An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery (Fig. 12.9). Calculate

(a) the total resistance of the circuit,

(b) the current through the circuit, and

(c) the potential difference across the electric lamp and conductor.

Slide17.JPG

Slide18.JPG
Slide19.JPG Slide20.JPG Slide21.JPG

  1. Class 10
  2. Chapter 12 Class 10 - Electricity

Transcript

Example 12.7 An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery (Fig. 12.9). Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor. Resistance of electric lamp = R1 = 20 Ω Resistance of conductor = R2 = 4 Ω Potential difference = V = 6 V Finding total Resistance (R) Since the resistors are connected in series R = R1 + R2 R = 20 + 4 R = 24 Ω (b) Finding current (I) By ohm’s law, V = I R 𝑉/𝑅 = I I = 𝑉/𝑅 I = 6/24 I = 1/4 I = 0.25 A (c) Finding potential difference across lamp and conductor We know that, In series connection Current remains same across all resistors Potential difference is different across all resistors By ohm’s law, V = I R Electric lamp (R1 = 20 Ω) V1 = I R1 V1 = 0.25 × 20 = 25/100 × 20 = 1/4 × 20 = 5 V Electric lamp (R2 = 4 Ω) V2 = I R2 V2 = 0.25 × 4 = 25/100 × 4 = 1/4 × 4 = 1 V Thus, Total resistance of circuit is 24 Ω Current flowing the circuit is 0.25 A Potential difference across Electric lamp is 5V Conductor is 1 V

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.