Download Tutorial 11

INNOVA JUNIOR COLLEGE
JC 1 Physics (2009/2010)
Tutorial 11: D.C. circuits
H2
Equivalent resistance of multiple resistors
1. N95/I/15; J2000/I/15
Four resistors are connected as shown.
1
P
Q
4
2
S
R
3
Between which 2 points is the resistance of the combination a maximum?
A P and Q
B Q and S
C R and S
D S and P
Suggested solution
By connecting across any two terminals (P, Q, R or S), the four resistors are effectively
connected in parallel (with some connected in series first). (Refer to the circuit diagram
below)
P and Q:
1
P
Q
4
3
1
Q and S:
2
R
S
P
4
Q
S
2
3
R
4
R and S:
1
S
2
Q
P
R
3
4
S and P:
S
P
3
2
R
1
Q
The effective resistance of resistors in parallel is less than the resistance in any of the
‘branches’. Thus for maximum effective resistance, the choice of the terminal should give
resistance of each branch to be approximately similar, i.e. choice B where each branch has
a subtotal of 5 Ω. And
Resistance between Q and S:
1
1
1


RQS
4 1 3  2
RQS = 2.5 Ω
(If not, you can work out all the configurations)
Ans: B
2. Five resistors of resistance 2.0  are connected as shown in the diagram below.
2.0 
2.0 
2.0 
2.0 
2.0 
P
Q
What is the equivalent resistance between points P and Q?
Suggested solution
2.0  X
2.0 
2.0 
2.0 
P
2.0 
Q
Y
Assumption: no current flows through middle resistor.
Thus the potential at X can be assumed to be only affected by the two resistors at the top
while the potential at Y can be assumed to be only affected by the two resistors at the
bottom. Hence VX and VY are the same value (using potential divider rule on each branch).
Therefore, the p.d. across the middle resistor is zero and the assumption of no current
flowing through the middle resistor is proven. Hence the circuit diagram can be considered
without the middle resistor.
2.0 
2.0  X
Q
P
2.0 
2.0 
1
1
1


R 22 22
R = 2.0 Ω
The equivalent resistance is 2.0 Ω.
Y
3. Given that each resistor has a resistance of R, determine the total resistance between the
points PQ in the circuits shown in Fig. 3(a), 3(b), and 3(c).
X
P
1
2
Y
P
1
2
3
Q
1
X
Q
Y
2
X
3
4
3
Fig. 3(b)
Fig. 3(a)
5
P
Fig. 3(c)
Suggested solution
Take note of how each resistor are being connected to another, i.e. terminals of each
resistor.
(a)
2
X
P
3
Q
1
R PQ
 1 1
  
R R
1
R
3
R
2
(b) Note: Point X has the same potential as Point Q while Point Y has the same potential as
Point P.
P
Y/P
P
X/Q
1
X/Q
2
Y/P
Q
Q
3
1
1 1 1
  
R PQ R R R
RPQ =
1
R
3
(c)
X
2
1
P
Y
3
4
Q
5
R1 and R2 are connected in series in which they are connected in parallel to R3. These
resistors R1, R2 and R3 are then connected in series to R4. And the whole R1, R2, R3 and
R4 are connected in parallel to R5.
Q
Effective Resistance
1
1 1
1 1
RPQ = {[(

)  R 4 ] 1 
}
R1  R 2 R 3
R5
1
1
1
= {[(
 ) 1  R ] 1  } 1
RR R
R
5
= R
8
Electric circuits with one emf
4. (J2006/II/2)
The figure below shows an electrical circuit in which the internal resistance of the battery is
negligible.
5.0 A
X
B
12 V
5.0 
4.0 
A
C
Y
Complete the table below by giving the electrical quantities for each of the components in
the above circuit. You are advised to start by completing the column for component A.
Suggested solution
circuit component
A
B
C
whole circuit
potential difference / V
12
10
2
12
current / A
3.0
2.0
2.0
5.0
power / W
36
20
4
60
resistance / 
4.0
5.0
1.0
2.4
B and C are connected in series
 IB = IC
VXY = VB + VC
(B+C) is in turn connected in parallel to A
 VA = VXy = VB + VC
IT = IA + IB
or IT = IA + IC
A
B
C
whole circuit
potential difference / V
(1) VA = Vxy = E
(4) VB = IB × RB
(5) VA = VB + VC
12
current / A
(2) IA = VA / RA
(3) IT = IA + IB
(5) IB = IC
5.0
circuit component
P = V2 / R = I2 R = IV
power / W
resistance / 
4.0
5.0
(6) RC = VC / IC
(7) RT = E / IT
5. N03/II/2
The variation with current of the potential difference (p.d.) across a component X is shown in
Fig. 2.1.
(a) (i) State how the resistance of component X varies, if at all, with increase of current.
(ii) On Fig. 2.1, draw a line to show the variation with current of the pd across a resistor R of
constant resistance 3.0 .
(b) The component X and the resistor R of resistance 3.0  are connected in series with a
battery of negligible internal resistance, as shown in Fig. 2.2.
The current in the circuit is found to be 2.0 A.
(i) Use Fig. 2.1 to determine the p.d. across component X.
(ii) Determine
1. the p.d. across R,
2. the emf of the battery.
(c) The resistor R and the component X are now connected in parallel with the battery, as
shown in Fig. 2.3.
Use your answer to (b)(ii) and the graph of Fig. 2.1, determine the current from the battery.
Suggested solution
(a) (i) Resistance of X increases with an increase of current. (The best way is to physically
calculate the resistance of two different scenario with difference current)
(ii) For a resistance with 3.0 Ω, when the current is 2.0 A, the p.d. across the resistor of 6.0
V is required. Thus to show an ohmic resistor, draw a line passing through origin and
coordinates (2,6) on the graph.
a (ii)
b (ii) 1.
b (i)
c
(b) (i) When the current in X is 2.0 V, p.d. across X is 5.0 V [Read off from Fig 2.1(green
line)]
(ii) 1. Since X and resistor R is connected in series, the current flowing through both
component is the same. Thus, with a current of 2.0 A, a p.d. of 6.0 V is required.
[Read off from Fig 2.1 (blue line)]
2. Since X and resistor R is connected in series, E = VX + VR = 5.0 + 6.0 =11 V
(c) However when the X and resistor R is connected in parallel,
 E = VX = VR
 IT = IX + IR
Since the same battery is used, p.d. across X and R is 11 V (e.m.f. of battery).
Current through X, IX = 2.9 A (from Fig 2.1. (orange line)
V
11
Current through R, I R  R 
= 3.7 A
R
3.0
Total current drawn from battery, IT = 3.7 + 2.9 = 6.6 A
Potential divider circuits
6. In the figure below, the two lamps L1 and L2 labelled 3.0 V, 0.25 A and 6.0 V, 0.05 A,
respectively, light up with normal brightness.
R1
X
3 V, 0.25 A
+
L1
L2
12 V
_
6 V, 0.05 A
R2
Y
(a) Calculate the values of the resistance R1 and R2.
(b) What happens to L1 if L2 burns out first?
(c) What happens to L2 if L1 burns out first?
Suggested solution
L1 and R2 are connected in series. Thus the current through L1 and R2 is the same while the
sum of voltage across L1 and R2 is the voltage across XY.
 IL1 = IR1
 VXY = VL1 + VR2
IL1 = IR1
IR1 = 0.25 A
Since L2 lights up normally, the p.d. across it is 6.0 V which is equivalent to VXY.
VXY = VL1 + VR2
6.0 = 3.0 + VR2
VR2 = 3.0 V
R1 
VR1
3.0

 12 
I R1 0.25
R1 is connected in series with (L2 and L1 + R2 which are connected in parallel).
 E = VR1 + VL2
or
E = VR1 + VL1 + VR2
 IT = IR1 = IL1 + IL2 or
IT = IR1 = IR2 + IL2
E = VR1 + VL2
12 = VR1 + 6.0
VR1 = 6.0 V
IT = IR1 = IL1 + IL2
IR1 = 0.25 + 0.05
= 0.30 A
V
6.0
R2  R2 
 20 
I R 2 0.30
(b) If L2 burns out first, no current flows through L2, Thus only R1, L1 and R2 are connected in
series.
Total resistance in circuit, RT
= R1 + R2 + Resistance of L1
= 20 + 12 + 3.0
0.25
= 44 
E = ITRT
12 = IT × 44
IT = 0.27 A (> 0.25 A)
Hence, the lamp L1 shines more brightly
(c) If L1 burns out first, no current can flow through L1 thus no current will flow through R2 as
well. Thus the only R1 and L2 are connected in series.
Total resistance in circuit
= R1 + Resistance of L2
= 20 + 6.0
0.05
= 140 
E = ITRT
12 = IT × 140
IT = 0.086 A (> 0.05 A)
Hence, the lamp L2 shines more brightly.
7. A thermistor has a resistance 3900  at 0 °C and resistance 1250  at 30 °C. The
thermistor is connected into the circuit below in order to monitor temperature changes.
thermistor
1.50 V
R
V
The battery of e.m.f 1.50 V has negligible internal resistance and the voltmeter has infinite
resistance.
(a) The voltmeter is to read 1.00 V at 0 °C. Show that the resistance of resistor R is 7800 .
(b) The temperature of the thermistor is increased to 30 °C . Determine the reading on the
voltmeter.
(c) The voltmeter in the figure is replaced with one having a resistance of 7800 . Calculate
the reading on this voltmeter for the thermistor at a temperature of 0 °C.
Suggested solution
(a) Since the resistance of voltmeter is infinite, there is negligible current through the
voltmeter. Thus, the thermistor and resistor are connected in series.
Thus, using the potential divider principle, having the same current
VR
E

R R  R thermistor
1.00
1.50
At 0 °C:

R
R  3900
R = 7800 
(b) When the temperature is 30 °C , the resistance of the thermistor is 1250 .
Using the potential divider principle, still having the same current
VR
E

R
R  R thermistor
VR
1.50

7800 7800  1250
VR = 1.29 V
(c) Now, the voltmeter has a resistance of 7800 Ω.
Effective resistance of voltmeter and R
'1
1
1


R ' 7800 7800
R ‘ = 3900 Ω
This effective resistance R ‘ is still connected in series to the thermistor. Thus potential
divider rule is still applicable.
VR '
E

R ' R 'R thermistor
VR '
1.50

3900 3900  3900
VR ’ = 0.75 V
8. The galvanometer reading is zero in the circuit.
75 
galvanometer
X
75 
8.0 V
75 
What is the value of resistor X?
2.0 V
Suggested solution
75 
A
galvanometer
C
75 
X
8.0 V
75 
2.0 V
B
D
When there is no current flowing through the galvanometer, VA = VC (i.e. VAC = 0)
And VB = VD
Therefore VAB = VCD
The circuit can thus seen as the circuit diagram below:
75 
A
C
75 
X
8.0 V
75 
2.0 V
B
D
Looking at the right circuit, it is an open circuit. Thus VCD = E = 2.0 V
With
VAB = VCD
VX = 2.0 V
(Since VAB = VX)
Looking at the left circuit, the two resistors of 75 Ω and X are connected in series. Thus
using potential divider
VX
E

R X RT
2.0
8.0

R X R x  75  75
RX = 50 Ω
9. Fig. 9 shows a 120 V supply that is connected to two loads A and B through distribution
wires of resistance per unit length 1.00 x 10-4 Ω m-1. Load A is 200 m away from the supply,
while load B is a further 100 m away from load A.
+
I
30 A
50 A
120 V
_
A
B
I
100 m
200 m
Fig. 9
(a) Calculate current I.
(b) Determine the potential difference across load A and load B.
Suggested solution
+
I
0.02 Ω
0.01 Ω
30 A
50 A
120 V
_
A
I
0.02 Ω
200 m
B
0.01 Ω
100 m
The distribution wire can be considered as resistor that is connected along the path where
the current flow in the circuit.
(Look at the circuit diagram and try to imagine how each resistor is connected to each other)
(a) I = 50 + 30 = 80 A.
(b) For load A:
E = I(0.020) + IA RA + I(0.020)
120 = (80 × 0.020) + VA + (80 × 0.020)
VA = 117 V
For load B:
E = I(0.020) + IB(0.01) + IB RB + IB(0.01) + I(0.020)
120 = (80 × 0.020) + (30 × 0.010) + VB + (30 × 0.010) + (80 × 0.020)
VB = 116 V
Alternative: VA = IB(0.01) + IB RB + IB(0.01)
Potentiometer
10 (a) A uniform wire AB of resistance 1.5  is connected in series with a resistor of 5.0 , a
cell of e.m.f. of 9.0 V and internal resistance 1.0 , as shown in Fig. 2.1. Calculate the
potential difference (p.d.) across the wire AB.
9.0 V
1.0 
5.0 
A
B
80.0 cm
Fig. 2.1
Suggested solution
(a) There are a total of 3 resistors (i.e. 1.0 , 5.0  and 1.5 ) connected in series.
By using potential divider rule:
V AB
E

R AB RT
V AB
9.0

1.5 1.5  5.0  1.0
VAB = 1.8 V
(b) A cell C of e.m.f. 1.5 V and internal resistance 0.80  is connected to the circuit of
Fig. 2.1, as shown in Fig. 2.2.
The moveable contact can be connected to any point along the wire AB. At point D,
there is zero current in the galvanometer.
(i) Calculate the length of AD.
(ii) State and explain one advantage of using Fig. 2.2 to measure the e.m.f. of C
compared to using a laboratory voltmeter.
Suggested solution
(b) At point D, there is zero current in the galvanometer, there is no p.d. between D and F,
i.e. VD = VF.
and since point A and point E are essentially the same ‘point’ in the wire, VA = VE
Thus VAD = VEF.
The circuit can thus seen as the circuit diagram below:
Looking at the lower circuit, it is an open circuit.
Thus the p.d. of EF is the same as that of the e.m.f. of cell C, i.e. VEF = EC = 1.5 V
Thus with VAD = VEF
VAD = 1.5 V
V = IR and R 
l
A
I l
Therefore V 
A
Vαl
V AD l AD

V AB l AB
l
1.5
 1
1.8 0.80
lAD = 0. 667 m
(ii) Using Fig. 2.2 to measure e.m.f. of C is more accurate as there is no current flowing in
the lower circuitry, thus measuring the e.m.f. of cell C instead of measuring the terminal
potential difference due to the presence of internal resistance) if a laboratory voltmeter is
used.
11. A circuit is set up as shown below. When R1 = 10 and R2 = 90 there is no current through
the galvanometer G. What is the resistance X? The internal resistances of the cells are
negligible.
4.0 V
R1
A
10
C
R2
B
G
D
X
1.2 V
Suggested solution
When there is no current flowing through the galvanometer, VA = VC (i.e. VAC = 0)
And VB = VD
Therefore VAB = VCD
i.e. V1 = V10Ω
The circuit can thus seen as the circuit diagram below:
4.0 V
R1
A
10
C
R2
B
G
D
X
1.2 V
Looking at the top circuit, it is a closed circuit with the resistors R1 and R2 connected in
series.
V1 E

R1 RT
V1
4.0

10 10  90
V1 = 0.40 V
i.e. V10Ω =V1 = 0.40 V
Looking at the below circuit, it is an closed circuit with the resistors of X and 10 Ω connected
in series.
V10Ω
E

R10Ω RT
0.40
1.2

10
Rx  10
RX = 20 Ω
12. In the potentiometer shown in Fig. 12, battery A has an e.m.f. of 5.0 V and internal
resistance 2.0 Ω while battery B has unknown e.m.f. E and internal resistance r. The
resistance wire PQ is 1.00 m long with an overall resistance of 8.0 Ω.
A
5.0 V , 2.0 Ω
P
Q
B
E,
r
Fig. 12
The balance length is found to be 0.800 m for the set-up as shown in Fig. 12. When a 3.0 Ω
resistor R is added in parallel to battery B, the balance length becomes 0.600 m long.
Calculate the e.m.f. E and the internal resistance r of battery B.
Suggested solution
Scenario 1: without the resistor R where S is the balance point where no current flow in
galvanometer
A
5.0 V , 2.0 Ω
P
S
C
Q
D
B
E,
r
When there is no current flowing through the galvanometer, VS = VD (i.e. VSD = 0)
And VP = VC
Therefore VPS = VCD
With original set-up, p.d. across PQ:
VPQ
E

RPQ RT
5.0
8.0 8.0  2.0
VPQ = 4.0 V
VPQ

V = IR and R 
l
A
I l
Therefore V 
A
Vαl
VPS l PS

VPQ l PQ
VPS 0.800

4.0 1.00
VPS = 3.2 V
Since VPS = VCD
VCD = 3.2 V
Looking at the bottom circuit, it is an open circuit, i.e. E = VCD = 3.2 V
Scenerio 2:
with the resistor R where T is the balance point where no current flow in
galvanometer
A
5.0 V , 2.0 Ω
P
T
B
C
D
E,
r
R
With the new balance length at T ,
VPT = VCD
Q
VPT l PT

VPQ l PQ
VPT 0.600

4.0 1.00
VPT = 2.4 V
Therefore VCD = 2.4 V
With R in parallel to battery B to the original circuitry, R is in fact connected across the
battery (i.e. R and r are in series).
VCD = terminal p.d. of the battery = VR
VR E

R RT
2.4
3.2

3.0 3.0  r
r = 1.0 Ω
E = 3.2 V, r = 1.0 