Railway Science (Physics-Electric Current and its Effects) (Part-VI)Total Questions: 5031. Which of the following is an INCORRECT formula for electric power? [RRC Group D 13/09/2022 (Afternoon)](a) P = 𝑉/𝐼(b) P = 𝐼²R(c) P = IV(d) P = V²/RCorrect Answer: (a) P = 𝑉/𝐼Solution:According to ohm’s law V = IR ---(1) and The electric power is given by P = VI -----(2) where, V is the potential difference, I is the electric current and P is the electric power Putting value of V from equation 1 to equation 2, P = I²R Putting the value of I (I = V/R ) from equation (1) to equation (2) P = V²/R32. Two resistors, each of 20Ω, are connected in parallel. The combination, in turn, is connected in series to a third resistor of 20Ω and a battery of 3 V. The power supplied by the battery is: [RRC Group D 13/09/2022 (Evening)](a) 0.675 W(b) 0.15 W(c) 1.35 W(d) 0.30 WCorrect Answer: (d) 0.30 WSolution:33. If the current and potential difference are tripled in a circuit, keeping the resistance constant, then power will increase: [RRC Group D 14/09/2022 (Morning)](a) Three fold(b) six fold(c) One third(d) nine foldCorrect Answer: (d) nine foldSolution:Power = Rate at which work is done or energy is transferred per unit time. Power (P) = VI ; where v = potential difference, I = current; P’ = V’I’ = 3V × 3I = 9VI = 9P.34. If the induced current in the coil of a conductor of resistance 4 Ω is 1 mA, then EMF induced in the coil is: [RRC Group D 14/09/2022 (Morning)](a) 1 V(b) 4 V(c) 4 mV(d) 1 mVCorrect Answer: (c) 4 mVSolution:I = E/R, E = IR, E = 4 × 1 = 4mv, where E= Electromotive force, I = current, R = resistance.35. If a toaster is connected to 440 V AC instead of 220 V AC, then the heat produced per second will be _______ of the initial amount. [RRC Group D 14/09/2022 (Afternoon)](a) half(b) one-fourth(c) four times(d) twiceCorrect Answer: (c) four timesSolution:36. If a constant current of 2 A is maintained in a conductor of resistance 100Ω, then the potential difference across the conductor is: [RRC Group D 15/09/2022 (Afternoon)](a) 50 V(b) 20 V(c) 200 V(d) 0.02 VCorrect Answer: (c) 200 VSolution:Given, Resistance (R) = 100 Ω, Current (I) = 2 A, V - potential difference. By ohm’s law, V = IR, so, V = 100 × 2 ⟹ V = 200 V.37. The correct relation for heat produced by electric current is: [RRC Group D 15/09/2022 (Afternoon)](a) H = R/T(b) H = V.I.T(c) H = V.I/T(d) H = I².RCorrect Answer: (b) H = V.I.TSolution:According to Joule's law of Heating, the heat energy generated is proportional to the time when the electric current and electrical resistance are constant. We know that, Energy = Power × Time Heat Energy due to current = Electric Power × Time H= P × T H = V × I × T {V= Potential Difference, I= Current} H = VIT H = (I × R) × I × T (V= I R, Ohm's Law) H = I²RT H = VIT = V × v/R × T = V²/R T. The unit of heat is joule.38. Electric power consumed by two electric bulbs (each with 5 Ω resistance) connected in series with a 3 V battery is : [RRC Group D 15/09/2022 (Evening)](a) 0.3 W(b) 3 W(c) 0.9 W(d) 1 WCorrect Answer: (c) 0.9 WSolution:V = 3V, R = 5 Ω Equivalent Resistance = R1 + R2 (In series) = 5 + 5 = 10 Ω P = V²/R = 3 × 3/10 = 9/10 = 0.9 W39. When 60 KJ heat is produced by a heater in 1 minute and 20 seconds then its power is : [RRC Group D 15/09/2022 (Evening)](a) 750W(b) 7.5 kW(c) 1.5 kW(d) 150WCorrect Answer: (a) 750WSolution:Work = 60 KJ = 60 × 10⁻³ J Time = 1 Minute and 20 Second = 60 + 20 = 80 Second Power = work/time = 60 × 10³/80 = 750W40. A bulb draws 36W when connected to a 12V supply. If it is connected to a 6V supply then power consumed will be [RRC Group D 16/09/2022 (Morning)](a) 12W(b) 6W(c) 9W(d) 24WCorrect Answer: (c) 9WSolution:Given:- P = 36, V = 12Submit Quiz« Previous12345Next »