ELEC166 Tutorial Week 3 Solutions

Transcription

1 ELEC166 Tutorial eek 3 Solutions Q1 n ideal voltmeter gives a reading of 9 when measuring between the terminals of a (real) battery n ideal ohmmeter gives a reading of 900Ω when measuring between the ends of a resistor If the resistor is connected across the battery, will a current of 10m flow through the resistor, and, if not, why not? : The current will be slightly lower than 10m This is due to the total resistance in the circuit being larger than 900Ω when the internal resistance of the battery is connected in series with the 900Ω resistor (Note that the 9 measured originally at the battery terminals was when there was no current flowing Ohm s Law tells us that there was no voltage across the battery s internal resistance, so the 9 is an accurate measure of the only voltage source in the circuit that is, the Thévenin equivalent voltage) Q hen a 10 bulb is connected in parallel with a 0 bulb, the 0 bulb is brighter hich bulb is brighter if they are connected in series? Explain : Note that in all our calculations with light bulbs, we assume that the resistance of bulbs is the same whether they are lit or not; this is not correct, but simplifies the calculations hen the bulbs are connected in parallel, the same voltage is across each (this is the normal way that they would be connected) First find the relative resistances of the two bulbs; we don t need to know the voltage across them Using P =, we have that R0 = for the 0 bulb, and R 0 0 = for the 10 bulb Hence R 10 =R 0 10 There are two ways you might reach the final conclusion: (1) (Simplest method) Since they are connected in series, the same current flows through each bulb Power is related to current and resistance by P=I R, so for the same current, the power is proportional to resistance Since the resistance of the 10 bulb is higher, it dissipates more power (and therefore, presumably, is brighter) () lternatively, consider the two bulbs in series forming a potential divider If the total voltage across the two bulbs is S, then the voltage across the 10 bulb 0 R0 is 10 = S = S = S, and the power dissipated is R R R R P = = = = R 0 R0 R0 Similarly, the voltage across the 0 bulb turns out to be 0333S, and the power dissipated is P = = = Thus the 0 bulb dissipates only R R R 0 0 ELEC166 Tutorial eek 3 Solutions Page 1

2 half as much power as the 10 bulb; the 10 bulb is brighter Q3 Given a 40 battery and four resistors of values 1Ω, Ω, 4Ω and 5Ω, design a voltage divider circuit to produce an output voltage of 4 (you may not need to use all the resistors) : e want an output which is one-tenth of the input voltage Notice that 145=10, so by creating a voltage divider out of the 1Ω, 4Ω and 5Ω resistors the 1 voltage across the 1Ω resistor will be 40 = Q4 Find the Thévenin equivalent of the following circuit, as seen between terminals and (HINT: Determine the open-circuit voltage and the shortcircuit current) 10 3Ω 5 5Ω Output : dding up the voltage sources around the circuit, we see that the open-circuit voltage OC is 105=15 The short-circuit current I SC will thus be 15/(35)=1875 Following the procedure in the lecture notes, we find that the Thévenin voltage is 15 and the Thévenin resistance (= OC /I SC ) is 8Ω (lternatively, note that you can find the Thévenin resistance by simply looking back into the terminals and replacing all the voltage sources with short circuits) Q5 Find the Thévenin equivalent of the following circuit, as seen between terminals and 5 1Ω 5 5Ω Output 10Ω : This is the same as the previous question, except we now have two resistors in parallel which we must reduce to their equivalent resistance before going through the same steps as before The Thévenin resistance is 433Ω and the Thévenin voltage is 10 ELEC166 Tutorial eek 3 Solutions Page

3 Q6 For the following circuit, calculate (i) the voltage between terminals and C, and (ii) the voltage between terminals and hat do you notice about these two voltages? 15 1kΩ 3kΩ kω 5kΩ C : This circuit forms a two-resistor potential divider, with each resistor being a parallel combination of two others The upper combination is (1kΩ kω) = 0667kΩ, while the lower combination is (3kΩ 5kΩ) = 1875kΩ 1875kΩ Thus the voltage between and C is C = 15 = kΩ 0667kΩ The voltage between and is calculated in the same way, except that we now want the voltage across the upper resistor pair 0667kΩ This is = 15 = kΩ 0667kΩ The two voltages must add up to 15 Q7 For the following circuit, algebraically calculate its Thévenin equivalent circuit as seen between terminals and (do this in the standard way by calculating the open-circuit voltage, short-circuit current and so on) The equivalent resistance turns out to be a simple combination of and R what is it? s R out : Since this is a potential divider, the open-circuit voltage is given by R OC = S This is the Thévenin equivalent voltage R If and are shorted together, all the circuit current will flow through the short circuit, and none through R, and since the total resistance in the circuit is just, the short-circuit curent is given by I S SC = The Thévenin equivalent resistance is given by the ratio OC R R = S = This is an ISC R S R alternative form of the formula for the equivalent resistance of two resistors in parallel So the Thévenin equivalent resistance of a two-resistor potential divider is equal to the parallel combination of the two resistors ELEC166 Tutorial eek 3 Solutions Page 3

4 Q8 Find the Thévenin equivalent of the following circuit, as seen between terminals and s=1 1kΩ 15kΩ 0kΩ : First let s calculate the open-circuit voltage OC If we ignore the 15kΩ resistor for the moment, we see that the two left-hand resistors form a potential divider, 0kΩ so that the voltage across the 0kΩ is 0 k Ω = 1 = 7 5 Since 0kΩ 1kΩ the terminals and are open-circuited, no current flows through the 15kΩ, so there is zero voltage across it Hence the voltage between and is also 75 This is OC, and hence the Thévenin equivalent voltage Now for the short-circuit current I SC This will be equal to the current flowing in the 15kΩ when and are connected together, as shown in the circuit below 1kΩ s=1 0kΩ 15kΩ Isc In this case, the total resistance in the circuit will be 1kΩ(15kΩ 0kΩ)=8571kΩ Hence the total current in the circuit will be 1/8571kΩ=1400m However, this is not the current we want, since it is shared between the 15kΩ and 0kΩ, but we can find I SC by calculating the voltage across the 15kΩ Notice that the second circuit forms a potential divider The voltage across the 15kΩ (and the 0kΩ) is thus given by (0kΩ 15kΩ) 8571kΩ 15 k Ω = 1 = 1 = 5 00 Thus the current 1kΩ (0kΩ 15kΩ) 0571kΩ through the 15kΩ is just 5/15kΩ=0333m, which is I SC Now the Thévenin equivalent resistance is given by the ratio OC 75 = = 5kΩ I 0333m SC ELEC166 Tutorial eek 3 Solutions Page 4

5 Q9 Consider the circuit below y initially converting blocks and into their Thévenin equivalent circuits, calculate the current flowing in the 10kΩ resistor 5 kω 10kΩ 6kΩ 10 3kΩ 4kΩ : locks and are simply potential dividers e can use the result from a previous question to quickly calculate the Thévenin equivalent circuits of these blocks (recall that the value of the equivalent series resistor is equal to the parallel combination of the two original resistors in the divider) So the overall circuit can be reduced to the following: 1kΩ 10kΩ 4kΩ 3 4 ll the components except the 10kΩ resistor can then be combined into another Thévenin equivalent circuit (see another earlier question), so that the final circuit can be redrawn like this (notice also that the two voltage sources oppose each other, so that the resulting voltage is 4 3=1): 36kΩ 1 10kΩ It is then a relatively simple matter to see that the current in the 10kΩ is equal to 1/136kΩ = 735µ Q10 Instead of its Thévenin equivalent, a linear circuit can also be replaced by its Norton equivalent circuit, which consists of a current source in parallel with a resistor, as shown below For a given circuit, if the Thévenin voltage source and series resistor have values th and Rth respectively, calculate the values of the corresponding Norton equivalent current source IN and parallel resistor RN in terms of th and Rth (Hint: s for the Thévenin case, consider the open-circuit voltage and short-circuit current; they should be the same for both equivalent circuits) ELEC166 Tutorial eek 3 Solutions Page 5

6 Rth th IN RN Thévenin equivalent circuit Norton equivalent circuit : For the Thévenin equivalent circuit, the open-circuit voltage and shortcircuit current are equal to th and th/rth respectively, while for the Norton equivalent circuit they are equal to INRN and IN respectively (why?) Equating these gives IN = th/rth and RN = Rth ELEC166 Tutorial eek 3 Solutions Page 6