Download Exercise 14 Wave Motion

Exercise 2 (Electric circuits) Suggested Answers
1. (a) Q = It = 2 x 1 = 2C
(b) 12 J
(c) Electrical potential energy transferred in 1s = 12 J C-1 x 2 C s-1 = 24 J
(d) Time required = 360 / 24 = 15 s
2. (a) 1/R = 1/4 + 1/12 => R = 3 .
so equivalent resistance = 3 + 3 = 6 
(b) Current through 3  = main current = 6 / 6 = 1 
P.d. across parallel combination = 6 – 3 x 1 = 3 V
Current through 4  = 3 / 4 = 0.75 A
Current through 12  = 3 / 12 = 0.25 A
(c) P.d. across 3  = 3 V
P.d. across 4  = p.d. across 12  = 3 V
3. (a)
(b) The slider should be set at P so that the total resistance is a maximum, and the current is a minimum.
(c) The ammeter’s reading is wrong. It is because the ammeter records the sum of current through the resistor
R and the voltmeter.
(d) The readings remain unchanged.
(e)
4. (a) (i) V = IR => 12 = I (300+600) => I = 0.0133 A
(ii) Required p.d. = IR = 0.0133 x 600 = 8V
(b) When S is closed,
combined resistance of 600  and voltmeter = 300 
Required p.d. = 12 / (300+300) x 300 = 6 V
(c) Statement 1 is correct because the voltmeter is connected across 600  resistor.
Statement 2 is incorrect because the ammeter is measuring the total current through the 600  resistor as
well as the voltmeter.
(d) The resistance of the unknown resistor (~1000 ) is of comparable resistance as the voltmeter (~600 ),
hence the current through the resistor is comparable with that through the voltmeter. The ammeter measures
the total current through the resistor as well as the voltmeter, not just the current through the resistor, so the
error will be significant.
5. (a) The CRO of practically infinite resistance reads the emf of the cells (4.5 V).
However, the voltmeter only reads the p.d. across PQ as there is current flowing in the circuit, some of the p.d.
drops across the 10 k resistor.
(b) 4.5 [ RV / (RV + 10 )] = 4.2
RV = 140 k
Reading = 4.5 V
6. (a)
x / cm
10
20
30
40
50
I/A
0.5
0.5
0.5
0.5
0.5
V/V
0.8
1.6
1.5
3.0
2.3
4.6
3.1
6.2
4.2
8.4
R/
All correct 1A
R/
Resistance of nichrome wire with length x x
8
x
6
r = 1.7x10-4m
r = 8.5x10-5m
x
4
x
2
x
0
10
20
30
40
labeled axes with units 1A
correct data points 1A
Correct straight line passing through origin 1A
Slope = R/l = 8/0.5 = 16 m-1


R
A  16  1.7  10 4
l
(Accept 1.3 –1.6 x 10-6  m)
   1.45  10
2
6
m 
50 x /cm
Table 1
(c) Since the radius is doubled, and area is proportional to r2, so the area is 4 times as before.
So the resistance is (1/4) of its initial value. So the line of slope 1/4 of original line
7.
(a)
Terminal p.d. V = E –Ir
If current increases, terminal p.d. V decreases.
Or when current increases, the p.d. across internal resistance increases /
energy loss in internal resistance increases.
(b)
(i) Emf = y-intercept = 1.52 V (1.51-1.53V)
(ii) Since V = E – Ir,
r = - slope = - 0.45 
MC
1-5 B E C E E
6-10 C D D A E
(0.43-0.47)
11-15 D B B C A
16-19 C C B D
Explanation
1. The circuit can be redrawn as follows:
2. Current through R = 2.4-1.8 = 0.6 A
P.d. across R = p.d. across 8  => 0.6 x R = 1.8 x 8
=> R = 24 
3. Current through 50  = 0.04 + 0.02 = 0.06 A
P.d. across 50  = IR = 0.06 x 50 = 3 V
4. If L1 burns out, there is no currrent passing through L1 and L2. Current passing through L3 remains unchanged
because the p.d. across it is unchanged (the battery voltage).
5. Power = I2 R, since X, Y and Z have the same current, Z must have a smaller resistance, that’s why it is less
bright.
6. Equivalent resistance of voltmeter and 20 k
= 20/2 = 10 k
Therefore p.d. across voltmeter = 12/2 = 6V.
7. Let I be the current through the 10  resistor.
0.6 (5) = I (10) => I = 0.3 A
Current through 20 = 0.3 + 0.6 = 0.9
P.d. across battery = 0.6 (5) + 0.9 (20) = 21 V
8. Since I is greater because it reads the total current through R and voltmeter, so R = V / I is smaller.
9. If equivalent resistance is large, the current is small.
10. The circuit can be redrawn as follows:
11. The circuit can be redrawn as follows.
Current I1 = 2 A, since I1 (R) = 1 (R + R)
Current I = 1 + 2 = 3A
12. length is halved and cross-sectional area is doubled.
R  l / A.
13. Required p.d. = p.d. across the LHS 10 
= 9 (10/15) = 6 V
(Note : there is no current through the RHS 10 )
14. If the voltmeter connected across a and b reads 4 V, since the voltmeter has very high resistance, so essentially
there is negligible current passing through voltmeter. In other words, current only flows through all the three 2 
resistors.
I = V/R = 4/2 = 2A
Emf of battery = IR = 2 (2+2+2) = 12V
Now a very low resistance ammeter is connected across a and b, in this question, we are forced to assume that
the ammeter has negligible resistance compared with the other resistors (although it is not practical because usually
an ammeter has resistance of several ohms).
Then the equivalent resistance of the parallel branches made by 2  and two 1  resistors is 1 .
Current delivered by battery = E/R = 12/(2+2+1) = 2.4 A
Required current = 2.4/2 = 1.2 A
15. Refer to the circuit below.
I1 = 0.3 – 0.2 = 0.1 A
P.d. across 5  = IR = 0.2 x 5 = 1 V
P.d. across 1  = IR = 0.1 x 1 = 0.1 V
So VC – VA= 1V, VC – VB = 0.1 V
Thus VB - VA = 0.9 V
Since potential of B is higher than A, so current flows in upward direction as shown.
I2 = V/R = 0.9 / 3 = 0.3 A
Since I2 flows upward, I3 must be flowing towards B and I1+ I3 = I2.
I3 = 0.3 – 0.1 = 0.2 A
16. Fastest method: p.d. between X and Y = 0.2 x 2 = 0.4 V
17. Since the resistance of the voltammeter is comparable with the resistor, the ammeter is connected in series with
the resistor to ensure the reading of the ammeter is the current flowing through the resistor.
18. Terminal p.d. = E - Ir
Statement 1, connecting another resistor in series with R will decrease the current, as a result, terminal p.d.
increases.
Statement 2, connecting another resistor in parallel with R will increase the current, as a result, terminal p.d.
decreases.
Statement 3, decreasing internal resistance increases the terminal p.d.
19. R 
l
A
l
4 l
 2
2
d
d 
 
2
 (2l )
32 l

 8  10  80
Now R 
2
d 2
 0.5d 


 2 
10 