to give 0.4 A ( 0.1); 4 max (iii) lamp A will have greater power dissipation; since current the same, but it takes greater share of p.d.; 2 max [25] 7. (a) (i) 1 EI; (ii) 1 I2r; (iii) 1 VI; (b) (from the conservation of energy), = + EI I2r VI; therefore, = / = 2 V E – Ir E V + Ir; (c) correct position of voltmeter; correct position of ammeter; correct position of variable resistor; 3 (d) (i) = when = 0; E V I so = 1.5 2 E V; (ii) recognize this is when = 0; V intercept on the = 1.3 (0.1) A; 2 x-axis (iii) is the slope of the graph; r sensible choice of triangle, at least half the line as hypotenuse; = ; = 1.2 (±0.1) or when = 0 = V E Ir; = r = ; = 1.2 3 (e) = 1.2; R = = 0.63A; I = = (0.63)2 × 1.2 = 0.48W / 0.47 W; 3 P I2R [18] 8. (a) [1] for each appropriate and valid point eg thermal energy is the KE of the component particles of an object;
to give 0.4 A ( 0.1); 4 max (iii) lamp A will have greater power dissipation; since current the same, but it takes greater share of p.d.; 2 max [25] 7. (a) (i) EI; 1 (ii) I2r; 1 (iii) VI; 1 (b) (from the conservation of energy), EI = I2r + VI; therefore, V = E – Ir / E = V + Ir; 2 (c) correct position of voltmeter; correct position of ammeter; correct position of variable resistor; 3 (d) (i) E = V when I = 0; so E = 1.5 V; 2 (ii) recognize this is when V = 0; intercept on the x-axis = 1.3 (0.1) A; 2 (iii) r is the slope of the graph; sensible choice of triangle, at least half the line as hypotenuse; = ; = 1.2 (±0.1) or when V = 0 E = Ir; r = = ; = 1.2 3 (e) R = 1.2; I = = 0.63A; P = I2R = (0.63)2 × 1.2 = 0.48W / 0.47 W; 3 [18] 8. (a) [1] for each appropriate and valid point eg thermal energy is the KE of the component particles of an object;
ib-electricity-ws-01-as.docx