Resistance in meter bridge’s two arms are 5 ohms and R ohms. When resistance R is shunted with equal resistance, new balance point becomes 1.6l1. Calculate R.

Resistance can be defined as the measurement of opposition of current flow in an electrical circuit. Therefore, in order to determine the value of resistance (R), we need to:

  • Known resistance, R = 5Ω
  • Initial Unknown Resistance = RΩ
  • New Balancing Point, L = 1.6l1

Thus, R/L = S/100 − L … [1]

Here,

  • R = Known Resistance
  • S = Unknown Resistance

In the first stage, the balancing point is at l1.

Now, after replacing the values as:

R = 5Ω and L = l1 in equation [1], we can obtain,

⇒ 5/l1 = S/100−l1 … [2]

Now, after resistance R is shunted with equal resistance R, the resistance in that arm can be R2

Then,

⇒ 5L = R/2/100−L … [3]

Replace the value of L1 in equation 3. We can obtain,

⇒ 5/1.6l1 = R/2/100−1.6l1 … [4]

Hence, divide the given equation 2 by 4, in order to get,

1.6/2 = (100 − 1.6l1)/100 − l1

⇒ 160 − 1.6l1 = 200 − 3.2l1

⇒ 1.6l= 40

⇒ l= 401.6

= 25 cm

Replace l1=25cm in the above equation 1, we get,

5/25 = R/100 − 25

 R = 15Ω


Related Questions

  1. If R, C And L Are Fundamental Quantities In A Circuit Like Resistance, Capacitance And Inductance In W, Then Find Dimensional Formula For Resistance And Capacitance.
  2. In A Graph Between Current I And Voltage V, Find The Portion Corresponding To Negative Resistance.
  3. A Closed Coil Has 500 Turns Across Rectangular Frame Of Area 4.0 Cm2 With Resistance Of 500 Ohms. The Coil Is Plane Perpendicular To A Uniform Magnetic Field Of 0.2wb/M2. Find Amount Of Charge Through Coil If Turned Over (180 Degrees Rotation).
  4. 1.0 M Rectangular Loop With A Sliding Connector Is In Uniform Magnetic Field 2t Perpendicular To Plane Of Loop. Resistance Is 2 Ohms. Two Resistances, 6 Ohms And 3 Ohms, Are Connected.
  5. 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.
  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. For The Resistor Combination, Find The Equivalent Resistance Between M and N.
  8. Two Concentric Coplanar Circular Loops Of Wire (Resistance Per Unit Length 10 4 Ohms M-1) Have Diameters 0.2 M And 2 M. With Time-Varying Potential Difference Of (4+2.5t) Applied To Larger Loop, Find Current In Smaller One.
  9. 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).

Read More:

CBSE CLASS XII Related Questions

  • 1.
    A charged particle $+q$ in an electric field $\vec{E}$ experiences a force in the direction of the electric field. As a result, its kinetic energy changes. Similarly, the charged particle also experiences a force when it moves in a magnetic field $\vec{B}$. But this magnetic force is perpendicular to both velocity $\vec{v}$ of the charged particle and the magnetic field $\vec{B}$, so it cannot change the kinetic energy of the charged particle. Consider two charged particles 1 and 2 of masses $m$ and $\frac{m}{2}$ having charges $-q$ and $+2q$ respectively. They are accelerated from rest through the same potential difference $V$ and acquire kinetic energy $K_1$ and $K_2$. Then they enter in a region of uniform magnetic field $\vec{B}$ perpendicular to their velocities.


      • 2.
        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 $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2) (r_1 + r_2)}{r_1 r_2}$

        • 3.
          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.


            • 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.
                      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.

                        CBSE CLASS XII Previous Year Papers

                        Comments


                        No Comments To Show