A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ohm when connected to a 10 V battery. Calculate the resistance of the electric lamp.

Now if the resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in the current flowing through the 5 Ω conductor and potential difference across the lamp will take place? Give reason.

Let R1 be the resistance of the electric lamp and R2 the resistance of a conductor.

In series, the total resistance R = R1 + R2 = 5 + R1

Current (I) = 1 A

Voltage (V) = 10 V

Using Ohm’s law

I = V/R

1 = 10/(5+R)

5 + R = 10

R = 5Ω

Now, a 10 Ω resistance is connected in parallel with the series combination. Therefore, the total resistance of the circuit is calculated as – 

1/Rp = 1/(R1 + 5) + 1/10

Rp = 5Ω

Therefore, the current flowing in the circuit is equal to I = V/R

I = 10/5

I = 2A

Read More: Ohm’s Law and its Limitations


Related Questions

  1. What is the necessary condition for a conductor to obey Ohm's Law?
  2. State Ohms Law. Express It Mathematically. Define Si Unit Of Resistance.
  3. How do you find the resistance to Ohm's law?
  4. Why is the curve representing Ohm's law linear?
  5. State Ohms law. How can it be verified experimentally?
  6. What are the 3 forms of Ohm's law
  7. What are the limitations of Ohm's Law?
  8. What are the applications of ohm's law used in daily life?
  9. Is resistance constant in Ohm's law?
  10. Draw a circuit diagram to verify ohm’s law.

Read More:

CBSE CLASS XII Related Questions

  • 1.
    Two metal spheres of radii $r_1$ and $r_2$ ($> r_1$) having charges $q_1$ and $q_2$ respectively kept in air, are brought in contact. Which of the following statements is not correct ?

      • The total charge of the two spheres is conserved.
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        • 3.
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            • 4.
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                • 5.
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                        CBSE CLASS XII Previous Year Papers

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