1.1. A resitance of 1.5ohm is connected in parallel with a resistance of 3 ohm. This combination is connected in series with a resistance of unknown resistance of value R. The circuit is then connected acress a 10 V supply.
Calculate:
1.1.1. The value of the resistor R when a 2A current is drawn from the supply
1.1.2. The powe sdissipated in the circuit.
1.2. The field coil of a motor has a resistance of 100 ohms at 10 degrees C. By how much will the resistance increase of the motor attains a temperature of 35 degrees C when running? Take the temperature coefficient as 0.004 per degrees C at 10 degrees C.
3. Define the "ampere"
4. An aluminum conductor, 1 km long, is connected in parallel with a copper conductor having the same length. When a current of 220A passed through the combination it is found that the current through the copper conductor is 120A. The diameter of the aluminum is 10mm.
Calculate:
4.1. The diameter of the copper conductor if the specific resistivity of copper is 0.015 coefficient ohm/m and the aluminuim is 0.028 coefficient ohm/m.
4.2. The voltage drop across conductors.
5. A DC generator of EMF 60V and internal resistance of 0.5 ohm are connected in parallel with a battery of EMF 30V and internal resistance 0.6ohm. The combination is used ro supplt a load having a resistance of 10ohm.
Using kirchhoff's laws to determing:
5.1. The value and direction of current through the generator.
5.2. The value and direction of current through the batter
5.3. The terminal voltage across the load.
6. A resistance of 6ohm is connected in parallel with resistance of 9ohm. The combination is connected in series with a third resistance of 2ohm. If the whole circuit is connected across a battery having an EMF of 18V and an interbal resistance of 0.4ohm, calculate:
6.1. The terminal voltage of the battery.
6.2. The current through each resistor
1. E = 10 Volts.
R1 = 1.5 Ohms.
R2 = 3 Ohms.
R = ?
1.1.1 R1//R2 = R1*R2/(R1+R2) = (1.5*3)/(1.5+3) = 1 Ohm.
Rt = E/I = 10/2 = 5 Ohms = Total resistance.
R = 5-1 = 4 Ohms.
1.1.2 Pd = E*I = 10 * 2 = 20 Watts.
1.2 0.004/C*(35-10)C * 100 = 10 Ohm Increase.
6. Rt = (R1*R2/(R1+R2) + R3 + Ri
Rt = (6*9)/(6+9) + 2 + 0.4 = 6 Ohms.
It = E/Rt = 18/6 = 3 Amps.=Total current
6.1 Et = E-I*Ri = 18 - 3*0.4=16.8 Volts
6.2 I1 = (Et-V3)/R1 = (16.8-6)/6 = 1.8 Amps.
I2 = (16.8-6)/9 = 1.2 Amps.
I3 = It = 3 Amps.
5. E1 = 60 V.
E2 = 30 V.
R1 = 0.5 Ohms.
R2 = 10-Ohm Load.
R3 = 0.6 Ohms.
R2*I2 + R1*I1 = 60 V.
I1 = I2 + I3
R2*I2 + R1*(I2+I3) = 60
10I2 + 0.5I2 + 0.5I3 = 60
Eq1: 10.5I2 + 0.5I3 = 60
R2I2 - R3I3 = 30 V.
Eq2: 10I2 - 0.6I3 = 30
Multiply Eq1 by 0.6 and Eq2 by 0.5. Then
add the Eqs:
6.3I2 + 0.3I3 = 36
5.0I2 - 0.3I3 = 15
Sum: 11.3I2 = 51
I2 = 4.51A. = Current through R2.
In Eq2, replace I2 with 4.51A. and solve
for I3:
10*4.51 - 0.6I3 = 30
45.1 - 0.6I3 = 30
I3 = 25.2A = Current through R3 entering
E2.
I1 = I2 + I3 = 4.51 + 25.2 = 29.7A =
Current through R1 = Current drawn from
E1.
V2 = I2 * R2 = 4.51 * 10 = 45.1 V. = Voltage across R2.
5. E1 = 60V., E2 = 30V.
10I2 - 0.6I1 = 30
Eq1: - 0.6I1 + 10I2 = 30.
30 + 0.6I1 + 0.5(I1+I2) = 60
Eq2: 1.1I1 + 0.5I2 = 30.
Multiply Eq1 by 1.1 and Eq2 by 0.6:
-0.66I1 + 11I2 = 33
0.66I1 + 0.3I2 = 181
sum: 11.3I2 = 51
I2 = 4.51A. = Load current.
In Eq1, replace I2 with 4.51 and solve for I1:
-0.6I1 + 10*4.51 = 30
I1 = 25.2A = Current entering E2.
I1 + I2 = 25.2 + 4.51 = 29.7A = Current from E1.
V2 = I2*R2 = 4.51 * 10 = 45.1V. = Voltage across load.