \] \[ (1452.9 - 1438.371) R_s = 1438.371 \times 1452.9 \] \[ 14.529 R_s = 2088056.3059 \] Solve for \( R_s \): \[ R_s = \frac{2088056.3059}{14.529} \approx \boxed{143682.6\, \Omega} \] **(Rounded to 1 decimal: \(\boxed{143,682.6\, \Omega}\))** --- ## **Summary Table** | Step | Calculation | Result | |------|-------------|--------| | 1 | Target parallel resistance \( R_{eq} \) | \( 1438.371\, \Omega \) | | 2 | Shunt resistor \( R_s \) needed | \( 143,682.6\, \Omega \) | --- **Let me know if you want the solutions for Questions 2 and 3!**

Certainly! Let's carefully solve all three problems step-by-step with detailed explanations, following a clear approach. --- ## **Question 1** **Given:** - Original resistor \( R_a = 1452.9\, \Omega \) - Desired reduction in resistance by 1% (i.e., the equivalent resistance should be 1% less than \( R_a \)). ### **Step 1: Find the target equivalent resistance** The goal is to reduce \( R_a \) by 1%, so: \[ R_{eq} = R_a \times (1 - 0.01) = 0.99 R_a \] Calculate: \[ R_{eq} = 0.99 \times 1452.9 = 1438.371\, \Omega \] ### **Step 2: Find the shunt resistor \( R_s \) value** When resistors are in parallel: \[ R_{eq} = \frac{R_a R_s}{R_a + R_s} \] Plug in the known \( R_a \) and desired \( R_{eq} \): \[ 1438.371 = \frac{1452.9 \times R_s}{1452.9 + R_s} \] Multiply both sides by \( 1452.9 + R_s \): \[ 1438.371 \times (1452.9 + R_s) = 1452.9 \times R_s \] Expand: \[ 1438.371 \times 1452.9 + 1438.371 R_s = 1452.9 R_s \] Bring all \( R_s \) terms to one side: \[ 1438.371 \times 1452.9 = 1452.9 R_s - 1438.371 R_s \] Calculate the numerator: \[ 1438.371 \times 1452.9 \approx 2,088,056.31 \] Calculate the difference: \[ 1452.9 R_s - 1438.371 R_s = (1452.9 - 1438.371) R_s = 14.529 R_s \] Now, solve for \( R_s \): \[ 14.529 R_s = 2,088,056.31 \] \[ R_s = \frac{2,088,056.31}{14.529} \approx \boxed{143,682.6\, \Omega} \] **_Answer for Q1:_** **The shunt resistor \( R_s \) should be approximately 143.68 kΩ to reduce the resistor value by 1%.** --- ## **Question 2** **Given:** - Original resistor \( R_a = 5\, \Omega \) - The combined resistor value needs to be **reduced by 5%** (i.e., to 95% of original). ### **Step 1: Find the target equivalent resistance** \[ R_{eq} = 0.95 \times R_a = 0.95 \times 5 = 4.75\, \Omega \] ### **Step 2: Find the shunt resistor \( R_s \)** Using the parallel formula: \[ R_{eq} = \frac{R_a R_s}{R_a + R_s} \] Plug in \( R_a = 5\, \Omega \): \[ 4.75 = \frac{5 \times R_s}{5 + R_s} \] Multiply both sides by \( 5 + R_s \): \[ 4.75 (5 + R_s) = 5 R_s \] Expand: \[ 4.75 \times 5 + 4.75 R_s = 5 R_s \] \[ 23.75 + 4.75 R_s = 5 R_s \] Bring all \( R_s \) terms to one side: \[ 23.75 = 5 R_s - 4.75 R_s = 0.25 R_s \] Solve for \( R_s \): \[ R_s = \frac{23.75}{0.25} = 95\, \Omega \] **_Answer for Q2:_** **The resistor \( R_s \) should be 95Ω to reduce the combined resistance by 5%.** --- ## **Question 3** **Given:** - Original resistor \( R_a = 678\, \Omega \) - Need to **increase** the resistor value by 10% by adding a resistor in series. ### **Step 1: Find the target resistance** Since adding a resistor in series increases total resistance: \[ R_{total} = R_a + R_s \] We want: \[ R_{total} = R_a \times (1 + 0.10) = 1.10 R_a \] Calculate: \[ R_{total} = 1.10 \times 678 = 745.8\, \Omega \] ### **Step 2: Find the series resistor \( R_s \)** \[ R_s = R_{total} - R_a = 745.8 - 678 = 67.8\, \Omega \] **_Answer for Q3:_** **Add a resistor of approximately 67.8Ω in series to increase the resistor value by 10%.** --- # **Summary of All Solutions** | Question | Action | Value of Resistor \( R_s \) | Explanation | |------------|---------|----------------------------|--------------| | **1** | Parallel to reduce by 1% | **143.68 kΩ** | To achieve a 1% reduction in resistance, find the parallel resistor \( R_s \) that yields the desired equivalent resistance. | | **2** | Parallel to reduce by 5% | **95Ω** | Similar process for 5% reduction. | | **3** | Series to increase by 10% | **67.8Ω** | Adding series resistor increases total resistance by the desired percentage. | --- If you'd like any further clarification or additional problems solved, feel free to ask!