Statements:
1. A 12V potential difference is applied across a copper conductor with a resistance of 6 kilo-ohm. Find:
The current passing through it.
The charge passing through any cross-sectional area in one hour.
2. During lightning, a current of 15,000A is established when 40C charge passes through a tree due to a potential difference of 1000V. Find:
The duration of the lightning bolt.
The power of the lightning bolt.
3. A copper wire has a diameter of 7.360 mm. Find the resistance of a 5 km long wire used for power transmission.
(Specific resistance of copper = ohm meter)
4. Two resistors of 15 ohm and 75 ohm are connected in series. If the voltage across the 15-ohm resistor is 604. V, find:
The current passing through the 75-ohm resistor.
5. You are given three resistors of 5 kilo-ohm, 10 kilo-ohm, and 15 kilo-ohm. Find:
The greatest resistance possible by connecting them.
The smallest resistance possible by connecting them.
6. Two resistors of 20 ohm and 40 ohm are connected in series with a 50V battery. Find:
The current and power for each resistor.
Repeat the calculation if they are connected in parallel.
7. Two resistors of 4 kilo-ohm and 6 kilo-ohm are connected in parallel with a 10V battery. Find:
The equivalent resistance.
The current passing through each resistor.
The potential difference across each resistor.
Repeat for the case when they are connected in series.
8. A house is installed with the following electrical appliances: 30 LED bulbs of 12W each, running 10 hours daily. 5 fans of 50W each, running 18 hours daily.
Find:
The total electrical units (kWh) consumed per month (30 days).
The monthly electricity cost if the rate is PKR 15 per unit
Problem 1:
Given:
- Voltage, V=12V
- Resistance, R=6kΩ=6000Ω
(a) Current Passing Through It
Using Ohm’s Law:
I=V/R=12/6000=0.002A=2mA
(b) Charge Passing Through in One Hour
Charge is given by:
Q=I×t
Time = 1 hour = 3600 seconds
Q=0.002×3600=7.2C
Problem 2:
Given:
- Current, I=15000A
- Charge, Q=40C
- Voltage, V=1000V
(a) Duration of the Lightning Bolt
Using:
I=Q/t
t=Q/I=40/15000=0.00267s=2.67ms
(b) Power of the Lightning Bolt
Power:
P=VI=1000×15000=15,000,000W=15MW
Problem 3:
Given:
- Diameter, d=7.360mm =7.36×10−3m
- Length, l=5km=5000m
- Resistivity of Copper, ρ=1.68×10−8Ωm
Resistance formula:
R=ρl/A
Area:
A=πd2/4 =3.1416×(7.36×10−3)2/4
Now,
R=(1.68×10−8)×5000/4.25×10−5 R=1.98Ω
Problem 4:
Given:
- R1=15Ω
- Voltage across R1V1=604V
Since in series:
I=V1/R1=604/15=40.27A
Since the current is the same in series, the current passing through the 75Ω resistor is also 40.27A.
Problem 5:
Given:
- R1=5kΩ,R2=10kΩ,R3=15kΩ
(a) Greatest Resistance (Series)
Rtotal=R1+R2+R3 =5+10+15=30kΩ
(b) Smallest Resistance (Parallel)
1/Req=1/5000+1/10000+1/15000
Req=1/0.0003667 =2727.3Ω
Problem 6:
Given:
- R1=20Ω,R2=40Ω
- Voltage, V=50V
(a) Series Connection
Total resistance:
Rtotal=20+40=60Ω
Current:
I=V/Rtotal=50/60=0.833A
P1=I2R1=(0.833)2×20=13.89W
P2=I2R2=(0.833)2×40=27.78W
(b) Parallel Connection
1/Req=1/20+1/40=0.05+0.025=0.075
Req=1/0.075=13.33Ω
I=V/Req=50/13.33=3.75A
Current through each resistor:
I1=V/R1=50/20=2.5A
I2=V/R2=50/40=1.25A
Power:
P1=V×I1=50×2.5=125W
P2=V×I2=50×1.25=62.5W
Problem 7:
Given:
- R1=4kΩ
- Voltage, V=10V
(a) Parallel Connection
1/Req=1/4000+1/6000 = 0.00025+0.000167=0.000417
Current through each:
I1=V/R1=10/4000=2.5mA
I2=V/R2=10/6000=1.67
Total current:
I=I1+I2=4.17mA
(b) Series Connection
Rtotal=4000+6000=10000ΩR_{total} = 4000 + 6000 = 10000
I=V/Rtotal=10/10000=1mA
Voltage across each:
V1=IR1=1mA×4000=4V
V2=IR2=1mA×6000=6V
Problem 8:
(a) Total Electrical Units (kWh)
LED bulbs:
Power=30×12=360W
Daily consumption=360×10/1000=3.6kWh
Monthly consumption=3.6×30=108kWh
Fans:
Power=5×50=250W
Daily consumption=250×18/1000=4.5kWh
Monthly consumption=4.5×30=135kWh
Total = 108 + 135 = 243 kWh
(b) Monthly Cost
Cost=243×15=PKR3645