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) the frequency f, 2 120 60 Hz f f       (ii) the root-mean-square (r.m.s.) voltage [60 Hz, 6.4 V] 0 rms 9.0    6.4 V 2 2 VV    (b) Calculate the peak current from the power supply. [0.56 A] 0 0 6.0 12effective resistance of external circuit, 12 12 4.0 16.0 6.0 12 9.0peak current, 0.56 A 16.0 R V R         I (c) Calculate the mean power dissipated in the resistor of resistance 6.0 . [0.42 W] Using potential divider principle, peak voltage across the 6.0 Ω resistor =  4.0 9.0 2.25 V16.0  mean power dissipated in the 6.0 Ω resistor,   22 2 rms 0 2.25 0.42 W2 2 6.0 V VP R R     9 (a) The r.m.s. value of an a.c. is the value of the steady direct current which would dissipate heat at the same average rate in a given resistor. (b) (i) For a sinusoidal voltage, 0 rms 170 120 V 2 2 VV    (ii) 377 60 Hz2 2f      (iii) 2 2 01 1 170 249 W2 2 58 VP R   (c) 1 1 0 017 s60T . f  

DHS Physics H2 2025 Topic 18: Alternating Currents Tutorial Answers 18-4 For Internal Use Only 2 2499 sinVP t R   A graph of P vs t shows a positive sine square curve with a time equal to two periods of the alternating potential difference: MCQ Answer Key 10        0 02 1 0 2 0 1 1 180 11.25 V16 16 16 V V VV        2 2 2 2 rms 0 0 11.25R 1.98 2.0 (to 2 s.f.)R 2R 2 2 32 V V VP P        11 from rms, 2 rms, 12 1 rms, 1 rms, 2 turn ratio VN N V   I I , 1 rms, 1 1 rms, 1 600 9.0 250 1we have 750 3 3 600 1800 and 3 9.0 27 V N V N V           12 from rms, 2 rms, 12 1 rms, 1 rms, 2 turn ratio VN N V   I I , 2 1if is doubled and stays the same, then the new turn ratioN N  2 ps p s V V  I I 1 (halves) and 2 (doubles)2 p p s s V V   I I 10 11 12 B D B 0 T = 0.0166 2T = 0.0333 t / s 499 600 400 200

DHS Physics H2 2025 Topic 18: Alternating Currents Tutorial Answers 18-5 For Internal Use Only 13 (a) Use Faraday’s law to explain why an alternating current in the primary coil gives rise to an alternating e.m.f. in the secondary coil. [4] When a current flows through the primary coil, it produces a magnetic field. [B1] If the current is alternating (due to an alternating e.m.f. source), the magnetic flux linkage ΦP through the primary coil also alternates continuously. [B1] The iron core serves to strengthen the magnetic flux and ensure efficient linkage of the flux from the primary coil to the secondary coil. [B1] In the secondary coil, the alternating magnetic flux linkage induces an alternating e.m.f., in accordance with Faraday’s Law, which states that the induced e.m.f. is proportional to the rate of change of magnetic flux linkage. [B1] (b) In practice, the core of some transformer is made of laminated soft iron. (i) State two reasons why soft iron, which is easily magnetized and demagnetized, is used as the core. [2] Soft iron has high magnetic permeability, which greatly increases the magnetic flux density in the core. This amplifies the changes in magnetic flux linkage, resulting in a stronger induced e.m.f. in the secondary coil. Consequently, it enhances the efficiency of the transformer by improving the transfer of magnetic flux between the primary and secondary coils. [B1] Soft iron has low hysteresis loss because it can be easily magnetized and demagnetized by the alternating magnetic field caused by the current in the primary coil. This minimizes energy loss as heat in the core, which occurs when the energy required to magnetize the core is not entirely recovered during demagnetization. [B1] (ii) Explain how the lamination of the core reduces energy losses. [2] Alternating magnetic flux in the core induces eddy currents, which are circulating currents in the core material. These eddy currents cause resistive heating, leading to energy losses. [B1] Lamination of the core involves using thin sheets of iron separated by an insulating varnish, which significantly increases the electrical resistance of each lamina's surface. This confines the eddy currents to individual laminae, narrowing their possible paths. The increased resistance of these paths reduces the s ize of the eddy currents, thereby minimizing their heating effects and energy losses. [B1] [Refer to Lecture Notes Annex D on Page 18-19.]

DHS Physics H2 2025 Topic 18: Al