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Here is the solution to question 8 in English: 8. (a) (i) Ohm's Law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature and other physical conditions remain constant. Mathematically, it is expressed as V = IR. 8. (a) (ii) To plot the graph of potential difference, V, against current, I: • Draw an x-axis (horizontal) for current I (in amperes, A) and a y-axis (vertical) for potential difference V (in volts, V). • Choose appropriate scales for each axis so that all data points can be represented. • Plot the given data points: (0.0, 0.0), (1.1, 1.5), (2.1, 3.0), (3.2, 4.5), (4.4, 6.0), (5.7, 8.0), (6.4, 9.0), (7.1, 10.0). • Draw the best straight line passing through the origin and the plotted points. 8. (a) (iii) Step 1: Calculate the gradient of the V-I graph. The gradient m is given by the formula m = ( V)/( I). Using the points (0.0 A, 0.0 V) and (7.1 A, 10.0 V) from the table (representing the best straight line passing through the origin): m = 10.0 V - 0.0 V7.1 A - 0.0 A = (10.0)/(7.1) ≈ 1.408 V/A Step 2: State the significance of the gradient. The gradient of a potential difference (V) versus current (I) graph represents the resistance of the conductor. The gradient is approximately 1.41 . Its significance is the resistance of the conductor. 8. (a) (iv) Step 1: Use Ohm's Law and the calculated resistance. From Ohm's Law, V = IR. We calculated the resistance R ≈ 1.408 from the gradient. We want to find the current I when V = 7.5 V. I = (V)/(R) Step 2: Substitute the values and calculate the current. I = 7.5 V1.408 ≈ 5.3267 A The current in the circuit when the potential difference is 7.5 V is approximately 5.33 A. 8. (b) (i) Electromotive force (e.m.f.) is the energy supplied by a source (such as a battery) per unit charge to drive the charge around a complete circuit. It is measured in volts (V) and represents the potential difference across the terminals of the source when no current is flowing. 8. (b) (ii) The resistance of a wire is given by the formula R = (L)/(A), where is resistivity, L is length, and A is cross-sectional area. If the wire is replaced by a thinner wire of the same material and length, its cross-sectional area A decreases. A decrease in A leads to an increase in resistance R. According to Ohm's Law (I = (V)/(R)), if the resistance R increases and the battery voltage V remains constant, the current I will decrease. 8. (c) (i) The function of a fuse is to protect electrical circuits from overcurrents. It melts and breaks the circuit if the current exceeds a safe level, preventing damage to appliances and reducing fire hazards. It is connected in series with the live wire of the circuit. 8. (c) (ii) A fuse contains a thin wire made of a material with a low melting point. When the current flowing through the circuit exceeds the fuse's rated value, the wire heats up due to the Joule heating effect (P = I^2 R). This heat causes the fuse wire to melt, creating an open circuit. This interruption stops the flow of current, thereby protecting appliances and preventing electrical hazards. That's 2 down. 3 left today — send the next one.