Check sibling questions

An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected to a 6 V battery as shown in the circuit. Calculate:

(a) the total resistance of the circuit.

(b) the current through the circuit.

(c) the potential difference across the

(i) electric lamp and

(ii) conductor, and

(d) power of the lamp.

An electric lamp of resistance - Teachoo.jpg

 

Answer:

Given,

  • Resistance of electric lamp, R 1 = 20 ohm
  • Resistance of conductor, R 2 = 4 ohm
  • Potential difference, V = 6 V

(a)

Since R 1 and R 2 are connected in series,

Total resistance, R = R 1 + R 2

                                  = 20 + 4

                                   = 24 ohm

Therefore, total resistance of the circuit is 24 ohm.

(b)

Let the current through the circuit be I.

According to ohm’s law,

V = I*R

I = V / R

  = 6 / 24

  = 0.25 A

Therefore, the current through the circuit is 0.25 A

(c)

(i) For electric lamp,

      V = I*R 1

         = 0.25 * 20

          = 5 V

Therefore, the potential difference across the electric lamp is 5 V.

(ii) For conductor,

       V = I*R 2

         = 0.25 * 4

          = 1 V

Therefore, the potential difference across the conductor is 1 V.

(d)

Let the power of the lamp be P.

We know that,

P = V*I

    = 5 * 0.25

     = 1.25 W

Therefore, power of the lamp is 1 . 25 W

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 12 years and a teacher from the past 16 years. He teaches Science, Economics, Accounting and English at Teachoo