hkcee mc(current and voltage)

1. The figures in the diagram are not clear. I presume the battery voltage is 6v
Equivalent resistance of pararllel circuit = 4 x (3+1)/(4+(3+1)) ohms = 2 ohms
Total resistance = (4+2) ohms = 6 ohms
Main current = 6/6 A = 1 A
Current through 3-ohm resistor = 1/2 A = 0.5 A
Hence, potential difference across XY = 0.5 x 3 v = 1.5 v

3. Use: power = voltage^2/resistance
New power delivered by the heater P = Po(100/200)^2
where Po is the power when the heater is connected to 200v source.
hence, P = Po/4
Energy needed to boil the wayer = (Po).T
Hence, new time required = (Po).T/P = (Po).T/(Po/4) = 4T

7. After S is closed, total resistance of circuit = (10+20) ohms = 30 ohms
current = 3/30 A = 0.1 A
Hence, voltage across 20-ohm resistor = 0.1 x 20 v = 2 v
Therefore, reading of V1 = 0 v (because voltmeter V1 is not connected across any resistor) , reading of V2 = 2v

1. B
V = (6-4)* 3/(3+1)
= 2 * 0.75
= 1.5 V

3. E
P = E / t
V^2 / R = E / t
E = 40000 / R * T
As energy required to boiled the water and the resistance of the kettle remain unchanged
40000 / R * T = 100^2 / R * t (t is the time required to boil the water for 100V source)
40000 T = 10000t
t = 4T

7. A
When switch S is closed, no current flow through V1 as the resistance of the path V1 is higher then that of the path with switch S.
Current in the circuit = V / R = 3 / (10+20) = 0.1 A
Reading of V2 = Voltage across 20 ohm resistor = IR = 20 * 0.1 = 2V

#32
V=IR
6=I [4+(1/4 + 1/4)^-1]
I=2 A
V=IR
V=(1)(2)
=2
V=2 x(3/4)
=1.5V
ans:B

#35
E=Pt
=V^2 T/R
=200^2T/R
E=V'^2 T'/R
=100^2 T'/R
T'=4T
ans:E

#33
V1=0
V=IR
3=I(10+20)
I=0.1 A
V=IR
=0.1(20)
=2 V
ans:A