A 20.0- lamp and a 5.0- lamp are connected in series and placed across a potential difference of 50.0 V. What is
a. the equivalent resistance of the circuit?
b. the current in the circuit?
c. the voltage drop across each lamp?
d.the power dissipated in each lamp?
a. the equivalent resistance of the circuit?
20.0 + 5.0 = 25.0
b. the current in the circuit?
I = V / R = 50V / 25.0 = 2.00 A
c. the voltage drop across each lamp?
V = IR = (2.00 A)(20.0 ) = 4.00 x 101 V
V = IR = (2.00 A)(5.0 ) = 1.0 x 101 V
d.the power dissipated in each lamp?
P = IV = (2.00 A)(4.00 x 101 V) = 8.00 x 101 W
P = IV = (2.00 A)(1.0 x 101 V) = 2.0 x 101 W
R equivalent= 20+5 ohm= 25 ohm
Current = V/R= 50/25=2amp
Voltage drop across 20 ohm= 50*(5/25)=10V
Voltage drop across 5 ohm= 50-10= 40 V
Power in 20 ohm= VI= 10*2= 20watt
Power in 5 ohm= VI= 40*2=80 watt
A 20.0- lamp and a 5.0- lamp are connected in series and placed across a potential...
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