How many 2,300-W appliances can be plugged into the sockets without blowing the fuse?

Question

All of the electrical outlets in a room are connected in a single parallel circuit. The circuit is equipped with a 20-A fuse, and the voltage is V = 120 V.

What is the maximum total power that can be supplied by the outlets without blowing the fuse?

How many 2,300-W appliances can be plugged into the sockets without blowing the fuse?

Answer 1

P = VI = 120* 20 = 2400 watt.

which is greater than 2300w.

Hence you cannot use more than one appliance.

Answer 2

ELECTRICITY: Wattage

Q: A parallel electrical circuit connects the electrical outlets located within a room. A 20-A fuse is put into place to protect the circuit from unexpected surges of current ( I ). The voltage drop across each circuit element is V = 120 V. What is the maximum power ( W ) output that can be provided by the outlets without blowing the fuse? How many 2,300-W appliances can be plugged into the sockets without blowing the fuse?

A: Power is the amount of energy in joules ( J ) provided per second ( s ) by an electrical device. The SI unit for Power ( P ) in ( J / s ) is Watts ( W ). Mathematically, power can be expressed in any number of ways: 

P = IV, where I is the current in Amperes ( A ), and the voltage ( V ) is expressed as the quantity of energy provided in Joules ( J ) per coulomb ( C ) of charge. A single electron has a charge of 1.602 x 10-19 C. A Coulomb of electrons equal to ( 1 / 1.602 x 10-19 C ) = 6.25 x 1018 e-. An ampere of electrical current is the number of coulombs per second that pass a point of reference within a circuit.

 

Since P = IV, and V = IR, power may also be expressed as P = ( I )( IR ) = ( I2 )( R ). Power losses as a function of current ( I ) are often calculated using this expression. Power may also be expressed as P = ( V / R )( V ) = ( V2/ R ).

 

The question states that V = 120 V, and a current of I = 20A will short circuit the fuse. Therefore, the maximum power cannot exceed P = ( I )( V ). P = ( 20A )( 120V ) = 2,400 W, or 2.4 kW. Therefore, only one 2,300 W ( 2.3 kW ) appliance can be plugged into a socket without blowing the fuse.

 

Note: A kilowatt-hour ( kWh ) is a unit of energy in joules ( J ) equal to one kilowatt of power sustained for an hour ( 3,600 s ). Energy ( J ) = ( W )( t ). Since wattage is expressed in ( J / s ), 1kWh = ( 1kW )( 3,600s ) = 3,600 kJ = 3.6 MJ.

 

Similarly, an electron-volt is a unit of energy in joules ( J ) required to move a single electron against a 1V electrical gradient. 1eV = ( 1V )( q ) = ( 1V )( 1.602 x 10-19 C ) = 1.602 x 10-19 J.