152709
Four $2 \Omega$ resistors are connected together along the edges of a square. A $10 \mathrm{~V}$ battery of negligible internal resistance is connected across a pair of the diagonally opposite corners of the square. The power dissipated in the circuit is
1 $500 \mathrm{~W}$
2 $50 \mathrm{~W}$
3 $5 \mathrm{~W}$
4 $10 \mathrm{~W}$
Explanation:
B $\mathrm{R}_{\mathrm{eq}}=\frac{4 \times 4}{4+4}=2 \Omega$ Power dissipated in circuit, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{(10)^{2}}{2}=50 \mathrm{~W}$
EAMCET 1992
Current Electricity
152710
A house, served by a $220 \mathrm{~V}$ supply line, is protected by a 9 A fuse. What is the maximum number of 60 watt bulbs that can be turned on in parallel?
1 11
2 22
3 33
4 44
Explanation:
C Total power, $\mathrm{P}=\mathrm{V} \times \mathrm{I}=200 \times 9$ Number of $60 \mathrm{~W}$ bulb required, $\mathrm{n}=\frac{220 \times 9}{60}=33$
NDA (II) 2009
Current Electricity
152711
How many sixty watt $(60 \mathrm{~W})$ bulbs may be safely used in a 240-V supply with 4-ampere fuse?
1 4
2 8
3 12
4 16
Explanation:
D Given, $\mathrm{P}=60 \mathrm{~W}, \mathrm{~V}=240 \mathrm{~V}, \mathrm{I}=4 \mathrm{~A}$ Total power, $\mathrm{P}=\mathrm{VI}=240 \times 4=960 \mathrm{~W}$ Number of $60 \mathrm{~W}$ bulb required $\mathrm{n}=\frac{960}{60}=16$
NDA (I) 2010
Current Electricity
152714
An electric heater is rated 1500 watt. If electric power costs Rs. 2 per kilo-watt hour, then the cost of power for $\mathbf{1 0}$ hours running of the heater is
1 ₹ 30
2 ₹ 15
3 ₹ 150
4 ₹ 25
Explanation:
A Given, $\mathrm{P}=1500 \mathrm{~W}$, Rs. $=2$ per kilo-watt, time $=10$ hours Energy consume $=\mathrm{P} \times \mathrm{t}=1500 \times 10$ $\mathrm{H}=15$ kilo-watt-hour Total cost $=2 \times 15=30$ rupees
NDA (II) 2013
Current Electricity
152718
$1 \mathrm{kWh}$ is equal to
1 $3.6 \times 10^{6} \mathrm{~J}$
2 $3.6 \times 10^{4} \mathrm{~J}$
3 $3.6 \times 10^{3} \mathrm{~J}$
4 $6 \times 10^{-4} \mathrm{~J}$
Explanation:
A Kilo-watt hour is a unit of energy used to denote the energy consumption of large machinery and the electricity consumption in homes and industries. It is also a common unit representation in electrical power engineering. One kilo watt Mean 1000 watts of power and 1 hour contains 3600 second in it. So, if we try to express one kilo-watt hour is terms of joules we get, $\mathrm{I} \mathrm{kW} \mathrm{h}=(1000 \mathrm{~W}) \times(3600 \mathrm{sec})$ $1 \mathrm{~kW} \mathrm{~h}=3.6 \times 10^{6} \mathrm{~J}$
152709
Four $2 \Omega$ resistors are connected together along the edges of a square. A $10 \mathrm{~V}$ battery of negligible internal resistance is connected across a pair of the diagonally opposite corners of the square. The power dissipated in the circuit is
1 $500 \mathrm{~W}$
2 $50 \mathrm{~W}$
3 $5 \mathrm{~W}$
4 $10 \mathrm{~W}$
Explanation:
B $\mathrm{R}_{\mathrm{eq}}=\frac{4 \times 4}{4+4}=2 \Omega$ Power dissipated in circuit, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{(10)^{2}}{2}=50 \mathrm{~W}$
EAMCET 1992
Current Electricity
152710
A house, served by a $220 \mathrm{~V}$ supply line, is protected by a 9 A fuse. What is the maximum number of 60 watt bulbs that can be turned on in parallel?
1 11
2 22
3 33
4 44
Explanation:
C Total power, $\mathrm{P}=\mathrm{V} \times \mathrm{I}=200 \times 9$ Number of $60 \mathrm{~W}$ bulb required, $\mathrm{n}=\frac{220 \times 9}{60}=33$
NDA (II) 2009
Current Electricity
152711
How many sixty watt $(60 \mathrm{~W})$ bulbs may be safely used in a 240-V supply with 4-ampere fuse?
1 4
2 8
3 12
4 16
Explanation:
D Given, $\mathrm{P}=60 \mathrm{~W}, \mathrm{~V}=240 \mathrm{~V}, \mathrm{I}=4 \mathrm{~A}$ Total power, $\mathrm{P}=\mathrm{VI}=240 \times 4=960 \mathrm{~W}$ Number of $60 \mathrm{~W}$ bulb required $\mathrm{n}=\frac{960}{60}=16$
NDA (I) 2010
Current Electricity
152714
An electric heater is rated 1500 watt. If electric power costs Rs. 2 per kilo-watt hour, then the cost of power for $\mathbf{1 0}$ hours running of the heater is
1 ₹ 30
2 ₹ 15
3 ₹ 150
4 ₹ 25
Explanation:
A Given, $\mathrm{P}=1500 \mathrm{~W}$, Rs. $=2$ per kilo-watt, time $=10$ hours Energy consume $=\mathrm{P} \times \mathrm{t}=1500 \times 10$ $\mathrm{H}=15$ kilo-watt-hour Total cost $=2 \times 15=30$ rupees
NDA (II) 2013
Current Electricity
152718
$1 \mathrm{kWh}$ is equal to
1 $3.6 \times 10^{6} \mathrm{~J}$
2 $3.6 \times 10^{4} \mathrm{~J}$
3 $3.6 \times 10^{3} \mathrm{~J}$
4 $6 \times 10^{-4} \mathrm{~J}$
Explanation:
A Kilo-watt hour is a unit of energy used to denote the energy consumption of large machinery and the electricity consumption in homes and industries. It is also a common unit representation in electrical power engineering. One kilo watt Mean 1000 watts of power and 1 hour contains 3600 second in it. So, if we try to express one kilo-watt hour is terms of joules we get, $\mathrm{I} \mathrm{kW} \mathrm{h}=(1000 \mathrm{~W}) \times(3600 \mathrm{sec})$ $1 \mathrm{~kW} \mathrm{~h}=3.6 \times 10^{6} \mathrm{~J}$
152709
Four $2 \Omega$ resistors are connected together along the edges of a square. A $10 \mathrm{~V}$ battery of negligible internal resistance is connected across a pair of the diagonally opposite corners of the square. The power dissipated in the circuit is
1 $500 \mathrm{~W}$
2 $50 \mathrm{~W}$
3 $5 \mathrm{~W}$
4 $10 \mathrm{~W}$
Explanation:
B $\mathrm{R}_{\mathrm{eq}}=\frac{4 \times 4}{4+4}=2 \Omega$ Power dissipated in circuit, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{(10)^{2}}{2}=50 \mathrm{~W}$
EAMCET 1992
Current Electricity
152710
A house, served by a $220 \mathrm{~V}$ supply line, is protected by a 9 A fuse. What is the maximum number of 60 watt bulbs that can be turned on in parallel?
1 11
2 22
3 33
4 44
Explanation:
C Total power, $\mathrm{P}=\mathrm{V} \times \mathrm{I}=200 \times 9$ Number of $60 \mathrm{~W}$ bulb required, $\mathrm{n}=\frac{220 \times 9}{60}=33$
NDA (II) 2009
Current Electricity
152711
How many sixty watt $(60 \mathrm{~W})$ bulbs may be safely used in a 240-V supply with 4-ampere fuse?
1 4
2 8
3 12
4 16
Explanation:
D Given, $\mathrm{P}=60 \mathrm{~W}, \mathrm{~V}=240 \mathrm{~V}, \mathrm{I}=4 \mathrm{~A}$ Total power, $\mathrm{P}=\mathrm{VI}=240 \times 4=960 \mathrm{~W}$ Number of $60 \mathrm{~W}$ bulb required $\mathrm{n}=\frac{960}{60}=16$
NDA (I) 2010
Current Electricity
152714
An electric heater is rated 1500 watt. If electric power costs Rs. 2 per kilo-watt hour, then the cost of power for $\mathbf{1 0}$ hours running of the heater is
1 ₹ 30
2 ₹ 15
3 ₹ 150
4 ₹ 25
Explanation:
A Given, $\mathrm{P}=1500 \mathrm{~W}$, Rs. $=2$ per kilo-watt, time $=10$ hours Energy consume $=\mathrm{P} \times \mathrm{t}=1500 \times 10$ $\mathrm{H}=15$ kilo-watt-hour Total cost $=2 \times 15=30$ rupees
NDA (II) 2013
Current Electricity
152718
$1 \mathrm{kWh}$ is equal to
1 $3.6 \times 10^{6} \mathrm{~J}$
2 $3.6 \times 10^{4} \mathrm{~J}$
3 $3.6 \times 10^{3} \mathrm{~J}$
4 $6 \times 10^{-4} \mathrm{~J}$
Explanation:
A Kilo-watt hour is a unit of energy used to denote the energy consumption of large machinery and the electricity consumption in homes and industries. It is also a common unit representation in electrical power engineering. One kilo watt Mean 1000 watts of power and 1 hour contains 3600 second in it. So, if we try to express one kilo-watt hour is terms of joules we get, $\mathrm{I} \mathrm{kW} \mathrm{h}=(1000 \mathrm{~W}) \times(3600 \mathrm{sec})$ $1 \mathrm{~kW} \mathrm{~h}=3.6 \times 10^{6} \mathrm{~J}$
152709
Four $2 \Omega$ resistors are connected together along the edges of a square. A $10 \mathrm{~V}$ battery of negligible internal resistance is connected across a pair of the diagonally opposite corners of the square. The power dissipated in the circuit is
1 $500 \mathrm{~W}$
2 $50 \mathrm{~W}$
3 $5 \mathrm{~W}$
4 $10 \mathrm{~W}$
Explanation:
B $\mathrm{R}_{\mathrm{eq}}=\frac{4 \times 4}{4+4}=2 \Omega$ Power dissipated in circuit, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{(10)^{2}}{2}=50 \mathrm{~W}$
EAMCET 1992
Current Electricity
152710
A house, served by a $220 \mathrm{~V}$ supply line, is protected by a 9 A fuse. What is the maximum number of 60 watt bulbs that can be turned on in parallel?
1 11
2 22
3 33
4 44
Explanation:
C Total power, $\mathrm{P}=\mathrm{V} \times \mathrm{I}=200 \times 9$ Number of $60 \mathrm{~W}$ bulb required, $\mathrm{n}=\frac{220 \times 9}{60}=33$
NDA (II) 2009
Current Electricity
152711
How many sixty watt $(60 \mathrm{~W})$ bulbs may be safely used in a 240-V supply with 4-ampere fuse?
1 4
2 8
3 12
4 16
Explanation:
D Given, $\mathrm{P}=60 \mathrm{~W}, \mathrm{~V}=240 \mathrm{~V}, \mathrm{I}=4 \mathrm{~A}$ Total power, $\mathrm{P}=\mathrm{VI}=240 \times 4=960 \mathrm{~W}$ Number of $60 \mathrm{~W}$ bulb required $\mathrm{n}=\frac{960}{60}=16$
NDA (I) 2010
Current Electricity
152714
An electric heater is rated 1500 watt. If electric power costs Rs. 2 per kilo-watt hour, then the cost of power for $\mathbf{1 0}$ hours running of the heater is
1 ₹ 30
2 ₹ 15
3 ₹ 150
4 ₹ 25
Explanation:
A Given, $\mathrm{P}=1500 \mathrm{~W}$, Rs. $=2$ per kilo-watt, time $=10$ hours Energy consume $=\mathrm{P} \times \mathrm{t}=1500 \times 10$ $\mathrm{H}=15$ kilo-watt-hour Total cost $=2 \times 15=30$ rupees
NDA (II) 2013
Current Electricity
152718
$1 \mathrm{kWh}$ is equal to
1 $3.6 \times 10^{6} \mathrm{~J}$
2 $3.6 \times 10^{4} \mathrm{~J}$
3 $3.6 \times 10^{3} \mathrm{~J}$
4 $6 \times 10^{-4} \mathrm{~J}$
Explanation:
A Kilo-watt hour is a unit of energy used to denote the energy consumption of large machinery and the electricity consumption in homes and industries. It is also a common unit representation in electrical power engineering. One kilo watt Mean 1000 watts of power and 1 hour contains 3600 second in it. So, if we try to express one kilo-watt hour is terms of joules we get, $\mathrm{I} \mathrm{kW} \mathrm{h}=(1000 \mathrm{~W}) \times(3600 \mathrm{sec})$ $1 \mathrm{~kW} \mathrm{~h}=3.6 \times 10^{6} \mathrm{~J}$
152709
Four $2 \Omega$ resistors are connected together along the edges of a square. A $10 \mathrm{~V}$ battery of negligible internal resistance is connected across a pair of the diagonally opposite corners of the square. The power dissipated in the circuit is
1 $500 \mathrm{~W}$
2 $50 \mathrm{~W}$
3 $5 \mathrm{~W}$
4 $10 \mathrm{~W}$
Explanation:
B $\mathrm{R}_{\mathrm{eq}}=\frac{4 \times 4}{4+4}=2 \Omega$ Power dissipated in circuit, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{(10)^{2}}{2}=50 \mathrm{~W}$
EAMCET 1992
Current Electricity
152710
A house, served by a $220 \mathrm{~V}$ supply line, is protected by a 9 A fuse. What is the maximum number of 60 watt bulbs that can be turned on in parallel?
1 11
2 22
3 33
4 44
Explanation:
C Total power, $\mathrm{P}=\mathrm{V} \times \mathrm{I}=200 \times 9$ Number of $60 \mathrm{~W}$ bulb required, $\mathrm{n}=\frac{220 \times 9}{60}=33$
NDA (II) 2009
Current Electricity
152711
How many sixty watt $(60 \mathrm{~W})$ bulbs may be safely used in a 240-V supply with 4-ampere fuse?
1 4
2 8
3 12
4 16
Explanation:
D Given, $\mathrm{P}=60 \mathrm{~W}, \mathrm{~V}=240 \mathrm{~V}, \mathrm{I}=4 \mathrm{~A}$ Total power, $\mathrm{P}=\mathrm{VI}=240 \times 4=960 \mathrm{~W}$ Number of $60 \mathrm{~W}$ bulb required $\mathrm{n}=\frac{960}{60}=16$
NDA (I) 2010
Current Electricity
152714
An electric heater is rated 1500 watt. If electric power costs Rs. 2 per kilo-watt hour, then the cost of power for $\mathbf{1 0}$ hours running of the heater is
1 ₹ 30
2 ₹ 15
3 ₹ 150
4 ₹ 25
Explanation:
A Given, $\mathrm{P}=1500 \mathrm{~W}$, Rs. $=2$ per kilo-watt, time $=10$ hours Energy consume $=\mathrm{P} \times \mathrm{t}=1500 \times 10$ $\mathrm{H}=15$ kilo-watt-hour Total cost $=2 \times 15=30$ rupees
NDA (II) 2013
Current Electricity
152718
$1 \mathrm{kWh}$ is equal to
1 $3.6 \times 10^{6} \mathrm{~J}$
2 $3.6 \times 10^{4} \mathrm{~J}$
3 $3.6 \times 10^{3} \mathrm{~J}$
4 $6 \times 10^{-4} \mathrm{~J}$
Explanation:
A Kilo-watt hour is a unit of energy used to denote the energy consumption of large machinery and the electricity consumption in homes and industries. It is also a common unit representation in electrical power engineering. One kilo watt Mean 1000 watts of power and 1 hour contains 3600 second in it. So, if we try to express one kilo-watt hour is terms of joules we get, $\mathrm{I} \mathrm{kW} \mathrm{h}=(1000 \mathrm{~W}) \times(3600 \mathrm{sec})$ $1 \mathrm{~kW} \mathrm{~h}=3.6 \times 10^{6} \mathrm{~J}$
