Question

A 20.0- $\Omega$ lamp and a 5.0- $\Omega$ 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?

Answer 1

a. the equivalent resistance of the circuit?

20.0 $\Omega$ + 5.0 $\Omega$ = 25.0 $\Omega$

b. the current in the circuit?

I = V / R = 50V / 25.0 $\Omega$ = 2.00 A

c. the voltage drop across each lamp?

V = IR = (2.00 A)(20.0 $\Omega$ ) = 4.00 x 10¹ V

V = IR = (2.00 A)(5.0 $\Omega$ ) = 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

Answer 2

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