Three incandescent bulbs (each 100 W) are attached in series. In another circuit, three more bulbs of same wattage are attached parallelly to an equal source. So, give reasons for the following.

A circuit can only be connected in series if same amount of current passes via the resistors. In every circuit, the amount of voltage across the resistors will be different. Simply put, in a series combination, if even one of the resistors is broken or turns out faulty, then the entire circuit turns off.

Will the bulb in the two circuits have the same brightness?

The bulbs’ resistances in series combination will have to be three times the resistance of one bulb alone. Therefore, the current that passes in the series combination will have to be about one-third in opposition to the current in each bulb in a parallel combination. Hence, it can be said that in parallel combination, the bulbs will glow brighter.

With one bulb fused, will the rest continue to glow in each circuit? 

In series combination, the bulbs will gradually stop glowing due to the broken or faulty circuit and will have zero current. However, if the bulbs continue in parallel combination, then they will glow with the same brightness as earlier.


Related Questions

  1. In An Arrangement Of Resistances, Find Effective Resistance Between Points A and B.
  2. What Is Effective Resistance?
  3. What is Null Voltage?
  4. X and Y, which are two resistors, with resistances of 2 Ω and 3 Ω respectively are first connected in parallel and then in series. In both cases, the voltage that is supplied is 5 V. (i) Illustrate a circuit diagram to show the combination of resistors. (ii) Calculate the amount of voltage across 3 Ω resistor in the series combination of resistors.
  5. Draw An Electric Circuit With A Cell, Key, Ammeter, A Resistor (Series) Of 2 Ohm With Combination Of Two Resistors (4 Ohm Each) In Parallel And A Voltmeter Across Parallel Combination.
  6. Two Identical Resistors With Resistances 15 Ohm Are Connected In Series And Parallel To A Battery Of 6 V. Calculate Ratio Of Power Consumed.
  7. A Circuit Consists Of A Battery Of 3 Cells (2 V Each), A Combination Of Three Resistors, 10 Ohm, 20 Ohm And 30 Ohm, Attached Parallelly, With Plug Key And Ammeter (In Series).

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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.
      • Both spheres attain the same potential.
      • The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2)}{(r_1 + r_2)}$
      • The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2) (r_1 + r_2)}{r_1 r_2}$

    • 2.
      Consider the nuclear reaction \( X \to Y + Z \). Let \( M_x \), \( M_y \), and \( M_z \) be the masses of the three nuclei X, Y, and Z respectively. Then which of the following relations hold true?

        • \( (M_x - M_z)<M_y \)
        • \( (M_x - M_y)<M_z \)
        • \( M_x>(M_y + M_z) \)
        • \( M_x<(M_y + M_z) \)

      • 3.
        Two air-filled capacitors of capacitances $C_1$ and $C_2$ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.


          • 4.
            An electric field $\vec{E}$ is established across the ends of a cylindrical conductor of length L and area of cross-section A. Discuss how electrons attain an average velocity, independent of time. Hence, obtain a relation between current in the conductor and this ‘average velocity’ of electrons.


              • 5.
                This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?


                  • 6.
                    Read the following paragraph and answer the questions that follow.
                    A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. The motion of majority charge carriers causes diffusion current across the junction while the barrier electric field causes motion of minority carriers for drift current. In case of unbiased diode, the diffusion and drift currents are equal. This equilibrium is disturbed by the biasing batteries. Diodes, therefore, allow currents in one direction. This property of diode is used in making rectifiers.

                      CBSE CLASS XII Previous Year Papers

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