DHS 18 Alternating Current (Tutorial Solutions)

DHS Physics H2 2025 Topic 18: Alternating Currents Tutorial Answers 18-1 For Internal Use Only MCQ Answer Key 1 2 2 rms 0 rms 0 rmsfor sinusoidal alternating current: and 22 V V V V PR R R R    I 0 rmsif is doubled, doubles, and increases by four timesV P I 2   2 2 2 0 rms 5.0 10 125 W2 2P R R    II 3 half-wave sinusoidal rectified current:     0 rms 22 rms 7.6 3.8 A2 2 3.8 9.4 135 W 140 W (to 2 s.f.)P R        II I 4 full-wave sinusoidal rectified voltage:     0 rms rms rms 22 rms 224 158.39 V 2 2 158.39 V 2.26 A70.0 and 2.263 70.0 358 W VV V R P R            I I 5 A sinusoidal current I through a 10.0  resistor varies with time t according to the equation I = 1.2 sin 100t. Determine the current and power dissipated at t = 0.018 s. I = 1.2 sin (100 × 0.018) =  0.705 A P = I2R = ( 0.705)2 × 10 = 4.98 W 6 A steady current I dissipates a certain power in a variable resistor. The resistance has to be halved to obtain the same power when a sinusoidal alternating current is used. What is the r.m.s. value of the alternating current?   2 2 rms 2 2 rms rms and 2 Since 2 2 dc ac dc ac RP R P P P RR              I I I I I I 7 Determine the r.m.s. current in each case. (a) A sinusoidal current of peak value 2.0 A. For a sinusoidal current, o rms 2 1.41 A 2 2   II (b) A full-wave rectified sinusoidal current of peak value 3.0 A. o rms 3.0 2.12 A 2 2   II 1 2 3 4 B C C A

DHS Physics H2 2025 Topic 18: Alternating Currents Tutorial Answers 18-2 For Internal Use Only (c) A square wave current with a frequency of 1.0 Hz and has a value of 0.10 A for one half cycle and 0.10 A for the next half cycle. I2 would yield a constant value of 0.010 A2. Hence <I2> = 0.010 A2 and Irms = 0.10 A. rms o 0.10 A I I (d) An uneven square wave voltage as shown below. T T TV dt V V V T T RR                    2 2 2 2 0 rms rms 2 14 13 3 3.0 V 1.73 V 1.73 A where (in ) is the resistance in th e a.c. circuitI I/A t/s T 2T I0 - I0 I02 V2 voltage 2 1 2 3 4 5 6

DHS Physics H2 2025 Topic 18: Alternating Currents Tutorial Answers 18-3 For Internal Use Only 8 An a.c. power supply is connected to three resistors as shown. The variation with time t of the voltage V of the power suply is given by the expression V = 9.0 sin 120t (a) For the power supply, determine (i