Download Solution of Exercise 20 (Electric Circuits)

Solution of Exercise 14 (Electric Circuits)
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) a4 < a2 < a3 < a1
(b) RP : RR = lP /AP : lR /AR = ( lP / lR ) (AR / AP) = (3/4) x (1/2) = 3:8
(c) a4 = a2 = 0. a1 and a3 increase.
A3 would be ruined if S were absent.
14.1
Multiple Choice
1-5 B E C E E
6-10
CDDAE
11-15 D B B B C 16-17 A E
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. You will learn in the next chapter that power = I2 R, so 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
14.2
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. Current through each branch = 1A (since each branch has the same resistance 6 )
P.d. across LHS 2  = 1 x 2 = 2V
P.d. across LHS 4  = 1 x 4 = 4 V
So p.d. between X and Y = 4 – 2 = 2 V (X is 2 V higher than that of Y)
17. Current through 3R = 6 / 3R = 2/R
Equivalent resistance across the parallel branches = R / 2
Required p.d. = 2/R x R/2 = 1 V
14.3