Chapter 14 Current Electricity (Numerical Problems)
14.1. A current of 3 mA is flowing through a wire for 1 minute. What is the charge flowing through the wire?
14.2. At 100,000 Ω, how much current flows through your body if you touch the terminals of a 12 V battery? If your skin is wet, so that your resistance is only 1000 Ω, how much current would you receive from the same battery?
14.3. The resistance of a conductor wire is 10 MΩ. If a potential difference of 100 volts is applied across its ends, then find the value of current passing through it in mA.
14.4. By applying a potential difference of 10 V across a conductor, a current of 1.5 A passes through it. How much energy would be obtained from the current in 2 minutes?
14.5. Two resistances of 2 kΩ and 8 kΩ are joined in series, if a 10 V battery is connected across the ends of this combination, find the following quantities:
(a) The equivalent resistance of the series combination.
(b) Current passing through each of the resistances.
(c) The potential difference across each resistance.
14.6. Two resistances of 6 k and 12 kΩ are connected in parallel. A 6 V battery is Ω connected across its ends, find the values of the following quantities:
(a) Equivalent resistance of the parallel combination.
(b) Current passing through each of the resistances.
(c) Potential difference across each of the resistance.
14.7. An electric bulb is marked with 220 V, 100 W. Find the resistance of the filament of the bulb. If the bulb is used 5 hours daily, find the energy in kilowatt-hour consumed by the bulb in one month (30 days).
14.8. An incandescent light bulb with an operating resistance of 95 Ω is labelled “150 W.” Is this bulb designed for use in a 120 V circuit or a 220 V circuit?
14.9. A house is installed with
(a) 10 bulbs of 60 W each of which are used 5 hours daily.
(b) 4 fans of 75 W each of which run 10 hours daily.
(c) One T.V. of 100 W which is used for 5 hours daily.
(d) One electric iron of 1000 W which is used for 2 hours daily.
If the cost of one unit of electricity is Rs.4. Find the monthly expenditure of electricity (one month =30 days).
14.10. A 100 W lamp bulb and a 4 kW water heater are connected to a 250 V supply.
Calculate (a) the current which flows in each appliance (b) the resistance of each appliance when in use.
14.11. A resistor of resistance 5.6 Ω is connected across a battery of 3.0 V by means of a wire of negligible resistance. A current of 0.5 A passes through the resistor. Calculate
(a) Power dissipated in the resistor.
(b) Total power produced by the battery.
(c) Give the reason of difference between these two quantities.
14.1. A current of 3 mA is flowing through a wire for 1 minute. What is the charge flowing through the wire?
Solution: Given data:
Current I = m3 mA = 3.0 x 10-3 A
Time t = 1 min = 60 sec.
Q = ?
Formula I = V …… by definition
I = Q/t
Q = It
Q = 3.0 x 10-3 x 60
Q = 180 x 10-3
Q = 180mC
14.2. At 100,000 Ω, how much current flows through your body if you touch the terminals of a 12 V battery? If your skin is wet, so that your resistance is only 1000 Ω, how much current would you receive from the same battery?
Solution: Given Data:
Resistance R = 100,000Ω = 1 MΩ
Current I = ? If battery is V = 12 Volts
Find current I = ? If For same battery but R = 100Ω
Ohm’s Law V = IR …… by definition
⇒ I = V / R
Putting value we get
⇒ I = 12 / 100
⇒ I = 1.2 x 10-4 A
Again Ohm’s law V = IR …… by definition
⇒ I = V / R
Putting value we get
⇒ I = 12 / 1000
14.3. The resistance of a conductor wire is 10 MΩ. If a potential difference of 100 volts is applied across its ends, then find the value of current passing through it in mA.
Solution: Give data
Resistance R = 10MΩ = 1.0 x 107 Ω ∴ 1M = 106
Current I = ? If battery is V = 100 Volts
By Ohm’s law V = IR …… by definition
⇒ I = V / R
Putting value we get I = 100 / 10000000
I = 1.0 x 10-5 A
14.4. By applying a potential difference of 10 V across a conductor, a current of 1.5 A passes through it. How much energy would be obtained from the current in 2 minutes?
Solution. Give Data:
Potential difference V = 10 Volts
Current I 1.5 A
Energy = ?
IF current passes for time t = 2 mins = 120 seconds
∴ 1min = 60 seconds
First we find resistance R = ?
By Ohm’s law V = IR …… by definition
⇒ R = V / I
Putting the values we get
R = 10 / 1.5
R = 6.7Ω
By definition W = I2Rt
W = (1.5)2 x 6.7 x 120
W = 1809 joules
14.5. Two resistances of 2 kΩ and 8 kΩ are joined in series, if a 10 V battery is connected across the ends of this combination, find the following quantities:
(a) The equivalent resistance of the series combination.
(b) Current passing through each of the resistances.
(c) The potential difference across each resistance.
Solution. Given Data:
Resistance R1 = 2kΩ and R2 = 8 kΩ
(a) Equivalent resistance Re = ?
(b) Current through each resistor I = ?
(c) Potential across each resistor V1 = ? and V2 = ?
(a) As the resistors are connected in series, so we know Formula:
Re = R1 + R2
Re = 2kΩ + 8 kΩ
Re = 10 kΩ
(b) For series current will be same through all resistors, By Ohm’s Law
V = IR
I = V / Re
Putting the values we get
I = 10 / 10000
I = 1.0 x 10-3 A = 1mA
(c) The potential difference across R1 is V1 = IR1
Putting the value we get
V1 = (10-3) x (2 x 103 Ω)
V1 = 2 Volts
Similarly the potential difference across R2 is
V2 = IR2
Putting the value we get
V2 = (10-3) x (8 x 103 Ω)
14.6. Two resistances of 6 k and 12 kΩ are connected in parallel. A 6 V battery is Ω connected across its ends, find the values of the following quantities:
(a) Equivalent resistance of the parallel combination.
(b) Current passing through each of the resistances.
(c) Potential difference across each of the resistance.
Solution: Given Data:
The resistance R1 = 6 kΩ and R2 = 12 kΩ
Potential V = 6 Volts
(a) Re = ? (b) I1 = ? I2 = ? I3 = ? I = ?
(a) As the resistance are in parallel combination, we use formula
1/Re = 1 / R1 + 1 / R2
1/Re = 1 / (6 x 103) + 1 / (12 x 103 )
1/Re = (2 + 1) / (12 x 103)
1/Re = 3 / (12 x 103)
Re = (12 x 103) / 3
Re = 4kΩ
Total current can be found V = IRe
I = V / Re
I = 6 / 4 x 103
I = 1.5 x 10-3 A
I = 1.5 mA
Current through R1 is I1 = V / R1 = 6 / 6 x 103 = 1 mA
Current through R2 is I1 = V / R2 = 6 / 12 x 103 = 0.5 mA
(c) As the resistance are in parallel combination so all the resistances are at the same potential V = 6 Volts
14.7. An electric bulb is marked with 220 V, 100 W. Find the resistance of the filament of the bulb. If the bulb is used 5 hours daily, find the energy in kilowatt-hour consumed by the bulb in one month (30 days).
Solution: Given data
Potential V = 220 Volts
Power P = 100 Watts
Resistance R = ?
It is used daily for time t = 5 hours
Energy = ? (kwh) we know that 1 kwh = 3.6 x 106 Joules
First we find current I = ?
By power formula P = IV = (V/R).V = V2/R
R = V2 / P
Putting the values we get
R = (220)2 / 100
R = 484Ω
By definition W = Pt
∴ 3.6 x 106 Joules = 1 kwh
Putting the value we get
W = 100 x 5 x 30 x 3600
W = 15 x 3600,00 Joules
14.8. An incandescent light bulb with an operating resistance of 95 Ω is labelled “150 W.” Is this bulb designed for use in a 120 V circuit or a 220 V circuit?
Solution: Given Data
Resistance R = 95Ω
Power = P = 150 Watts
Is designed for 120 Volts or 220 Volts = ?
By power formula P = V2 / R
⇒ V2 = PR
Putting values we get
V2 = 150 x 95
V2 = 14250
Taking under root V = 120 Volts
14.9. A house is installed with
(a) 10 bulbs of 60 W each of which are used 5 hours daily.
(b) 4 fans of 75 W each of which run 10 hours daily.
(c) One T.V. of 100 W which is used for 5 hours daily.
(d) One electric iron of 1000 W which is used for 2 hours daily.
If the cost of one unit of electricity is Rs.4. Find the monthly expenditure of electricity (one month =30 days).
Solution: Given data
Appliances of different power are given
Let we find the total energy consumed by the appliances.
Ea = 10 x 60 x 5 = 3000 Wh
Eb = 4 x 75 x 10 = 3000 Wh
∴ 3.6 x 106 Joules = 1 kwh
1 KW = 1000 W
Ec = 250 x 2 = 500 Wh
Ed = 1000 x 2 = 2000 Wh
Et = Ea + Eb + Ec + Ed = 3000 + 3000 + 500 + 2000
Et = 8500 Wh
Total energy consumed for 30 days Et = 8500 x 30 = 255000 Wh
Et = 225 KWh
We know 1 KWh is termed as 1 unit of electricity commercially
Total cost = 255 units x 4 (Rupees)
14.10. A 100 W lamp bulb and a 4 kW water heater are connected to a 250 V supply.
Calculate (a) the current which flows in each appliance (b) the resistance of each appliance when in use.
Solution: Given Data
Power of lamp P1 = 100 W
Power of heater P2 = 4 kW = 4000 Watts
Voltage V = 250 V
(a) Current in lamp I1 = ? and Current in heater I2 = ?
Formula for power P = IV
Current through lamp I1 = P1/V
Putting the values we get I1 = 100/250
I1 = 0.4 A
Current through heater I2 = P2/V
I2 = 4000/250
I2 = 8 A
For resistance we use formula V = IR
Resistance through lamp R = V / I1
Putting values we get
R = 250 / 0.4
R = 625Ω
Resistance through heater R = V / I2
Putting values we get
R = 250 / 8
R = 31.25Ω
14.11. A resistor of resistance 5.6 Ω is connected across a battery of 3.0 V by means of a wire of negligible resistance. A current of 0.5 A passes through the resistor. Calculate
(a) Power dissipated in the resistor.
(b) Total power produced by the battery.
(c) Give the reason of difference between these two quantities.
Solution: Give Data
Resistance R = 5.6Ω
Potential difference V = 3.0 Volts
Current I = 0.5 A
(a) Power dissipated P = ?
(b) Total power produced by battery P =?
(c) Reason
(a) Formula for power P = I2R ( It is power dissipated in resistor)
Putting value we get P = (0.5)2 x 5.6
P = 1.4 Watt ———–(1)
P = IV ( It is power delivered by the battery)
P = (0.5) x 3.0
P = 1.5 Watt ————(2)
From equation (1) and (2) the difference is 1.5 – 1.4 = 0.1 Watt
This power is lost by the internal resistance of the battery.