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Electrical Appliances

Learn maximum power transfer theorem and electrical appliances, calculate equation power delivered to a resistor and dissipated by a resistor. Practice to calculation of electricity bill when more than one electrical appliances are used.

Power Delivered to a Load

  • The power delivered to a resistor when current  \(I\) is flowing through it and having potential difference \(\Delta V\) between the terminals, is expressed as,

\(P=I^2R\)

  • Voltage across resistor\(\Delta V\) is given as,

\(\Delta V=IR\)

or, \(I=\dfrac {\Delta V}{R}\)

  • Power delivered to load,

\(P=\left ( \dfrac {\Delta V}{R} \right )^2\;R\)

\(P=\dfrac {(\Delta V)^2}{R} \)

Power Rating

  • Consider a bulb that has power rating of 100 W, 110 V. It means that the bulb uses 100 W power at 110 V maximum.
  • Rating : 100 W, 110 V

Illustration Questions

Calculate the value of resistance of the bulb rated 100 W, 100 V.

A \(50\) \(\Omega\)

B  \(200\) \(\Omega\)

C \(120\) \(\Omega\)

D \(100\) \(\Omega\)

×

Power delivered to load is given as,  \(P=\dfrac {V^2}{R} \)

 

Given : P = 100 W, V = 100 V

\(\therefore\,100=\dfrac {(100)^2}{R}\)

or, \(R=\dfrac {(100)^2}{100}\)

or,  \( R = 100\,\Omega\)

Hence, option (D) is correct.

Calculate the value of resistance of the bulb rated 100 W, 100 V.

A

\(50\) \(\Omega\)

.

B

 \(200\) \(\Omega\)

C

\(120\) \(\Omega\)

D

\(100\) \(\Omega\)

Option D is Correct

Energy Consumed by Resistor

  • Power across the resistor or dissipated by resistor

\(P_R=I\;\Delta V_R\)

or, \(P_R=\dfrac {(\Delta V_R)^2}{R}\)

  • The energy dissipated by a resistor during time interval \(\Delta t\)

\(E_{th}=P_R\;\Delta t\)

S.I. Unit of Energy

Energy = Watts × seconds

  • The SI unit of energy is Joules.

Meaning of kWh

kW \(\rightarrow\)Kilowatt

h \(\rightarrow\) Hour

  • The energy dissipated by a load when it consumed power in kW in hours, is given as 

\(E=P_R\;\Delta t\,\,kWh\)  

Here, \(P_R\rightarrow\) power in kW and \(\Delta t\) in hours

Example-

  • Consider a fan rated at 200 W on a 120 Volt line is used for 5 hrs.

  • To Calculate energy consumed by it

Given :Power = 200 W, Power in kW = 0.2 kW, Time in hrs = 5 hrs

  • Energy in kWh = PR in kW × t in hrs

= 0.2  kW × 5  hrs

= 1 kWh

Illustration Questions

Calculate the amount of energy consumed in kWh by an electric heater rated P = 1000 W  in t = 4 hrs.

A 400 kW

B 4000 kW

C 4 kW

D 40 kW

×

Given : 

Power (P) = 1000 W

Power in kW = 1 kW

Time in hrs = 4 hrs

Energy consumed in kWh is given as

E = P in kW × t in hrs

= 1  kW × 4  hrs

= 4 kWh

Hence, option(C) is correct.

Calculate the amount of energy consumed in kWh by an electric heater rated P = 1000 W  in t = 4 hrs.

A

400 kW

.

B

4000 kW

C

4 kW

D

40 kW

Option C is Correct

Dependency of Electricity Bill on Power Consumption

  • Electricity Bill specifies the number of kWh consumed in a month.
  • This also represents the amount of energy delivered by electric company to the consumers and transformation of this energy into light using appliances by consumers.
  • 1 kWh indicates consumption of 1000 W in 1 hour.

Example-

  • Consider a single bulb of 100 W which consumes 30 kWh in a month and the cost of electricity is given as $ 0.10 kWh.
  • The electricity bill for a month is calculated as
  • Electricity Bill = Number of kWh × cost

    = 30 × $ 0.10 =  $ 3.00

Illustration Questions

A light bulb of power P=100 W lightens for t = 3 hrs a day. How much it will cost in month of April, if electric company cost (c) = $ 20 per kWh.

A $ 180

B $ 2

C $ 1

D $ 120

×

Month of April = 30 days

Number of hrs bulb lightens in a day = 3 hrs

Total number of hrs in month = 90 hrs

Given : Power \( (P) = 100\,\, W\)

Power in kW = \(\dfrac {100}{1000}=0.1\,\,kW\) 

Number of kWh = Power in kW × hrs used

Number of kWh = 0.1 × 90 = 9

Electricity bill for April = Cost / kWh × Number of kWh

= $ 20 × 9 = $ 180

Hence, option (A) is correct.

A light bulb of power P=100 W lightens for t = 3 hrs a day. How much it will cost in month of April, if electric company cost (c) = $ 20 per kWh.

A

$ 180

.

B

$ 2

C

$ 1

D

$ 120

Option A is Correct

Power Dissipation in Resistor

  • Consider a resistor R, connected with a battery with emf \(\mathcal{E}\) and internal resistance  \(r\), as shown in figure.

  • Current flowing in the circuit is \(I\)

\(I=\dfrac{\text {emf of battery}}{R_{eq}}\)

\(I=\dfrac {\mathcal{E}}{R+r}\)

  • Power dissipated in resistor  \(R\)

\(P_R=I^2R\)

\(P_R=\left ( \dfrac {\mathcal{E}}{R+r} \right )^2\cdot R\)

Illustration Questions

A load resistor of resistance  \(R=8\,\Omega\) is connected with a practical battery with emf  \(\mathcal{E}=40\,V\)  and internal resistance  \(r=2\,\Omega\). Calculate the power dissipated by the resistor.  

A 48 W

B 120 W

C 128 W

D 96 W

×

Current flowing in the circuit

\(I=\dfrac {\mathcal{E}}{R+r}\)

\(I=\dfrac {40}{8+2}=\dfrac {40}{10}=4A\)

image

Power dissipated in resistor 

\(P = I^2 R\)

or, \(P = (4)^2 × 8\)

or, \(P = 16 × 8\)

or, \(P = 128\, W\)

image

Hence,option (C) is correct.

image

A load resistor of resistance  \(R=8\,\Omega\) is connected with a practical battery with emf  \(\mathcal{E}=40\,V\)  and internal resistance  \(r=2\,\Omega\). Calculate the power dissipated by the resistor.  

image
A

48 W

.

B

120 W

C

128 W

D

96 W

Option C is Correct

Calculation of Electricity Bill when More than One Electrical Appliances are Used

  • Electricity Bill specifies the number of kWh consumed in a month.
  • This also represents the amount of energy delivered by electric company to the consumers and transformation of this energy into light using appliances by consumers.
  • 1 kWh indicates consumption of 1000 W in one hour.

Case 1 :

  • Consider, two instruments I1 and I2 of power rating P1 and P2 kW respectively, are simultaneously being averagely used t hours a day.
  • The total number of kWh consumed by them in a month of November,

Number of units in kWh = (P1 + P2) t × 30

  • Electricity bill in month of November

Electricity Bill = Number of kWh × Cost / kWh

Electricity Bill = (P1 + P2) × t × 30 × Cost / kWh

Case 2 :

  • Consider, two instruments I1 and I2 of power rating P1 and P2 kW respectively, but not operated simultaneously.
  • Let instrument I1 is used for t1 hours a day on an average a month and instrument I2 is used for t2 hours a day on an average a month.
  • Total electricity bill in the month of November

Electricity Bill = (P1t1 + P2t2) × 30 × Cost / kWh

Illustration Questions

A water heater of P1 = 1500 W is averagely used for t1 = 4 hrs daily and a hair dryer of P2 = 1000 W is averagely used for t2 = 2 hrs daily. Calculate electricity bill for month of July if cost is C = $ 0.20 / kWh.

A $ 49.6

B $ 40

C $ 80

D $ 60

×

For Water Heater :

Power (P1) = 1500 W

Power (P1) in kW = \(\dfrac {1500}{1000}=1.5\) kW

Used for number of hours/day t=  4 hrs

Number of kwh used in a day = P1 × t1

=1.5 × 4

= 6

For Hair Dryer :

Power (P2) = 1000 W

Power (P2) in kW = \(\dfrac {1000}{1000}=1\) kW

Used for number of hours/day t=  2 hrs

Number of kwh used in a day = P× t2

= 1 × 2

= 2

Total number of kWh used by both water heater and hair dryer in a day

Number of kWh = 6 + 2

= 8 

Month of July = 31 days

Number of kWh used in 1 day = 8

Total number of kWh in month of July = 8 × 31

= 248

Cost per kWh = $ 0.20 / kWh

Total number of kWh = 248

Electricity Bill = Total number of kWh × Cost / kWh

= 248 × 0.20

= $ 49.6

 

Hence,option (A) is correct.

A water heater of P1 = 1500 W is averagely used for t1 = 4 hrs daily and a hair dryer of P2 = 1000 W is averagely used for t2 = 2 hrs daily. Calculate electricity bill for month of July if cost is C = $ 0.20 / kWh.

A

$ 49.6

.

B

$ 40

C

$ 80

D

$ 60

Option A is Correct

Power Dissipation across Bulbs Connected in Series

  • Consider two bulbs such that bulb B1 with rating P1,V1 and bulb B2 with rating P2,V connected in series with a battery of emf \(\mathcal{E}\), as shown in figure:

  • Since B1 and B2 are connected in series so, current will be same. Let the current be I.
  • Power dissipation by bulb is given by \(P=\dfrac {V^2}{R}\)
  • The resistance of bulb B1

\(P_1=\dfrac {V_1^2}{R_1}\)

or, \(R_1=\dfrac {V_1^2}{P_1}\)

  • The resistance of bulb B2

\(P_2=\dfrac {V_2^2}{R_2}\)

or, \(R_2=\dfrac {V_2^2}{P_2}\)

  • Total current \(I=\dfrac {\mathcal{E}}{R_1+R_2}\)

\(I=\dfrac {\mathcal{E}}{\dfrac {V_1^2}{P_1}+\dfrac {V_2^2}{P_2}}\)

  • Power across B1

\(P_1'=I^2R_1\, \)

where  \(P'_1\) is the power dissipated by bulb \(B_1\)

or, \(P_1'=\left ( \dfrac {\mathcal{E}}{R_1+R_2} \right)^2×R_1\)

  • Power across B2

\(P_2'=I^2R_2\)

where \(P_2'\) is the power dissipated by bulb \(B_2\)

or, \(P_2'=\left ( \dfrac {\mathcal{E}}{R_1+R_2} \right)^2×R_2\)

When both Bulb of Same Voltage Rating

V1 = V2\(\mathcal{E}\)

Power across B1

\(P_1'=\left (\dfrac {\mathcal{E}}{\dfrac {\mathcal{E}^2}{P_1}+\dfrac {\mathcal{E}^2}{P_2}}\right)^2×\dfrac {\mathcal{E}^2}{P_1}\)

or, \(P_1'=\dfrac {1} {\left (\dfrac {1}{P_1} +\dfrac {1}{P_2}\right)^2}×\dfrac {1}{P_1}\)

or, \(P_1'=\dfrac {1} {\left (\dfrac {P_1+P_2}{P_1P_2} \right)^2}×\dfrac {1}{P_1}\)

or, \(P_1'=\dfrac {P_1^2\;P_2^2} {(P_1+P_2)^2} ×\dfrac {1}{P_1}\)

or, \(P_1'=\dfrac {P_1\;P_2^2} {(P_1+P_2)^2} \)

Similarly,

Power across B2

\(P_2'=\dfrac {P_1^2\;P_2} {(P_1+P_2)^2}\)

Illustration Questions

Two bulbs, bulb B1 with power rating P1 = 60 W and bulb B2 with power rating P2 = 40 W are connected in series. The rating voltage of both the bulbs is V = 120 V. Calculate power across both the bulbs if, they are connected with a battery of emf \(\mathcal{E}=\)150 V.

A 15 W, 30 W

B 15 W, 20 W

C 20 W, 15 W

D 15 W, 22.5 W

×

Resistance of bulb B1

\(R_1=\dfrac {V_1^2}{P_1}\)

or, \(R_1=\dfrac {120^2}{60}=240\,\Omega\)

image

Resistance of bulb B2

\(R_2=\dfrac {V^2}{P_2}\)

or, \(R_2=\dfrac {120^2}{40}=360\,\Omega\)

image

Total current in circuit = \(\dfrac {\mathcal{E}}{R_1+R_2}\)

\(I=\dfrac {150}{240+360}=\dfrac {1}{4}\)\(A\)

image

Power across bulb B1

\(P_1'=I^2R_1\)

or, \(P_1'=\left ( \dfrac {1}{4}\right)^2×240\)

or, \(P_1'=\)\(15 \,W\)

image

Power across bulb B2

\(P_2'=I^2R_2\)

or, \(P_2'=\left ( \dfrac {1}{4}\right)^2×360\)

or, \(P_2'=\)\( 22.5\, W\)

image

Hence,option (D) is correct.

image

Two bulbs, bulb B1 with power rating P1 = 60 W and bulb B2 with power rating P2 = 40 W are connected in series. The rating voltage of both the bulbs is V = 120 V. Calculate power across both the bulbs if, they are connected with a battery of emf \(\mathcal{E}=\)150 V.

image
A

15 W, 30 W

.

B

15 W, 20 W

C

20 W, 15 W

D

15 W, 22.5 W

Option D is Correct

Maximum Power Output Theorem

  • Consider a circuit consisting of a battery of emf \(\mathcal{E}\) with internal resistance r and load resistor of resistance R, as shown in figure.

  • Current in the circuit    \(I=\dfrac {\mathcal{E}}{r+R}\)
  • Power across load resistor    \(P=\left ( \dfrac {\mathcal{E}}{r+R} \right)^2\cdot R\)
  • For maximum Power

\(\dfrac {dP}{dR}=\dfrac {d}{dR} \left [ \left ( \dfrac {\mathcal{E}}{r+R} \right)^2 \cdot R \right]\)

or, \(\dfrac {dP}{dR}=\mathcal{E}^2\; \dfrac {d}{dR} \left [ \dfrac {R}{(R+r)^2} \right]\)

For power to be maximum, \(\dfrac {dP}{dR}=0\)

\(0=\dfrac {(R+r)^2×1-R×2(R+r)} {(R+r)^4}\)

or,\( 2 R (R + r) = (R + r)^2\)

or, \(R = r\)

Illustration Questions

A load resistor of resistance R is connected in series with real battery of emf \(\mathcal{E}\) and internal resistance r. If R < r, the power across load is 

A maximum

B less than maximum

C zero

D None of these

×

For maximum power across load  r = R 

Hence,option (B) is correct.

A load resistor of resistance R is connected in series with real battery of emf \(\mathcal{E}\) and internal resistance r. If R < r, the power across load is 

A

maximum

.

B

less than maximum

C

zero

D

None of these

Option B is Correct

Comparison of Brightness in Series and Parallel Connection

Case 1 : Series Connection

  • Consider four bulbs, B1, B2, B3, B4 each with rating P1, P2, P3 and P4 respectively are connected in series as shown in circuit.

  • If all the bulbs have same voltage rating, the relation among P1, P2, P3 and P4 is given as;

P< P< P3 < P4 

  • Resistance of B1\(\dfrac {V^2}{P_1}=R_1\)

 

  • Resistance of B2 = \(\dfrac {V^2}{P_2}=R_2\)

 

  • Resistance of B3 = \(\dfrac {V^2}{P_3}=R_3\)

 

  • Resistance of B4 = \(\dfrac {V^2}{P_4}=R_4\)
  • Since R is inversely proportional to P, so if the voltage rating of all bulbs are same, then the order of resistance is 

R> R> R3 > R4 

  • The brightness is determined by the power dissipated across bulb.

\(P=I^2R\)

or, \(P_{\text { Dissipated}}=I^2R\)

  • Power dissipated across B1

\(P_{\text { D}_1}=I^2R_1\)

  • Power dissipated across B2

\(P_{\text { D}_2}=I^2R_2\)

  • Power dissipated across B3

\(P_{\text { D}_3}=I^2R_3\)

  • Power dissipated across B4

\(P_{\text { D}_4}=I^2R_4\)

  • Since, the current is constant in the circuit, it is concluded that power dissipated is directly proportional to load resistance.

\(P\,_{D_1} \propto\;R_1\)

\(P\,_{D_2} \propto\;R_2\)

\(P\,_{D_3} \propto\;R_3\)

\(P\,_{D_4} \propto\;R_4\)

CONCLUSION:  The higher the resistance of bulb, the higher the power dissipated and the brighter the bulb glows.

The order of brightness is 

B1 > B2 > B3 > B4

Case 2 : Parallel Connection

  • In parallel connection, current is divided into all bulbs.
  • The bulb having high resistance will have less current and hence has less brightness.
  • Order of brightness is  B1 < B2 < B3 < B

Illustration Questions

Five bulbs are connected in series with power rating P1 = 20 W, P2 = 30 W, P3 = 40 W, P4 =60 W and P5 = 100 W respectively. What will be the order of brightness when all bulbs have same voltage rating V = 120 V.

A  B1 = B2 > B3 > B4 > B5

B  B1 < B2 < B3 < B4 < B5

C  B1 > B2 > B3 > B4 > B5

D  B1 = B2 = B3 = B4 = B5

×

Resistance of Bulb B1

\(R_1=\dfrac {V^2}{P_1}=\dfrac {(120)^2}{20}=720\;\Omega\)

Resistance of Bulb B2

\(R_2=\dfrac {V^2}{P_2}=\dfrac {(120)^2}{30}=480\;\Omega\)

 

 

Resistance of Bulb B3

\(R_3=\dfrac {V^2}{P_3}=\dfrac {(120)^2}{40}=360\;\Omega\)

 

 

Resistance of Bulb B4

\(R_4=\dfrac {V^2}{P_4}=\dfrac {(120)^2}{60}=240\;\Omega\)

 

 

Resistance of Bulb B5

\(R_5=\dfrac {V^2}{P_5}=\dfrac {(120)^2}{100}=144\;\Omega\)

 

 

Since power dissipation is directly proportional to resistance of bulb for series connection. Hence, bulb with high resistance has more brightness.

The order of resistance is 

R1 > R2 > R3 > R4 > R5

Therefore, order of brightness is

B1 > B2 > B3 > B4 > B5

Hence,option (C) is correct.

Five bulbs are connected in series with power rating P1 = 20 W, P2 = 30 W, P3 = 40 W, P4 =60 W and P5 = 100 W respectively. What will be the order of brightness when all bulbs have same voltage rating V = 120 V.

A

 B1 = B2 > B3 > B4 > B5

.

B

 B1 < B2 < B3 < B4 < B5

C

 B1 > B2 > B3 > B4 > B5

D

 B1 = B2 = B3 = B4 = B5

Option C is Correct

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