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Electric Energy and Power

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Advanced Knowledge of Electric Energy and Power

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Q: An electric lamp of resistance \$20 \, \Omega\$ and a conductor of resistance \$4\, \Omega\$ are connected to a \$6\, V\$ battery as shown in the circuit. Calculate:(a) The total resistance of the circuit(b) The current through the circuit(c)The potential difference across the (i) Electric lamp and (ii) Conductor, and (d) Power of the lamp.

a). the total resistance of the circuit =\$20\Omega +4 \Omega = 24\Omega \$b). The current through the circuit= current through the bulb= current through the conductor.The current through the circuit= \$\dfrac{voltage\ applied}{Total\ Resistance}.\$the current through the circuit \$=\dfrac{6}{24}amp\$The current through the circuit = \$0.25amp\$c). The potential difference across the lamp and the conductor.(i) The potential difference across the electrical lamp \$=0.25\times 20= 5 volt\$(ii) The potential difference conductor \$=0.25 \times 4= 1 \ volt\$d). Power can be calculated as \$\dfrac{voltage \ across\ the \ lamp}{current\ through\ the\ lamp}\$ power=\$\dfrac{5}{0.25}\rightarrow 20 W\$.

Q: An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

Hint : A fuse wire should be able pass that amount of current which the appliance requiresStep 1 : Find the maximum current that the appliance requires Given, Power \$ = 1KW= 1000W\$ & Voltage \$= 220\ V\$ \$P= V \times I\$ So, \$I=\dfrac{P}{V}\$ \$I=\dfrac{1000}{220}= 4.545 A\$ The fuse wire should be able to pass at least \$4.545\ A\$ current.So, the rating should be \$5A\$.

Q: Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Hint: We have to use concept of ohms law as values of voltage & current are given,Solution: Step1: Find resistance of each bulb,Given that,Voltage of the electric supply line, \$V = 220{\text{ }}V\$, Power on each bulb, \$P = 10W\$, Maximum current allowed, \$I = 5A\$So, Resistance of each bulb, \$R = \frac{{{V^2}}}{P} = \frac{{{{\left( {220} \right)}^2}}}{{10}}\Omega \$ Resistance of each bulb, \$R = 4840\Omega \$Step2: Find equivalent resistance ‘\${R_P}\$’Assume ‘\$n\$’ bulbs are to be connected in parallel with each other. Therefore, their equivalent resistance ‘\${R_P}\$’ is given by,\$\frac{1}{{{R_P}}} = \frac{1}{{4840}} + \frac{1}{{4840}} + \ldots \ldots \$\$n\$ Factors\$ \Rightarrow \frac{n}{{4840}}\$So,\${R_P} = \frac{{4840}}{n}{\text{ }}\Omega \$Step3: Apply ohms law,\${R_P} = \frac{V}{I}\$ \$ \Rightarrow \frac{{4840}}{n} = \frac{{220}}{5}\$\$n = \frac{{4840 \times 5}}{{220}} = 110\$ Therefore ‘110’ lamps can be connected.

Q: An electric iron has a rating of \$750\ W, 220\ V\$. Calculate the (i) current flowing through it, and (ii) its resistance when in use.

Given : Power rating \$P = 750\ W \$Rated voltage \$ V = 220\ V \$ \$(i)\$ We know \$ P = VI \$ \$750 = 220 \times I \$ \$ \Rightarrow \$ \$I = \dfrac{750}{220} = 3.40\ A \$ \$(ii)\$ Also, power can also be wriiten as \$P = \dfrac{V^2}{R} \$ \$ \Rightarrow \$ \$R = \dfrac{V^2}{P} = \dfrac{220^2}{750}\$So resistance of electric iron \$R = 64.53\ \Omega \$

Q: If two bulbs of \$25W\$ and \$100W\$ rated at \$200\$ volts are connected in series across a \$440\$ volts supply, then

\${ R }_{ 1 }=\cfrac { { v }^{ 2 } }{ P } =\cfrac { { 200 }^{ 2 } }{ 25 } =1600\Omega \$\${ R }_{ 2 }=\cfrac { { 200 }^{ 2 } }{ 100 } =400\Omega \$When both bulbs are connected in series\${ V }_{ 1 }=440\times \cfrac { { R }_{ 1 } }{ { R }_{ 1 }+{ R }_{ 2 } } =440\times \cfrac { 1600 }{ 1600+400 } =352V;{ V }_{ 2 }=440\times \cfrac { 400 }{ 2500 } =88V\$Voltage across \$25W\$ bulb is greater than specified voltage, hence it will fuse

Q: Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 volt a.c.supply line. The power drawn by the combination in each case respectively will be

Resistance of the 220V, 100W bulb is given by \$R=\quad \frac { { V }^{ 2 } }{ P } =\frac { 220^{ 2 } }{ 100 } \\ \quad =484\Omega \quad \$Thus, when connected in series, total resistance \$ R_{ s }=484+484\quad\Omega \quad\\=968\Omega \quad\$Power drawn in case of series combination, \${ P }_{ s }=\quad \frac { { V }^{ 2 } }{ R_{ s } } =\frac { 220^{ 2 } }{ 968 } \\ \quad =50W\$ When connected in parallel, total resistance \$ R_{ p }=\frac { 484 }{ 2 }\Omega \quad\\=242\Omega \quad\$Power drawn in case of series combination, \${ P }_{ p }=\quad \frac { { V }^{ 2 } }{ R_{ p } } =\frac { 220^{ 2 } }{ 242 } \\ \quad =200W\$

Q: An electric bulb is rated 220 volt and 100 watt, Power consumed by it when operated on 110 volt is :

Answer is A.The power dissipated by an electric bulb is given as P = VI.In this case, Power P = 100 W and Voltage V = 220 V.So, current I is I = P/V = 100/220 = 0.45 A.When the supplied voltage is 110 V, the power consumed will be P=VI = 110 * 0.45 = 50 W.Hence, the power consumed when operated at 110 V is 50 W.

Q: A \$100W, 200V\$ bulb is connected to a \$160V\$ supply. The actual power consumption would be

Resistance of the bulb \$R = \dfrac{V^2}{P}\$where \$P = 100 W\$ and \$V = 200\$ V.\$\therefore\$ \$R = \dfrac{(200)^2}{100} = 400\Omega\$Actual power consumption \$P' = \dfrac{V_1^2}{R}\$where \$V_1 = 160\$ V\$\therefore\$ \$P' = \dfrac{(160)^2}{400} = 64 W\$

Q: A domestic electric circuit has a fuse of \$5A\$. What is the maximum number of \$100 W ( 220 V )\$ bulbs that can be safely used in the circuit?

Power (P)= \$voltage(v) \times current (i)=100W\$\$i=\$\$\cfrac{100}{220}=0.45 A\$Number of bulbs that can be safely operated through a \$5A\$ fuse \$=\$\$\cfrac{5}{0.45}\$\$=11\$

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Important Questions

An electric lamp of resistance $20 Ω$ and a conductor of resistance $4 Ω$ are connected to a $6 V$ battery as shown in the circuit. Calculate:
(a) The total resistance of the circuit
(b) The current through the circuit
(c)The potential difference across the (i) Electric lamp and (ii) Conductor, and
(d) Power of the lamp.

Revise with Concepts

Advanced Knowledge of Electric Energy and Power

Learn with Videos

Quick Summary With Stories

Introduction to Electrical Energy and Power

Power in Electric Circuit

An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

An electric iron has a rating of 750 W,220 V. Calculate the (i) current flowing through it, and (ii) its resistance when in use.

If two bulbs of 25W and 100W rated at 200 volts are connected in series across a 440 volts supply, then

Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 volt a.c.supply line. The power drawn by the combination in each case respectively will be

An electric bulb is rated 220 volt and 100 watt, Power consumed by it when operated on 110 volt is :

A 100W,200V bulb is connected to a 160V supply. The actual power consumption would be

Write relation between heat energy produced in a conductor when a potential difference V is applied across its terminals and a current I flows through it for time t.?

A domestic electric circuit has a fuse of 5A. What is the maximum number of 100W(220V) bulbs that can be safely used in the circuit?

An electric iron has a rating of $750 W, 220 V$ . Calculate the (i) current flowing through it, and (ii) its resistance when in use.

If two bulbs of $25 W$ and $100 W$ rated at $200$ volts are connected in series across a $440$ volts supply, then

A $100 W, 200 V$ bulb is connected to a $160 V$ supply. The actual power consumption would be

Write relation between heat energy produced in a conductor when a potential difference $V$ is applied across its terminals and a current $I$ flows through it for time $t$ .?

A domestic electric circuit has a fuse of $5 A$ . What is the maximum number of $100 W (220 V)$ bulbs that can be safely used in the circuit?