OpenStax College Physics for AP® Courses, Chapter 23, Problem 6 (Test Prep for AP® Courses)

This is College Physics Answers with Shaun Dychko. An ideal step-up transformer has perfect efficiency and that's what ideal means, it has a turn ratio of 1 to 30 so the number of turns in the primary divided by number of turns in the secondary is 1 over 30 in other words, it's supplied with an input power of 120 watts so we'll call that P P for primary so the input side of the transformer is the primary side—120 watts there— and the output voltage is 210 volts so voltage in the secondary, in other words, is 210 volts. The question is what is the output power and the input current? So the output power is going to be 120 watts just as the input power is because with perfect efficiency, we should expect there to be no power loss and there's no way this could be more because that would require energy to be spontaneously created somehow and conservation of energy is the principle that you can't have a greater rate of energy production out than you have the rate of energy input. Okay! So let's calculate the input current. So we know the power in the primary is the voltage in the primary multiplied by the current in the primary and we're going to solve for this I P but first we need to make substitutions for V P because we are not told what the voltage in the primary is but we're told the voltage in the secondary and the ratio of turns. So this voltage in the primary divided by voltage in the secondary equals the number of turns in the primary divided by number of turns in the secondary and we can solve for V P by multiplying both sides by V S. So V P then is V S times N P over N S and we make this substitution for V P here in our expression for power in the primary and then we'll solve this for I P. We'll multiply both sides by N S over N PV S so the current in the primary then is N S over N P times the power in the primary divided by voltage in the secondary. Now N S over N P is the reciprocal of what we have been given in the question so let's flip this fraction over and call it N P over N S and we can do that so long as we sort of undo the flipping by introducing this exponent negative 1 here. Okay! So this N P over N S is this number, which is 1 over 30 and we will raise that to the power of negative 1 so it's going to be 30 in other words and multiply that by power in the primary— 120 watts—divided by voltage in the secondary of 210 volts and this works out to 17 amps. And down here I give a sort of a long-winded answer for what is the power in the output side of the transformer; this answer here was given by the conservation of energy principle but down here, we can derive an expression to show that P S equals P P. So voltage in the secondary divided by voltage in the primary is N S over N P and we can solve for V S by multiplying both sides by V P so we have this expression that we're going to substitute here. So we are replacing these factors that have a subscript S with factors that have in terms of their corresponding values in the primary. So the voltage in the secondary is going to be expressed as the voltage in the primary multiplied by number of turns in the secondary divided by number of turns in the primary. And likewise for the secondary current, current in the secondary divided by current in the primary is number of turns in the primary divided by number of turns in the secondary and we can solve for I S by multiplying both sides by I P. So notice a pattern here, we have for voltage, it's S over P equals turns S over P so the S and the P are on the like side... on the top of both sides and on the bottom of both sides whereas for current, the S is on the top but for the number of turns, it's the primary that's on the top so they are opposites so the current in the primary is on the bottom on this side whereas on the other side—number of turns— you have the secondary on the bottom. Okay! So I S equals I P times N P over N S and when we make substitutions for each of these things here, we end up with the N P's canceling, the N S is canceling and we have V P times I P, which is powering the primary which is to say the power in the secondary is the same as the power in the primary and that is 120 watts.