38Heating Effect of Current

📚
HEATING EFFECT OF CURRENT
▢ Joule's Law of Heating:
❖ Definition: When current passes through a resistance, heat is produced; this is Joule's heating effect of electric current
❖ Statement: Heat produced in a conductor is directly proportional to square of current, resistance and time

Table 1: Joule's Law of Heating

Quantity
Formula
Heat produced
\(H=I^2Rt\)
Using voltage and current
\(H=VIt\)
Using voltage and resistance
\(H=\frac{V^2t}{R}\)
Heat produced in calories
\(H=\frac{I^2Rt}{J}\)
Mechanical equivalent of heat
\(J=4.2\ J/cal\)
Heat in calories
\(H=\frac{I^2Rt}{4.2}=\frac{VIt}{4.2}=\frac{V^2t}{4.2R}\)
❖ Cause: Transformation of electrical energy into heat energy when current flows through resistance
▢ Electric Power:
❖ Definition: Rate at which work is done by source of emf in maintaining current in electric circuit

Table 1: Electric Power Formulae

Condition
Formula
Basic power
\(P=VI\)
Using current and resistance
\(P=I^2R\)
Using voltage and resistance
\(P=\frac{V^2}{R}\)
With internal resistance
\(P=\left(\frac{E}{R+r}\right)^2R\)
Maximum power condition
\(R=r\)
Maximum power
\(P_{max}=\left(\frac{E}{2r}\right)^2r=\frac{E^2}{4r}\)

Table 2: Units of Power

Unit
Value
SI unit
Watt
Other unit
Ampere-volt
1 kW
1000 W
1 HP
746 W
❖ Power in Series:
◉ Condition: Current through each resistance is same
◉ Formula: \(P=I^2R\)
◉ Relation: \(P\propto R\)
◉ Conclusion: Larger resistance consumes more power and has greater potential difference
❖ Power in Parallel:
◉ Condition: Potential difference across each resistance is same
◉ Formula: \(P=\frac{V^2}{R}\)
◉ Relation: \(P\propto\frac{1}{R}\)
◉ Conclusion: Smaller resistance consumes more power and draws more current
❖ Effective Power of Appliances:

Table 1: Appliances Connected to Voltage Source

Connection
Effective power
Series
\(\frac{1}{P}=\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}+...\)
Parallel
\(P=P_1+P_2+P_3+...\)
▢ Electric Bulb:

Table 1: Rated Values of Bulb

Symbol
Meaning / Formula
\(P_0\)
Rated / maximum power
\(V_0\)
Rated / maximum voltage
\(I_0\)
Rated / maximum current
Rated current
\(I_0=\frac{P_0}{V_0}\)
Resistance of bulb
\(R=\frac{V_0^2}{P_0}\)
❖ Important Points:
  • •
    If applied voltage/current is more than rated value, bulb fuses
  • •
    If applied voltage/current equals rated value, bulb consumes rated power
  • •
    Total actual power consumption is sum of individual actual powers
  • •
    \(P_T=P_1+P_2+P_3+...\)
  • •
    \(P_T=I^2R_{eq}\)
  • •
    \(I=\frac{E_{eq}}{R_{eq}+r_{eq}}\)
❖ Series Combination of Bulbs:
◉ Condition: Only when \(V_{01}=V_{02}=E\) and \(r=0\)
◉ Total Power: \(\frac{1}{P_s}=\frac{1}{P_{01}}+\frac{1}{P_{02}}\)
◉ Individual Power: \(P_1=\frac{P_T\times R_1}{R_1+R_2}\)
❖ Parallel Combination of Bulbs:
◉ Condition: Only when \(V_{01}=V_{02}=E\) and \(r=0\)
◉ Total Power: \(P_p=P_{01}+P_{02}\)
◉ Individual Power: \(P_1=\frac{P_T\times R_2}{R_1+R_2}\)
▢ Brightness and Power of Bulb:
❖ Basic Relation: Brightness is directly proportional to rate of heat production / power consumed
❖ Relation: Brightness \(\propto H\propto P\)

Table 1: Brightness in Series and Parallel

Connection
Common quantity
Heat relation
Conclusion
Series
Current same
\(H=I^2Rt\Rightarrow H\propto R\)
Bulb having low power rating has high resistance, so it glows brighter
Parallel
Voltage same
\(H=\frac{V^2t}{R}\Rightarrow H\propto\frac{1}{R}\)
Bulb having more power rating has low resistance, so it glows brighter
❖ Rated Bulb Relation: \(R=\frac{V_0^2}{P_0}\)
▢ Electric Energy:
❖ Definition: Total work done or energy supplied by source of emf in maintaining current for a given time

Table 1: Electric Energy Formulae

Quantity
Formula
Energy
\(E=Pt\)
Using voltage and current
\(E=VIt\)
Using current and resistance
\(E=I^2Rt\)
Using voltage and resistance
\(E=\frac{V^2t}{R}\)

Table 2: Units of Electric Energy

Unit
Value
SI unit
Joule
1 joule
1 watt × 1 second
1 joule
1 volt × 1 ampere × 1 second
Commercial unit
Kilowatt-hour \((kWh)\)
1 kWh
\(1000\ Wh=3.6\times10^6\ J\)
1 unit electricity
1 kWh
❖ Unit Consumed: \(n=\frac{Total\ watt\times Total\ hour}{1000}\)
▢ Maximum Power Transfer Theorem:
❖ Statement: Output power of a source is maximum when internal resistance of source equals external resistance of circuit
❖ Condition: \(R=r\)
❖ Applicable To: All types of sources of emf
❖ Important Point: Related to output power, not total power dissipated

Table 1: Maximum Power Transfer Formulae

Quantity
Formula
Current
\(I=\frac{E}{R+r}\)
At maximum power
\(R=r\)
Current at maximum power
\(I=\frac{E}{2r}\)
Maximum output power
\(P_{max}=I^2R\)
Maximum output power
\(P_{max}=\frac{E^2}{4r}=\frac{E^2}{4R}\)
❖ Trick Formula:
◉ General:
  • •
    \(P_{max}=\frac{E_{eq}^2}{4R}\)
  • •
    \(I_{max}=\frac{E_{eq}}{2R}\)

Table 1: Maximum Power in Cell Grouping

Combination
\(E_{eq}\)
\(P_{max}\)
\(I_{max}\)
Series
\(nE\)
\(\frac{(nE)^2}{4R}\)
\(\frac{nE}{2R}\)
Parallel
\(E\)
\(\frac{E^2}{4R}\)
\(\frac{E}{2R}\)
Mixed
\(nE\)
\(\frac{(nE)^2}{4R}\)
\(\frac{nE}{2R}\)
❖ Mixed Circuit:
◉ Total Number of Cells: \(N=nm\)
◉ Maximum Power Condition: \(R=r_{eq}=\frac{nr}{m}\)

Table 1: Mixed Circuit Arrangement

Quantity
Formula
Number of cells in each row
\(n=\sqrt{\frac{RN}{r}}\)
Number of rows
\(m=\sqrt{\frac{rN}{R}}\)
Short circuit
Output power is zero
Short circuit energy loss
Power dissipates as heat inside battery due to internal resistance
▢ Efficiency of Source of EMF:
❖ Definition: Ratio of output power to input power

Table 1: Efficiency Formulae

Quantity
Formula
Efficiency
\(\eta=\frac{P_o}{P_i}\)
Using voltage and emf
\(\eta=\frac{VI}{EI}=\frac{V}{E}\)
Using resistance
\(\eta=\frac{IR}{I(R+r)}=\frac{R}{R+r}\)
At maximum power
\(R=r\)
Efficiency at maximum power
\(\eta=\frac{1}{2}=50\%\)
❖ Conclusion: At maximum power, half of total power drawn is useful and half is dissipated inside the cell
▢ Fuse:
❖ Definition: Safety device connected in series with electrical installation to protect it from strong current

Table 1: Fuse Wire

Point
Answer
Material
Tin-lead alloy
Composition
About 63–64% tin + 37% lead
Required property
High resistance and low melting point
Function
Melts when strong current flows and protects installation
Fuse current
Maximum current through fuse wire without melting
❖ Safe Current:
◉ Radius Relation: \(I\propto r^{3/2}\)
◉ Area Relation: \(I\propto A^{3/4}\)
◉ Length Effect: Safe current is independent of length of fuse wire
▢ Electrical Heater:
❖ Assumptions:
  • •
    Same amount of water means same heat required
  • •
    Same source means voltage remains constant

Table 1: Time Taken to Boil Same Amount of Water

Combination
Time relation
Series
\(t_s=t_1+t_2+...\)
Parallel
\(\frac{1}{t_p}=\frac{1}{t_1}+\frac{1}{t_2}+...\)
▢ Read and Digest:

Table 1: Important Points

Fact
Answer
Joule heating
Heat produced when current passes through resistance
Heat produced
\(H=I^2Rt\)
Heat in calories
\(\frac{I^2Rt}{4.2}\)
Electric power
Rate of electrical work done
Power
\(P=VI=I^2R=\frac{V^2}{R}\)
Power maximum condition
\(R=r\)
Maximum power
\(\frac{E^2}{4r}\)
Series resistors
Larger resistance consumes more power
Parallel resistors
Smaller resistance consumes more power
Bulb rated current
\(I_0=\frac{P_0}{V_0}\)
Bulb resistance
\(R=\frac{V_0^2}{P_0}\)
Bulb brightness
Directly proportional to power consumed
Series bulbs
Lower power-rated bulb glows brighter
Parallel bulbs
Higher power-rated bulb glows brighter
Electric energy
\(E=Pt=VIt=I^2Rt=\frac{V^2t}{R}\)
Commercial unit
kWh
1 kWh
\(3.6\times10^6\ J\)
1 unit electricity
1 kWh
Maximum power transfer theorem
Output power maximum when \(R=r\)
Efficiency at maximum power
50%
Fuse wire property
High resistance and low melting point
Fuse current
Maximum current without melting fuse
Fuse current-radius relation
\(I\propto r^{3/2}\)
Fuse current-length relation
Independent of length
Heaters in series
\(t_s=t_1+t_2+...\)
Heaters in parallel
\(\frac{1}{t_p}=\frac{1}{t_1}+\frac{1}{t_2}+...\)
▢ High-Yield Recall:

Table 1: Heating Effect of Current One-Liners

Fact
Answer
Joule's law of heating
\(H=I^2Rt\)
Heat using voltage
\(H=VIt=\frac{V^2t}{R}\)
Heat in calories
\(\frac{I^2Rt}{4.2}\)
Power
\(P=VI\)
Power using current
\(P=I^2R\)
Power using voltage
\(P=\frac{V^2}{R}\)
Power with internal resistance
\(P=\left(\frac{E}{R+r}\right)^2R\)
Maximum power condition
\(R=r\)
Maximum power
\(P_{max}=\frac{E^2}{4r}\)
Unit of power
Watt
1 HP
746 W
Series power relation
\(P\propto R\)
Parallel power relation
\(P\propto\frac{1}{R}\)
Appliances in series
\(\frac{1}{P}=\frac{1}{P_1}+\frac{1}{P_2}+...\)
Appliances in parallel
\(P=P_1+P_2+...\)
Rated current of bulb
\(I_0=\frac{P_0}{V_0}\)
Bulb resistance
\(R=\frac{V_0^2}{P_0}\)
Brightness
\(Brightness\propto P\)
Series bulbs brightness
Low power-rated bulb brighter
Parallel bulbs brightness
High power-rated bulb brighter
Electric energy
\(E=Pt\)
Commercial unit
kWh
1 kWh
\(3.6\times10^6\ J\)
Electricity units
\(n=\frac{Watt\times hour}{1000}\)
Maximum power transfer
External resistance = internal resistance
Efficiency
\(\eta=\frac{V}{E}=\frac{R}{R+r}\)
Efficiency at maximum power
50%
Fuse material
Tin-lead alloy
Fuse wire property
High resistance, low melting point
Fuse current
\(I\propto r^{3/2}\propto A^{3/4}\)
Fuse length effect
Independent
Heaters in series
\(t_s=t_1+t_2+...\)
Heaters in parallel
\(\frac{1}{t_p}=\frac{1}{t_1}+\frac{1}{t_2}+...\)
Q1.
If the power consumed through a 10 kΩ resistance is 1 W, then current is:
📅BP 2013
Q2.
A bulb rated 100 W; 200 V was supplied with 160 V line. What was power dissipated?
📅BP 2012
Q3.
A heater coil is cut into two equal parts and only one part is used in the heater. How will the heat generated vary?
📅BP 2011
Q4.
Two electric bulbs A and B are connected in parallel to a constant voltage source. A and B are rated as 60 Watt and P Watt respectively. If the resistance of A and B are in the ratio 2:3, P will be
📅MOE 2014
Q5.
Heater is 1000 W then, energy consumed in 2 hrs is
📅IOM 2012
Q6.
25 W - 220 V and 100 W - 220 V is joined in series with 220 V mains then power will be
📅IOM 2012
Q7.
A bulb is rated at 100 V, 220 W, when the voltage drops by 2%, then change in power of bulb is:
📅KU 2014
Q8.
If the strength of current increases by 1% then the power of the bulb will change by
📅KU 2014
Q9.
If the power of a heater is 1 W, 1 A ampere of current is passed through it. Then find the resistance
📅KU 2010
Q10.
Two electric bulbs have tungsten filaments of same length. If one of them gives 60W and other gives 100W
📅IE 2010
Q11.
The power and voltage of a bulb is 100 W and 220 V. When the voltage is made 110 V, the power would be
📅IE 2012
Q12.
The two bulbs at same voltage have power 200 watt and 100 watt respectively. Find out the ratio of their resistance
📅IE 2013
Q13.
Fuse wire should have
📅KU 2009
Q14.
Two identical electric bulbs 200 W, 250 V are connected in parallel across 250 V. Power consumed by the combination is
📅MOE 2011
Q15.
A bulb is rated as 60 watt and 120 volt. The current through the bulb when the bulb is lighted at the rated voltage is
📅Bangladesh Emb
Q16.
If the strength of the current is decreased by 4%, the power of the bulb will change by
📅MOE 2000
Q17.
A bulb with 220 volt consumes the power of 66 watt. If it will be connected with 160 volt then the power consumed by the bulb will be
📅IOM 1999
Q18.
The power of two heater coils is P1 & P2. If they are connected in series, the resultant power is
📅IOM 2002
Q19.
The power of a bulb is 100 watt at 220 V. When the voltage is 110 V, power of the bulb is:
📅IOM 2001
Q20.
If two bulbs whose resistances are in the ratio 1:2 are connected in series. Then their powers will be in the ratio
📅IE 2004
Q21.
Calculate energy dissipated when 0.3A of current is passed at 6V in 2 minutes
📅IE 2006
Q22.
Two resistances 10Ω and 20Ω are connected in parallel. If P is the power consumed by 10Ω, then power consumed by 20Ω is
📅MOE 2008
Q23.
Two resistance of 10Ω & 20Ω joined in series. The power in 10Ω resistor is P then what is the power in 20Ω
📅IE 2007
Q24.
A resistor operated at 100V generates joule heat at a rate of 20W. When placed across a 50V source, it will draw:
📅IE 2007
Q25.
Three identical bulbs are arranged in parallel across 220V supply and the current across each is 40A. The total energy consumed in 1hr is
📅BPKIHS 2001
Q26.
Power dissipated by R1 and R2 resistor are 100 watt and 200 watt respectively when connected to same voltage. Their relation between R1 and R2 is
📅BPKIHS 2002
Q27.
Two heater wires of equal length are first connected in series and then in parallel, the ratio of heat production in the two cases is
📅BPKIHS 2004
Q28.
Two wires of the same mass and material are drawn 1 mm and 2 mm thick. They are connected in series and a current is sent through them. The heat produced will be in the ratio
📅MOE Curriculum
Q29.
In order to light a 6W - 6V bulb at rated power a battery of emf 6V and internal resistance 2Ω is used. The bulb will light at power:
📅IOM 2005
Q30.
Five equal resistors when connected in series dissipated 5 watt power. If they are connected in parallel, the power dissipated will be
📅MOE 2009
Q31.
If R1 and R2 are respectively resistances of a 200W bulb and a 100W bulb, designed to operate on the same voltage, then
📅BPKIHS 2007
Q32.
In the current in an electric bulb drops by 2% the power decreases by
📅KU 2012
Q33.
Two bulb 25W, 220V and 100W, 220V are connected in parallel across 220V mains. The current is more through
📅IOM 2010
Q34.
A 25W, 220V bulb and a 100W, 220V bulb are joined in series and connected to the mains. Which bulb will grow brighter?
📅BPKIHS 2009
Q35.
A 25W, 220V bulb and a 100W, 220V bulbs are joined in parallel and connected to the 220V mains. Which bulb will glow more brightly?
📅MOE 2011
Q36.
Two resistances R and 2R are connected in series in an electrical circuit. The ratio of heat in 2R to that in R is:
📅KU 2013
Q37.
Two bulbs 100W, 250V and 200W, 250V are connected in series across a 500V line. Then:
📅BPKIHS
Q38.
A uniform wire when connected directly across a 200V-line produces heat H per sec. If the wire is divided into n parts and all parts are connected in parallel across a 200V-line. The heat per second will be
📅IOM 2008
Q39.
Two electric bulbs rated P1 watt, V volt and P2 watt V-volt are connected in parallel across V-volt supply. The total power consumed is:
📅MOE 2012
Q40.
Two head lamps of a car are in parallel. They together consume 48W with the help of a 6V battery. The resistance of each bulb is:
📅BPKIHS 2006
Q41.
A fuse wire with a radius 1mm blows at 1.5A. If the fuse wire of the same material should blow at 3.0A, the radius of the must be:
📅IOM 2007
Q42.
Two bulb of 25W, 200W has resistances in the ratio:
📅BPKIHS 2010
Q43.
A 500 watt heating unit is designed to operate from a 115 volt-line. If the voltage drops to 110 volt, the percentage drop in the heat output will be:
📅MOE 2013
Q44.
A bulb rated 220V, 100W is connected across 160V line. The power dissipated will be
📅IOM 2014
Q45.
Two electric bulbs, each designed to operate with a power of 500 watts and 220 volts line are connected in series in 110 volts line. The power generated by each bulb will be:
📅BPKIHS 2012
Q46.
A heating coil is labeled 100W, 220V. The coil is cut into two equal parts and two pieces are joined in parallel to the same source. The energy now liberated per second is
📅IOM 2009
Q47.
The electric bulbs have tungsten filaments of same length. If one of them gives 60 watts and the other 100 watts, then
📅MOE 2010
Q48.
A cell of emf E and internal resistance r supplies currents for same time through external resistance R1 and R2 separately. If the heat developed in the external resistance in the two cases is same, r is
📅IOM 2006
Q49.
An electric kettle has two coils. When one of these is switched on, the water in the kettle boils in 6 minutes. When the other coil is switched on, the water boils is 3 minutes. If the two coils are connected in series, the time taken to boil the water in the kettle is
📅BPKIHS 2005
Q50.
In an ordinary heater if the length of the coil is halved, then a given quantity of water will boil in
📅MOE 2007
Q51.
Five electric bulbs each of 40 watt burns 5 hours each day. Calculate the cost in the month of 30 days; if the cost of 1 unit is Rs. 8
📅IOM 2008
Q52.
In a room one bulb of 1000W burns 5 hour each day. In the same room a heater of 100W burns 4 hour each day. Calculate total energy consumption in 30 days
📅MOE 2009
Q53.
An electric heater rated as 500 W, 220V raises the temperature of 1kg of water at 15°C to the temperature of boiling point in 15 minutes. Calculate the heat efficiency of the heater
📅BPKIHS 2003
Q54.
The same mass of copper is drawn into two wires of thickness 1mm and 2mm. If two wires are connected in series and current is passed, then heat produced in the wires is in the ratio of
📅IOM 2004
Q55.
A bulb is rated as 6V, 6W is connected through a supply of 6V and 1Ω internal resistance. Find the power consumption of bulb
📅BPKIHS 2008
Q56.
A bulb is rated as 100W, 220V and 200W, 220V are connected in series through a supply of 220V. Find the total power consumption
📅MOE 2015
Q57.
The voltage across an electric bulb is increased by 50%. Find the percentage change in power
📅KU 2016
Q58.
How much energy is dissipated as heat during a two minute time interval by a 1.5kΩ resistor which has a constant 20V potential difference across its leads?
📅KU 2015