Let's solve the step-by-step calculation and data analysis based on the provided images and data. --- ## **Step 1: Understanding the Data** ### **Parallel Circuit (Fig. 3, Table 1)** - Voltage (V) and Current (I) are given. - Equivalent Resistance \( R_{eq} = \frac{V}{I} \). - The table already provides calculated resistances. ### **Series Circuit (Fig. 5, Table 2)** - Same as above, \( R_{eq} = \frac{V}{I} \). --- ## **Step 2: Calculate Equivalent Resistance (Slope method)** You are asked to plot **Voltage (y-axis)** vs **Current (x-axis)**: - The slope \( \frac{V}{I} \) gives the equivalent resistance. Let's do the calculations for both circuits: --- ### **A. Parallel Circuit** Let's use the data points to calculate the average equivalent resistance. #### Data: | Voltage (V) | Current (I) | Resistance (Ω) | |-------------|-------------|----------------| | 1.78 | -0.05 | -35.6 | | 3.56 | -0.10 | -35.6 | | 5.86 | -0.17 | -34.47 | | 8.31 | -0.24 | -34.625 | | 10.35 | -0.31 | -33.38 | #### Calculate Slope (Resistance): You can use any two points or do a linear fit, but the average is: \[ R_{eq,parallel} = \frac{-35.6 + (-35.6) + (-34.47) + (-34.625) + (-33.38)}{5} \] \[ R_{eq,parallel} = \frac{-173.675}{5} = -34.74\,\Omega \] --- ### **B. Series Circuit** #### Data: | Voltage (V) | Current (I) | Resistance (Ω) | |-------------|-------------|----------------| | 1.77 | -0.01 | -177 | | 3.44 | -0.02 | -172 | | 5.78 | -0.03 | -192.6 | | 8.05 | -0.05 | 161 | | 10.32 | -0.06 | -172 | #### Calculate Slope (Resistance): \[ R_{eq,series} = \frac{-177 + (-172) + (-192.6) + 161 + (-172)}{5} \] \[ R_{eq,series} = \frac{-552.6}{5} = -110.52\,\Omega \] (*Note: The value at 8.05 V seems positive (161 Ω); it might be a typo since others are negative. Double-check your data source.*) --- ## **Step 3: Theoretical Calculation** ### **Series:** \[ R_{eq(series)} = R_1 + R_2 \] ### **Parallel:** \[ R_{eq(parallel)} = \frac{R_1 R_2}{R_1 + R_2} \] To use these, you'd need the values of \(R_1\) and \(R_2\) (not given directly). --- ## **Step 4: Answering the Questions** ### **Q1: Advantages and Disadvantages** **Series:** - **Advantages:** Simple to connect, current is the same through all elements. - **Disadvantages:** If one resistor fails, the entire circuit is broken; resistance adds up, reducing current. **Parallel:** - **Advantages:** If one branch fails, others keep working; total resistance is lower, allowing more current. - **Disadvantages:** More complex wiring; current splits, which may not be desirable in every application. --- ### **Q2: Current Behavior** - **Series:** Current is the same at every point. - **Parallel:** Current splits among branches; total current is the sum of branch currents. --- ### **Q3: Measured vs Theoretical Resistance** - **Did they match?** Small differences are expected due to measurement error, wire resistance, contact resistance, and possible instrument calibration errors. - **Factors:** Instrument error, connection resistance, tolerance of resistor values, temperature effects. --- ## **Final Answers** ### **A. Calculated Equivalent Resistances (from data):** - **Parallel:** \( R_{eq,parallel} \approx -34.74\ \Omega \) - **Series:** \( R_{eq,series} \approx -110.52\ \Omega \) *(if you consider average, check for possible typo in the data)* ### **B. Theoretical Formulas:** - **Series:** \( R_1 + R_2 \) - **Parallel:** \( \frac{R_1 R_2}{R_1 + R_2} \) --- **If you have the actual values of \(R_1\) and \(R_2\), substitute them into the formulas to check against your measured/average values.** --- Would you like help plotting the data or with any specific calculation?