Chapter 20. --------------- 2). Current is the amount of charge per unit time. I = q/t The total charge that pass through the CDROM in 1 minute is q = I * t = 0.27 A * 60 s = 16.2 C The charge of one electron is 1.6*10^(-16) C, thus the number of electrons is given by a ratio of the total chagre to charge of one electron: N = q/1.6*10^(-19) C = 16.2 C/1.6*10^(-19) C = 10^(20) 3). Use the Ohm's law: I = V / R = 120 V / 14 Ohm = 8.57 A 6). The electric power is P = I * V The energy delivered to the battery is equal to work performed by using power for a certain time E = P * t = 6 A * 12 V * 5 hours = = 6 A * 12 V * 18000s = 1.296 * 10^6 J 35). Delivered electric energy is E = P * t We must express power in terms of quantities available in the problem P = R * I^2 Thus E = R * I^2 * t = 5.3 Ohm * (25 A)^2 * 31 days * 9 hours/day = = 9.2 * 10^5 W*hour = 9.2*10^2 kW*hour The total cost of running the furnace is Cost = 920 kWh * \$0.1 /kWh = \$92 42). First we need to calculate the current in the circuit I = V / R_tot The total resistance of the two resistors connected in series is equal to a sum of individual resistances R_tot = 36 Ohm + 18 Ohm = 54 Ohm I = 15 V / 54 Ohm = 0.28 A The voltage across the 18 Ohm resistor is V_18 = I * R = 18 Ohm * 0.28 A = 5 V The voltage across 36 Ohm resistor is V_36 = I * R = 36 Ohm * 0.28 A = 10 V 51). For N resistors each with resistance R and all connected in parallel the total resistance of a system is 1/R_tot = 1/R + 1/R + 1/R +....1/R = N/R ---> N = R / R_tot = 4 Ohm / (1/16 Ohm) = 64 We need 64 resistors. 77). Apply the Kirchhof's rules. The current in the circuit is given by the equation constructed by adding all potential drops following the loop. The starting point is arbitrary. I start at point A and follow the loop clockwise 30 V - 27 Ohm * I - 10 V - 12 Ohm * I - 8 Ohm * I - 5 Ohm * I = 0 20 V - 52 Ohm * I = 0 a) I = 20 V / 52 Ohm = 0.38 A The answer is positive and that means that the current flows in the direction in which we formed the equation i.e., clockwise. b) Potential difference between points A and B is given by the battery potential decreased by the potential drop across the 27 Ohm resistor V_B - V_A = 30V - 0.38 A * 27 Ohm = 19.7 V c) The potential difference is positive V_B - V_A > 0 ====> V_B > V_A The potential at point B is higher than potential at point A. 90). Total charge on a capacitors (see formula 19.8) is q = V * C_tot The total capacitance of two capacities connected in parallel is C_tot = C_1 + C_2 = 2*10^(-6) F + 2*10^(-6) F = 6*10^(-6) F For parallel connction q = 60 V * 6*10^(-6) F = 3.6*10^(-4) C The total capacitance of two capacities connected in series is 1/C_tot = 1/C_1 + 1/C_2 C_tot = 1/(1/C_1 + 1/C_2) = 1/(1/4 + 1/6) microF = = 2.4 microF => For connection in series q = 60 V * 2.4*10^(-6) F = 1.44*10^(-4) C