Power W = IV
first using the power formula find the current. in series connection current is common at any point.
60 = I(25)
I = 60/25
I = 2.4A
Voltage across Resistor is
( total voltage=120 v; across the lamp=25; here the assumption is source and resistor and lamp is connected in series)
therefore source voltage= voltage of resistor + voltage of lamp
voltage of source - voltage of lamp = voltage of resistor
120 - 25 = 95 V
so 95 v is the voltage across the resistor
now you know both resistor's voltage and current through it
now using ohms law find the resistance of that resistor
R = V/I
R = 95/2.4
R = 39.583 ohms
Power Of Lamp = 60 Watt
Means, the bulb consumes 60 J in 1 second.
Power = Voltage X Current
Current = Power/Voltage
= 60/25 A
= 2.4 A
You should know, the law of conservation of
voltage. It can niether be created nor be destroyed.
Since, of 120 V, 25 V is used by lamp, so voltage law
says voltage used by resistor is (120-25) V = 95 V
We know,
Resistance = Voltage/Current
= 95/2.4 Ohms
= 38.583 Ohms
--------
Additional Info :
Kirchoff's Voltage Law is the law of conservation of
energy.
Power W = IV
60 = I(25)
I = 60/25
I = 2.4A
Voltage across Resistor is
120 - 25 = 95 V
R = V/I
R = 95/2.4
R = 39.583 ohms
u should know that current is same in devices that are in series.
now .. power= voltage*current
60=25*current
current=2.4 A
now since these are in series , thus current will be the same in resistor also but the voltage will be divided between the lamp and resistor since they are in series.
hence, voltage across lamp=120-25 = 95V
therefore,by ohm's law , resistance=voltage/current
=95/2.4 = 39.583 ohm
I hope u'll understand .. GOOD LUCK!
Assume this bulb is rated 120V 60W
Lamp resistance = 120^2 / 60 = 240 ohm
current = 25 / 240 = 0.104A
Resistor = ( 120 - 25 ) / 0.104 = 913 ohm
A 60.0 watt lamp is placed in series with a resistor and a 120.0 V source. If the voltage across the lamp is 25 V,What is the resistance of the resistor?
