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 r