Name: ……………………………….… CG : …………… Date: __ Jul/Aug 09
Tutorial 10 : Current of Electricity (Solutions)
1.It has been argued that the expression “electric current flow” is redundant. Why? Recall the definition of electric current and suggest the more correct phrasing? See Hewiit. Pg 534.
Electric current is already defined as the rate of flow of charges. So to say “current flows” is redundant It is like the commonly used phrase “This is the reason why I didn’t come for class.” The ‘why’ quoted here is redundant.
The correct phrasing should be “charges flow”. Alternatively, you can say the ‘current in the wire’. But according to Hewitt, it is conceptually simpler to say “current flows” and forgot about the technicality of correct grammar. So we will still use the expression.
2.A power station has the following warning sign. What causes electric shock in the human body – current or voltage? In light of your answer, is there a need to change the sign?
The sign is correct because “High Voltage” is the cause or the reason for current flow. It is potentially lethal if you come into contact with it as it may cause huge current to pass through you. And it is the current that delivers the damage. The voltage required varies a lot because most of it is normally used up as the current passes through your skin. If you are cut, soaked in salt water, or have metal piercing your skin, it takes a lot less voltage to drive the current through the rest of our insides.
So the shock is due to the current passing through you, but voltage is required so that a current can flow in the first place. Putting a sign “High Current” would be inaccurate because current does not flow unless the circuit is complete.
3.Light bulbs A and B are identical in every way except that B’s filament is thicker than A’s.
(i)If screwed into a 110 V socket, which will glow more brightly?
(ii)If in series to the supply?
Brightness of the light bulb depends on the power that is dissipated from it.
(i)When connected to a 110 V socket on its own, the two light bulbs can be seen as connected in parallel with a power source of 110 V. Hence the voltage across the two light bulb is the same. According to, power loss in the load is inversely proportional to its resistance. As the filament wire in B is thicker (i.e. larger cross-sectional area), its resistance is smaller. Consequently, power dissipation in the bulb would be larger. Thus, the bulb Bglows more brightly.
Misconceptions for parallel circuit:
- when we use the formula, P = I2R, assuming the current is the same for both scenario.However, It isV that is same for the two cases and not current. Hence, Pis not directly proportional to R.
- the larger the resistance, the greater the power dissipated. However, if the resistance is increased to a very large value, the current would be reduced to almost zero, bringing the power down to almost zero as well.
(ii)When two light bulbs A and B are connected in series and attached to the 110 V supply, both bulbs share the same current since there is only one path for the current to flow through. Accordingly to P = I2 R, theR for bulb A is greater than that for bulb B, power transfers for bulb A is greater than that for bulb B. Thus, the bulb Aglows more brightly.
This is a prelude to Q4b as the concept is the same.
4(a)What is the resistance and current through a light bulb labeled 220 V, 100 W when it is used according to this rating?
(b)The light bulb is connected to the supply by the connecting wires. As they are in series, they share the same current. Explain why the light bulb heats up and glows while the connecting wires do not.
(a)(i)The label of 220V, 100 W indicate the power given by the light bulb is 100 W when it has a voltage of 220 V.
V = 220 V, P = 100 W
R = 480
(ii)P = IV
100 = I (220)
I = 0.45 A
(b) The concept is the same as the follow-up question to Q3.
When connected in series, the current through the wire and light bulb is the same. Since P = I2R, P is directly proportional to R. The filament has a much finer wire so its resistance is higher - almost all the resistance in the circuit is there so almost all the voltage is across the filament. The resistance of the bulb is 480 compared to the resistance of the connecting wire which is just a few ohms. The temperature of the bulb rises to the point of making light. The other wire just gets a little bit warmer.
Trivial note:
Many fires in old shop-houses (eg in Chinatown) occur because high consumption appliances (eg air-cond) are connected to the old electrical wiring which is not designed to handle large current.
5 Four small conductors, on the edge of an insulating disc of radius r are each given a charge of Q. The frequency of rotation of the disc is f.
What is the equivalent electric current at the edge of the disc? [N98/P1/14]
A B C D
Ans: A
Current is defined as rate of flow of charge,
i.e.
For a rotation, a point on the edge will experience a total of 4Q within a period of T.
where
6.The electron beam current in a cathode-ray oscilloscope is 50A. The time-base causes the beam to sweep horizontally across the screen of 1.0104 cm s-1. What is the number of electrons arriving at the screen in one centimeter length of the horizontal trace?
[3.11010]
Using
All the variables are given except for N (the answer) and t.
The time taken is determined from the context of the distance where the number of electrons is referring to those that cover 1.0 cm of the horizontal trace.
Note: the electron beam is moving across the C.R.O. at a uniform velocity.
Hence Time taken to sweep 1.0 cm of horizontal trace
= (where constant Velocity = Distance / time)
= 104 s
Therefore
N = 3.11010
7.In the diagram of an oscilloscope below, the shaded area represents a section through the electron beam which is generated near P, deflected and accelerated at Q, and focused at R. The tube contains no gas.
How does the electric current of the beam vary along PQR? [87/P1/6]
AIt has the same value at P, Q and R
BIt decreases from P to Q, then increases to reach the same value at R as at P.
CIt increases from P to Q, then decreases to reach the same value at R as at P.
Ans : A
The current is constant in a complete circuit. The electron beam might diverge (from P to Q) and converge (from Q to P) but the rate of flow of the electrons is the same.
8. A sheet of copper of resistivity ρ = 1.70 10-8 Ω m is 2.00 mm thick. It is used to form a circular tube of height 24.0 cm and external radius 1.27 cm, what is the resistance between the ends?
[2.5610-5]
= 2.77 × 10-5 Ω
(The teacher’s solution is based on uniform sheet rolled into a loop where the cross-sectional area is taken to be a rectangle)
Note: The length is taken to be from the high potential to the low potential (i.e. where the two terminal where the component is connected to)
9. A conducting liquid fills a cylindrical metal case to a depth x as shown in the figure.
The resistance bet-ween the case and the metal rod is
A proportional to x2
B proportional to x
C inversely proportional to x
D inversely proportional to x2. [N88/P1/3]
Ans: C
The direction of current is from the metal rod at the centre to the casing (or vice-versa).
The connection for the terminal to this component must be across conducting liquid which has resistance. It cannot be connected across the rod or the metal case as the resistance of these component are low, thus the current will not pass through the conducting liquid.
Let r = radius of the cylindrical metal casing.
The current flows along the radius r (or l in the formula).
The cross-sectional area through which current flows is not constant; it increases with the distance from the centre. So to simplify matter, we take the average cross-sectional area of the curved surface area,
Thus,
10.Sketch the current-voltage (I-V) characteristics of
(i)a metallic ohmic conductor,
(ii)a filament lamp,
(iii)a thermistor and
(iv)a diode.
Explain for each component, how the resistance changes with increasing V.
Refer lecture notes on I-V characteristics.
Not necessary to explain from microscopic point of view.
11(a)A battery of e.m.f. 6.0 V and internal resistance 0.40 is connected to an external resistor of resistance of 2.9 .
What is the power supplied to the external resistance? [N2008/P1/27]
A1.3 WB5.3 WC9.6 WD12.4 W
(b)Explain what is meant by an e.m.f. of 6.0 V.
(a)The internal resistance can be seen as connecting in series with the resistor of 2.9 Ω. Thus the current flowing through both are the same.
E = I (R + r)
6.0 = I (2.9 + 0.40)
I = 1.818
= 9.59 W
(b)An e.m.f of 6.0 V means that 6.0 J is transferred to electrical energy from other forms per unit charge in the supply.
12(a)An electrical component C has an I-V characteristic as shown below.
(i)Calculate the resistance of component C when the p.d. of 6.0 V is applied across it.
(ii)Deduce the minimum value of the resistance of component C over the range 0-10V.
[4.2 , 3.1 ]
(i)From the graph,
when V = 6.0 V, I =1.43 A
R = V/ I = 6.0 / 1.43 = 4.20
(ii)The minimum value for the resistance of C occurs for the maximum value for the ratio I/V.
To find the min. R, we draw a straight line joining the origin and the ‘knee’ of the graph. The point of the knee is
when V=3.4 V, I = 1.1 A
Minimum resistance = 3.4 / 1.1 = 3.1
(b)Component C is then connected into a circuit with a supply of internal resistance 0.80 and a resistor of constant resistance 5.0 as shown below.
The current through the 5.0 resistor is found to be 0.85 A. Calculate
(i)the p.d. across component C
(ii)the total current from the supply
(iii)the e.m.f. of the supply
(iv)the energy supplied to component C in 20 minutes.
[4.3 V, 2.2 A, 6.0 V, 6600 J ]
[N2007/P2/3]
Note: The component C and the 5.0 resistor are connected in parallel which in turn is connected in series to the internal resistance. (They are not all connected in parallel)
(i)The p.d. across 5.0 = I5.0R5.0 = (0.85)(5.0) = 4.25 V
Since component C is connected parallel to the 5.0 resistor, the p.d. across component C and 5.0 Ω resistor is the same. Thus, the p.d. across component C is 4.25 V.
(Examiners’ report mentioned that many candidates left out the latter point,; pls emphasize this point)
(ii)Total current from supply = current flowing in 5.0 Ω resistor + current flowing in component C
(or E = I Rtotal. However e.m.f. of the supply is not given)
From the graph, at V=4.25V, current through component C is 1.3 A.
Total current from supply
= current flowing in 5.0 Ω resistor + current flowing in component C
= 0.85 + 1.3
= 2.15A
(iii) E = terminal p.d. + p.d. across the internal resistance
= V + Ir
= 4.25 + (2.15)(0.80)
= 5.97 V
(iv)E = VIt = (4.25)(1.3)(2060)
= 6630 J
13(a)Use energy considerations to distinguish between electromotive force (e.m.f.) and potential difference (p.d.)
E.m.f. is the energy converted from other forms to electrical forms per unit charge in the voltage source.
P.d. is the energy converted from electrical to other forms per unit charge across two points in a circuit.
(b) A cell of e.m.f. E and internal resistance r is connected to a resistor of resistance R as shown in the figure below.
A voltmeter of infinite resistance is connected in parallel with the resistor.
(i)Copy the figure above and indicate in the circuit a switch so that the voltmeter may be used to measure either the emf E of the cell or the terminal potential difference V.
The switch should be placed such that it can cut the current flowing through the external resistor R.
(ii)State whether the switch should be open or closed when measuring
1. e.m.f.
2. the terminal p.d.
When S is open, the voltmeter measures the e.m.f.
When S is closed, the voltmeter measures the terminal p.d.
(iii) Derive a relation between E, V, r and the current I in the circuit.
[J96/P3/4part]
Derivation from the first principle ie from the Principle of conservation of energy,
Total work done
= Work through internal resistance + work done through external resistance.
14.A variable resistor R is connected between the terminals of a battery of e.m.f. E and internal resistance r as shown.
The resistance of resistor R is varied. The potential difference across R is V and the power dissipated in R is P. The variation with V of P is as shown below.
(a) For the maximum value of P use above figure to
(i)calculate the current in the circuit.
(ii)show that the resistance of R is 3.6
[1.25 A]
The variable resistor and the internal resistance are connected in series, i.e. both internal resistance and variable resistor has the same current.
(i)From the graph, maximum P (5.62 W) occurs when V=4.50V
I = 1.25 A
(ii)
R = 3.6
(b) When R has resistance 2.03 , the current in the circuit is 1.60 A. Use these data and your answers to (a) to determine the internal resistance r of the battery.
[3.58 ]
[N2008/P2/4]
Based on (a), we have:
E = V + Ir = IR + Ir
= (1.25)(3.6) + (1.25)r
E = 4.50 + 1.25 r …..(1)
From data given in (b),
E = (1.60)(2.03) + 1.60 r
E = 3.25 + 1.60 r …. (2)
(1) = (2),
r = 3.58