Informative line

### 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, $$I=\dfrac {\Delta V}{R}$$

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

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

## 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

#### 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

#### 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

# 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

#### 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

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

#### 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, $$R_1=\dfrac {120^2}{60}=240\,\Omega$$

Resistance of bulb B2

$$R_2=\dfrac {V^2}{P_2}$$

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

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

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

Power across bulb B1

$$P_1'=I^2R_1$$

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

or, $$P_1'=$$$$15 \,W$$

Power across bulb B2

$$P_2'=I^2R_2$$

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

or, $$P_2'=$$$$22.5\, W$$

Hence,option (D) is correct.

### 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

Option D 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  $$R$$

$$P_R=I^2R$$

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

#### 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

$$I=\dfrac {\mathcal{E}}{R+r}$$

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

Power dissipated in resistor

$$P = I^2 R$$

or, $$P = (4)^2 × 8$$

or, $$P = 16 × 8$$

or, $$P = 128\, W$$

Hence,option (C) is correct.

### 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

Option C 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, $$R = r$$

#### 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

#### 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