The University of New South Wales
School of Electrical Engineering and Telecommunications
ELEC1111 Electrical and Telecommunications Engineering
Tutorial 2 Solutions
Q1. 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 1: 30 || 50 30x50 30 50 18.75 15 18.75 0.8 50 80x0.8 . 30 80x0.8 . Method 2: 15 30 . 15 50 .
Q2. 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 2
Q3. What is the value of a. I1 b. I2 c. V3 d. R3 e. V4 f. R4 g. IT
h. What is the power dissipated in each resistor?
i. What is the power supplied to the circuit by the 60V voltage source?
a. I1 = 60/300 = 0.2 A b. I2 = 42/14 = 3 A c. V3 = 60 - 42 = 18 V d. R3 = 18/3 = 6 e. V4 = 60 V f. R4 = 60/1.5 = 40 g. IT = 0.2 + 3 + 1.5 = 4.7 A h. P1 = (0.2)2 x 300 = 12 W P2 = (3)2 x 14 = 126 W P3 = (3)2 x 6 = 54 W P4 = (1.5)2 x 40 = 90 W i. P = 60 x 4.7 = 282 W (Note: P = P1 + P2 + P3 + P4)
Q4. In the figure below a current source and a voltage source are connected in parallel with a resistor. Calculate the current i1 in the resistor.
i1 = 5/5 = 1 A R1 R3 60V 14 + R2 R4 I4= 1.5A I1 IT I2 + V3 -+ V 2 = -42V --V 4 +
Q5. 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
v + -5 5V 2A i + i = 2 A
Q6. Calculate the voltage, V1 in each of the. following circuits.
a) I1 = 6/2 = 3A Current source Is = -2x3A = -6A Voltage V1 = 3xIs = -18V. b) V x x V V V x V A I s S 4 . 2 5 3 4 ) 2 3 ( 3 4 ) 2 ( 2 source Voltage controlled Current 2 ) 5 2 ( 2 7 1 1
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?
VT1 = 550 -250(0.02+0.01) = 542.5V VT2 = VT1 -200(0.02+0.01) = 536.5V VT3 = VT2 -100(0.04+0.02) = 530.5V PT1 = VT1 x IT1 = 542.5 x 50 = 27125W PT2 = VT2 x IT2 = 536.5 x 100 = 53650W PT3 = VT3 x IT3 = 530.5 x 100 = 53050W
Q8. An electric stove has a constant current of 10A entering the positive voltage terminal with a voltage of 240V. The stove operates for 2 hours.
a) Find the total charge that passes through the stove.
C s amp time I Q 10(26060) sec 72000
b) Find the power absorbed by the stove. Power = V x I = 240 x 10 = 2400W = 2.4kW
c) If electric energy costs 10 cents per kWh, determine the cost of operating the stove for 2 hours. Energy used during 2 hours = power x time = 2.4 x 2 kWh = 4.8 kWh
Cost of energy during 2 hours = 4.8 x 10 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 0<t<5secs. What is the total energy delivered to the element between t=0 and t=5 s?
P = v x i
Energy over the time t = 0 to 5 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 =0 to 1 secs: t i t v 3 15 t =1 to 3 secs: 3 10 i v t =3 to 5 secs: ) 5 . 7 2 3 ( 10 t i v
Ws t t t t dt t dt dt t dt t dt dt i t idt v Energy 105 5 . 157 5 . 187 60 15 )) 225 5 . 67 ( ) 375 5 . 187 (( ) 30 90 ( 15 ) 75 2 15 ( ) 30 ( ) 3 45 ( ) 75 15 ( 30 ) 45 ( ) 5 . 7 2 3 ( 10 3 10 ) 3 ( ) 15 ( 5 3 2 3 1 1 0 3 5 3 3 1 1 0 2 5 3 3 1 1 0 5 0
Method 2:Using area formula, 1
3x1x45 2x30 1
2x2x30
Q10. 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 E1 = 4 x(-6) = -24W Power absorbed by E2 = 10 x(+6) = +60W Power absorbed by E3 = -6 x(+2) = -12W Power absorbed by E4 = 2 x(-4) = -8W Power absorbed by E5 = 4 x(-4) = -16W
Total power absorbed = 0W so conservation of energy is confirmed.
Q11. A d.c. moving-coil meter is rated to carry 1.0mA for full scale deflection. The coil has a resistance of 100.
(a) What shunt resistance (in parallel across the meter) in the figure below is needed to convert this meter into an ammeter reading 2A full scale?
(b) Select the resistors R1 and R2 in the figure below to convert the meter into a voltmeter. The full scale readings are to be 50V and 100V.
(a) The current which flows through the meter for full scale deflection is Im = 1mA. The remainder of the current IS flows through the shunt.
So IS = 2.0 – 1.001 = 1.999A
As the meter and the shunt are in parallel, the potential difference across them is the same. Therefore: Vm = VS
Giving RC.Im = RS.IS or RS = RC.Im/ IS So RS =(100x0.001)/1.999 = 0.050025
(b) In order to convert the basic meter into a voltmeter with a full-scale reading of 50V, the current through the meter and attached series resistance must be 1mA. The total voltmeter resistance Rv must therefore be
Rv = 50/0.001 = 50,000
As meter is 100 then series resistance has to be R1 = 49,900 Similarly for 100V full scale Rv = 100/0.001 = 100,000 So R2 = 100,000 – 100 –R1
