Question

A 20 ohm lamp and a 5 ohm lamp are connected in series and placed across a potential difference of 50 V.

1. What is the equivalent resistance of the circuit?
2. What is the voltage drop across each lamp?
3. What is the power dissipated in each lamp

Answer 1

Solution:

Resistance of first lamp, R₁ = 20 Ω

Resistance of second lamp, R₂ = 5 Ω

Potential difference, V = 50 V

(a) Let the equivalent resistance of the circuit be R.

The lamps are connected in series,

∴ R = R₁ + R₂

  R = 20 Ω + 5 Ω = 25 Ω

Equivalent resistance of the circuit = R = 25 Ω

(b) Let the current through the circuit be I.

As we know, V = I R

I = V/R

I = (50/25) A

I = 2 A

The current flowing through each lamp is 2 amperes. (It is same for both lamps because they are connected in series).

Now,

Voltage drop across 20 ohm lamp, V₁ = I x R₁ = 2 x 20

                                                          = 40 V

Voltage drop across 5 ohm lamp, V₂ = I x R₁ = 2 x 5

                                                          = 10 V

(c) Power dissipated in 20 ohm lamp, P₁ = IV₁

                                                                    = 2 x 40 = 80 W

Power dissipated in 5 ohm lamp, P₂ = IV₂

                                                            = 2 x 10 = 20 W

#SPJ2

Answer 2

Given: $R_{1}$=20Ω

$R_{2}$=5Ω

V=50 V

Find: Answer the following questions.

Solution:

1) Equivalent resistance= $R_{1}+R_{2}$

                                      =20+5

                                       = 25Ω

2) Current I =V/R

                   =50/25

                   =2A

V1=IX$R_{1}$

   = 2X20

   = 40V

V2=IX$R_{2}$

    = 2X5

     = 10V

3) P1=IXV1

       =2X40

       =80W

P2=IXV2

    =2X10

     =20W

Final answer: 1) 25Ω

                       2) V1=40V

                           V2=10V

                       3) P1=80W

                           P2=20W

#SPJ3