Method 1: 30x Method 2: 15

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Winfred Ryan
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1 The University of New South Wales School of Electrical Engineering and Telecommunications ELEC Electrical and Telecommunications Engineering Tutorial Solutions Q. In the figure below a voltage source and two resistors are connected in parallel. Calculate the current in each resistor. Be careful to take into account the reference directions of the resistor currents. Method : x x.8. 8 x.8. 8 Method :.. Q. In the figure below a current source and two resistors are connected in series. Calculate the voltage across each resistor. Be careful of the reference directions of the resistor voltages. 8 x 6 x

2 Q. What is the value of I T a. I b. I c. V d. R e. V 4 6V + I R R I + V = -4V - R R V - I 4 =.A - V4 + f. R 4 g. I T h. What is the power dissipated in each resistor? i. What is the power supplied to the circuit by the 6V voltage source? a. I = 6/ =. A b. I = 4/4 = A c. V = 6-4 = 8 V d. R = 8/ = 6 e. V 4 = 6 V f. R 4 = 6/. = 4 g. I T = = 4.7 A h. P = (.) x = W P = () x 4 = 6 W P = () x 6 = 4 W P 4 = (.) x 4 = 9 W i. P = 6 x 4.7 = 8 W (Note: P = P + P + P + P 4 ) Q4. In the figure below a current source and a voltage source are connected in parallel with a resistor. Calculate the current i in the resistor. i = / = A

3 Q. In the figure below a current source and a voltage source are connected in series with a resistor. What is the current i flowing in the circuit and the voltage v across the resistor A + - V i + v - i = A Q6. Calculate the voltage, V in each of the. following circuits. a) I = 6/ = A Current source Is = -xa = -6A Voltage V = xis = -8V. b) I Current controlled Voltage source V V 7 A ( ) Vs x 4x.4V ( ) S x( ) 4V Q7. Three trams draw current from an overhead wire and return it through the rails as illustrated in the figure below. The effective resistance per kilometre is shown for each part of the circuit. What is the power taken by each tram from the system?

4 V T = -(.+.) = 4.V V T = V T -(.+.) = 6.V V T = V T -(.4+.) =.V P T = V T x I T = 4. x = 7W P T = V T x I T = 6. x = 6W P T = V T x I T =. x = W Q8. An electric stove has a constant current of A entering the positive voltage terminal with a voltage of 4V. The stove operates for hours. a) Find the total charge that passes through the stove. Q I time ( 6 6) amp sec s 7C b) Find the power absorbed by the stove. Power = V x I = 4 x = 4W =.4kW c) If electric energy costs cents per kwh, determine the cost of operating the stove for hours. Energy used during hours = power x time =.4 x kwh = 4.8 kwh Cost of energy during hours = 4.8 x cents = 48 cents Q9. The current through and voltage across an element as shown, vary with time as shown in the graphs below. Sketch the power delivered to the element for <t<secs. What is the total energy delivered to the element between t= and t= s?.

5 P = v x i Energy over the time t = to secs is given by: We have to integrate but because the functions of voltage and current are not continuous we have to describe the voltage and current over three separate periods: t = to secs: v t i t t = to secs: v i t = to secs: v i ( t 7.)

6 Energy (4t t (4 ) v idt ) dt dt Ws (t) Method : Using area formula, (t) (i) dt ( t 7) dt t ( 7t) dt xx4 x xx ( t 7.) dt (9 ) (( 87. 7) ( 67. )) Q. Conservation of energy requires that the sum of the power absorbed by all of the elements in a circuit be zero. For the circuit shown below, all of the element voltages and currents are specified. Are these voltage and currents correct? Justify your answer. (Suggestion: Calculate the power absorbed by each element.) Power absorbed by Element = v x(i) Power absorbed by E = 4 x(-6) = -4W Power absorbed by E = x(+6) = +6W Power absorbed by E = -6 x(+) = -W Power absorbed by E4 = x(-4) = -8W Power absorbed by E = 4 x(-4) = -6W Total power absorbed = W so conservation of energy is confirmed. Q. A d.c. moving-coil meter is rated to carry.ma for full scale deflection. The coil has a resistance of. (a) What shunt resistance (in parallel across the meter) in the figure below is needed to convert this meter into an ammeter reading A full scale?

7 (b) Select the resistors R and R in the figure below to convert the meter into a voltmeter. The full scale readings are to be V and V. (a) The current which flows through the meter for full scale deflection is I m = ma. The remainder of the current I S flows through the shunt. So I S =.. =.999A As the meter and the shunt are in parallel, the potential difference across them is the same. Therefore: V m = V S Giving R C.I m = R S.I S or R S = R C.I m / I S So R S =(x.)/.999 =. (b) In order to convert the basic meter into a voltmeter with a full-scale reading of V, the current through the meter and attached series resistance must be ma. The total voltmeter resistance R v must therefore be Rv = /. =, As meter is then series resistance has to be R = 49,9 Similarly for V full scale Rv = /. =, So R =, R R =,

Series and Parallel Circuits

Series and Parallel Circuits Components in a circuit can be connected in series or parallel. A series arrangement of components is where they are inline with each other, i.e. connected end-to-end. A parallel