9 (a) The output of a power supply is represented by: V = 9.0 sin 20t where V is the potential difference in volts and t is the time in seconds. Determine, for the output of the supply:
- (a)(i)the root-mean-square (r.m.s.) voltage, V_r.m.s. V_r.m.s. = ................................................................ V [1][1]
- (a)(ii)the period T. T = ................................................................................ s [2][2]
- (b)The variations with time t of the output potential difference V from two different power supplies are shown in Fig. 9.1 and Fig. 9.2. [Figure 9.1] [Figure 9.2] The graphs are drawn to the same scale. State and explain whether the same power would be dissipated in a 1.0Ω resistor connected to each power supply. [1][1]
- (c)(i)The power supply in (a) is connected to a transformer. The input power to the transformer is 80 W. The secondary coil is connected to a resistor. The r.m.s. voltage across the resistor is 120V. The r.m.s. current in the secondary coil is 0.64A. Calculate the efficiency of the transformer. efficiency = ................................................................ [3][3]
- (c)(ii)State one reason why the transformer is not 100% efficient. [1][1]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.
- [1] 9.0/√2 = 6.4 V
- [1] ω = 20
- [1] T = 2π / ω
- [1] T = 0.31 s
- [1] the r.m.s. voltages are different, so no
- [1] P = V_r.m.s. X I_r.m.s.
- [1] = 120 × 0.64 = 76.8 W
- [1] efficiency = (76.8/80) × 100 = 0.96 or 96%
- [1] heat losses due to resistance of windings / coils / heat losses in magnetising and demagnetising core / hysteresis losses in core / heat losses due to eddy currents in (iron) core / loss of flux linkage
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