VRAAG 8 (Begin op 'n nuwe bladsyn.) Die stroombanddiagram hieronder toon 'n battery met 'n emk van 12 V en 'n interne weerstand van 0,5 Ω wat aan drie weerstande, 'n gloeilamp, 'n skakelaar, 'n ammeter en 'n verbindingsdrade verbind is - NSC Physical Sciences - Question 8 - 2023 - Paper 1

To calculate the total external resistance of the circuit, we need to find the combined resistance of the resistors R1, R2 in series and R3 in parallel with their combination.

1. Calculate the series resistance: $R_{series} = R1 + R2 = 10 \, \Omega + 10 \, \Omega = 20 \, \Omega$

2. Calculate the total resistance using the parallel formula: $\frac{1}{R_{total}} = \frac{1}{R_{series}} + \frac{1}{R3} \implies \frac{1}{R_{total}} = \frac{1}{20} + \frac{1}{15}$ This leads to:
   $R_{total} = 7.5 \, \Omega$

Using Ohm's law, we calculate the total current in the circuit. The total voltage (12 V) divided by the total resistance (7.5 Ω) gives: $I = \frac{V}{R} = \frac{12}{7.5} = 1.6 \, A$ Thus, the ammeter will read approximately 1.6 A.

To find the power consumed by resistor R3, use the formula: $P = I^2 R$ First, calculate the current across R3 using the voltage across it: The voltage across R3 can be found by: $V_{R3} = I \times R_{total} = 1.6 \times 15 = 24 \, V$ Now apply it to find power: $P_{R3} = (1.6)^2 \times 15 = 38.4 \, W$

Die helderheid van die gloeilamp sal VERLAAG omdat die totale weerstand van die stroombaan toeneem as die skakelaar oopgemaak word, wat 'n afname in die totale stroomvlak tot gevolg het.

Die afname in totale stroom lei tot 'n afname in die stroom wat deur die gloeilamp vloei. Aangesien die gloeilamp afhanklik is van die stroom om te functioneer, sal 'n afname in stroom die helderheid van die gloeilamp verlaag. Soos die stroom afneem, sal die energie wat in die gloeilamp verbruik word, ook afneem, wat tot 'n verminderde helderheid lei.