A current through the thin filament wire of a bulb causes the filament to become Resistance 3 white hot, while the larger wires connected to the light bulb remain much cooler. This happens because (A) the larger connecting wires have more resistance than the filament. (B) the thin filament has more resistance than the larger connecting (C) the filament wire is not insulated. (D) the current in the filament is greater than that through the connecting (E) the current in the filament is less than that through the connecting In series circuits, larger resistors develop more power 97 Each member of a family of six owns a computer rated at 500 watts in a 120 V circuit. If Safety device all computers are Fuse plugged into a single circuit protected by a 20 ampere fuse, what is the maximum number Power of the computers can be operating at the same time? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 or more Each computer draws I = P/V = 4.17 A. 4 computers will draw 16.7 A, while 5 will draw over 20 A. 21 The product (2 amperes × 2 × 2 seconds) is equal to 2 (A) 8 coulombs (B) 8 newtons (C) 8 joules (D) 8 calories (E) 8 newton–amperes Amperes = I (current); Volts = V (potential difference); Seconds = t (time): IVt = energy
A current through the thin filament wire of a bulb causes the filament to become white hot, while the larger wires connected to the light bulb remain much cooler. This happens because (A) the larger connecting wires have more resistance than the filament. (B) the thin filament has more resistance than the larger connecting (C) the filament wire is not insulated. (D) the current in the filament is greater than that through the connecting (E) the current in the filament is less than that through the connecting In series circuits, larger resistors develop more power Resistance 3 97 Each member of a family of six owns a computer rated at 500 watts in a 120 V circuit. If all computers are plugged into a single circuit protected by a 20 ampere fuse, what is the maximum number of the computers can be operating at the same time? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 or more Each computer draws I = P/V = 4.17 A. 4 computers will draw 16.7 A, while 5 will draw over 20 A. Safety device Fuse Power 21 The product (2 amperes × 2 × 2 seconds) is equal to (A) 8 coulombs (B) 8 newtons (C) 8 joules (D) 8 calories (E) 8 newton–amperes Amperes = I (current); Volts = V (potential difference); Seconds = t (time): IVt = energy 2