This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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You're on a roll — Step 1: Address Question 3 - Determine the Emf of the battery. Question 3 asks to measure and record the electromotive force (Emf) of the battery with the circuit open. When the circuit is open, no current flows (I = 0 A). In this condition, the voltmeter measures the Emf (E) of the battery, as the voltage drop due to internal resistance (Ir) is zero. From the given data: • When Current (I) = 0.00 A, Potential Difference (V) = 3.90 V. Therefore, the electromotive force (E) of the battery is: E = 3.90 V E = 3.90 V Step 2: Address Question 4 - Use the data to determine the internal resistance (r). Question 4 instructs to record ammeter and voltmeter readings for different external resistances. The provided data is: | Reading | Current I (A) | Potential Difference V (V) | | :------ | :-------------- | :--------------------------- | | 1 | 0.00 | 3.90 | | 2 | 0.19 | 3.69 | | 3 | 0.23 | 3.65 | | 4 | 0.27 | 3.61 | | 5 | 0.33 | 3.55 | The relationship between terminal potential difference (V), Emf (E), current (I), and internal resistance (r) is given by: V = E - Ir To find the internal resistance (r), we can rearrange the formula or use the gradient of a V vs I graph. Since we already found E, we can use any data point (I, V) to calculate r. Let's use the last data point (I=0.33 A, V=3.55 V) and the calculated E=3.90 V. Substitute the values into the equation: 3.55 V = 3.90 V - (0.33 A) × r Step 3: Calculate the internal resistance (r). Rearrange the equation to solve for r: (0.33 A) × r = 3.90 V - 3.55 V (0.33 A) × r = 0.35 V r = 0.35 V0.33 A r ≈ 1.0606 Rounding to two decimal places, the internal resistance (r) is: r = 1.06 Got more? Send 'em.