H218 Alternating Currents - 2.1 Tutorial wSolutions (1718)

9749 H2 Physics Tutorial w/ Solutions Topic 18: Alternating Current Page 1 of 5 Topic 18 Alternating Current 1 A sinusoidal current flowing in a 10 Ω resistor varies with time according to the equation I = 1.2 sin (100 t). Calculate the instantaneous current and the instantaneous power dissipated at (a) t = 0.005 s ܫ= 1.2 sin(100ߨ× 0.005) = 1.2 A ܲ= ܫଶܴ= 1.2ଶܴ= 14.4 W (b) t = 0.010 s ܫ= 1.2 sin(100ߨ× 0.010) = 0 A ܲ= 0 W (c) t = 0.018 s ܫ= 1.2 sin(100ߨ× 0.018) = −0.705 A ܲ= ܫଶܴ= (−0.705)ଶ10 = 4.98 W 2 Determine the r.m.s. current in each case. (a) A sinusoidal current of peak value 2.0 A. For a sinusoidal current, ܫ௥.௠.௦. = ூ೚ √ଶ = ଶ √ଶ = 1.41 A (b) A full-wave rectified sinusoidal current of peak value 3.0 A. When a full-wave rectified sinusoidal current is squared, the power dissipated would be similar to the that of the regular sinusoidal current. Hence, ܫ௥.௠.௦. = ூ೚ √ଶ = ଷ √ଶ = 2.12 A (c) A square-wave current with a frequency of 1 Hz which is 0.1 A for one half cycle and 0.1 A for the next half cycle. I2 would yield a constant value of 0.01 A2. Hence <I2> = 0.01 A2 and Irms = 0.1 A. (d) An uneven square wave current as shown below. 〈ࡵ૛〉 = ∫ࡵ૛࢚ࢊ ࢀ ૙ ࢀ= ૝(૛) + ૚(૚) ૜ = ૜. ૙ ۯ૛ ∴ࡵ࢙࢓࢘= ඥ〈ࡵ૛〉 = ૚. ૠ૜ ۯ 1 2 3 4 5 6 7 8 9 current / A time / ms 0 2.0 – 1.0

9749 H2 Physics Tutorial w/ Solutions Topic 18: Alternating Current Page 2 of 5 3 An alternating current of r.m.s . value 2 A and a steady direct current I flowing through identical resistors dissipate heat at equal rates. Determine the value of current I. Since the r.m.s. value of an a.c. is the value of a steady d.c. which will dissipate energy at the same rate as the mean power dissipated by an a.c. in a given resistor, the value of I is 2 A. 4 A steady current I dissipates power P in a variable resistor. The resistance has to be halved to obtain the same power when a sinusoidal alternating current is used. Determine the r.m.s. value of the alternating current in terms of I. ܲௗ௖ =ܫଶܴ and ܲ௔௖ =ܫ௥.௠.௦.ଶ ோ ଶ Since ܲௗ௖ =ܲ௔௖ ܫଶܴ= ܫ௥.௠.௦.ଶܴ 2 ܫ௥.௠.௦. = √2ܫ 5 A sinusoidal current described by the equation I = 9.0 sin ωt flows through a 12.0 Ω resistor. Calculate (a) the r.m.s. value of the current through and potential difference across the resistor. ܫ௢ = 9.0 A For a sinusoidal current, ܫ௥.௠.௦. = ூ೚ √ଶ = ଽ.଴ √ଶ = 6.36 A ܸ௥.௠.௦. =ܫ௥.௠.௦.ܴ= 6.36 × 12 = 76.4 V (b) the maximum instantaneous power dissipated in the resistor. ܲ௢ =ܫ௢ଶܴ= 9ଶ × 12 = 972 W (c) the mean power dissipated in the resistor. ܲ௠௘௔௡ =ܫ௥.௠.௦.ଶܴ= 6.36ଶ × 12 = 486 W 6 An audio amplifier, represented by an a.c. source and the resistor R delivers alternating voltages at audio frequencies to the speaker. If the source puts out an r.m.s. alternating p.d. of 15 V, resistance R is 8.20 Ω and the speaker is equivalent to a resistance of 10.4 Ω, calculate the average power delivered to the speaker. By potential divider principle, ܸ௦௣௘௔௞௘௥ (௥.௠.௦.) = ܴ௦௣௘௔௞௘௥ ܴ௧௢௧௔௟ ܸ௧௢௧௔௟ (௥.௠.௦.) = 10.4 10.4 + 8.2 (15) = 8.39 V 〈ܲ〉 = ௏ೞ೛೐ೌೖ೐ೝ (ೝ.೘.ೞ.) మ ோೞ೛೐ೌೖ೐ೝ = ଼.ଷଽమ ଵ଴.ସ = 6.76 W

9749 H2 Physics Tutorial w/ Solutions Topic 18: Alternating Current Page 3 of 5 7 (a) A power station generates 100 kW of power which is transmitted at a potential difference of 10 kV. Given that the connecting cables have a total resistance of 20 Ω, determine the turn ratio required for an ideal step-down transformer to bring electrical energy to the home at 240 V. Current through transmission cables ܫ௖௔௕௟௘ = ௉೟ೝೌ೙ೞ೘೔೟೟೐೏ ௏೟ೝೌ೙ೞ೘೔ೞೞ೔೚೙ = ଵ଴଴×ଵ଴య ଵ଴×ଵ଴య = 10 A Potential difference across transmission cable ܸ௖௔௕௟௘ =ܫ௖௔௕௟௘ܴ= 10 × 20 = 200 V Potential difference at step-down transformer = 10 × 103 – 200 = 9800 V Hence turn ratio, ே೛ ேೞ = ଽ଼଴଴ ଶସ଴ = ଶସହ ଺ (b) Explain why, for the efficient transmission of electrical energy, it is necessary to (i) use an alternating supply and The process of stepping up (or down) utilizes the concept of electromagnetic induction which requires a changing magnetic linkage linking the secondary coil. This means that a changing magnetic flux has to be produced at the primary coil, which can only be achieved when an alternating supply is used. (ii) transmit at high voltage. When electrical energy is transmitted at high voltage, the transmission current can be reduced, which will result in less power loss through the transmission cables. 8 Electrical power of 4400 kW is supplied to an industrial consumer at a considerable distance from a generating station. This is represented below. In order to do this, the electricity supply company makes use of a circuit containing two transformers T and U. The transformers can be considered to be ideal and the supply cables to have negligible resistance. (a) The power is generated a t 11 kV r.m.s. and is supplied to the consumer at 11 kV r.m.s. Calculate, for the current supplied to the consumer, its (i) r.m.s. value, and ܫ௥.௠.௦. =ܲ ܸ௥.௠.௦. = 4400 × 10ଷ 11 × 10ଷ = 400 A (ii) peak value. For a sinusoidal current, ܫ௢ =ܫ௥.௠.௦.√2 = 400√2 = 566 A

9749 H2 Physics Tutorial w/ Solutions Topic 18: Alternating Current Page 4 of 5 (b) There is a potential difference of 275 kV r.m.s. between the supply cables. Calculate (i) the ratio p s N N for each transformer For transformer T, ேೞ ே೛ = ௏ೄ ௏೛ = ଶ଻ହ ଵଵ = 25 For transformer U, ேೞ ே೛ = ௏ೄ ௏೛ = ଵଵ ଶ଻ହ = ଵ ଶହ (ii) the r.m.s. value of current in the supply cables ܫ௥.௠.௦. =ܲ ܸ௥.௠.௦. = 4400 × 10ଷ 275 × 10ଷ = 16 A (c) Explain why, when the resistance of the supply cables cannot be neglected, this arrangement is preferable to a system which generates and transmits the power at the same voltage of 11 kV r.m.s. If there is resistance in the supply cables, there would be po wer loss ( I2R) when transmitting electrical energy across the cables. With a step -up and step -down transformer, the electrical energy wou ld be transmitted at a current lower than 400 A which would result in a much smaller power loss. 9 Determine the r.m.s. current Irms, in terms of its peak current Io, for a half-wave rectified sinusoidal current as shown below. Squaring the current: Hence, for half-wave rectification, <I2> = ¼ Io2 ⇒ Irms = ½ Io I t I o T 2T t I2/A2 IO2 T

9749 H2 Physics Tutorial w/ Solutions Topic 18: Alternating Current Page 5 of 5 Answer Key 1 (a) 1.2 A, 14.4 W (b) 0 A, 0 W (c) 0.705 A, 4.98 W 2 (a) 1.41 A (b) 2.12 A (c) 0.1 A (d) 1.73 A 3 2 A 4 2 I 5 (a) 6.4 A, 76.4 V (b) 972 W (c) 486 W 6 6.76 W 7 (a) 245:6 8 (a)(i) 400 A (ii) 566 A (b)(i) T:25, U: 25 1 (ii) Irms=16 A 9 Io / 2