11. A resistor of $5\,\Omega$ carries a current of $4\,\text{A}$. Calculate the power dissipated using $P = I^2R$. [2]
Mark Scheme
1. Electric current is the rate of flow of electric charge [1]; measured in amperes (A) [1] [2]
2. $Q = It$ [1]; unit: coulomb (C) [1] [2]
3. $Q = It = 0.8 \times 50 = 40\,\text{C}$ [2]
4. Convert: 4 min = 240 s [1]; $I = Q/t = 120/240 = 0.5\,\text{A}$ [1] [2]
5. $E = QV$ [1]
6. $E = QV = 2 \times 6 = 12\,\text{J}$ [2]
7. $E = QV = 1800 \times 5 = 9000\,\text{J}$ [2]
8. $P = IV$ [1]; $P = I^2R$ [1]; $P = V^2/R$ [1] [3]
9. a) $P = IV = 3 \times 10 = 30\,\text{W}$ [2]; b) $E = Pt = 30 \times 120 = 3600\,\text{J}$ [2] [4]
10. $I = P/V = 60/12 = 5\,\text{A}$ [1]; $Q = It = 5 \times 30 = 150\,\text{C}$ [2] [3]
11. $P = I^2R = 4^2 \times 5 = 16 \times 5 = 80\,\text{W}$ [2]
12. $I = P/V = 2000/230 \approx 8.7\,\text{A}$ [2]