Does high resistance mean low current?

From Ohm’s law:

The relationship between voltage, current, and resistance: V = IR

The current is inversely proportional to the resistance. This means high resistance will cause the current to be low.


Related Questions

  1. What Is The Formula Of Ammeter?
  2. Why is the series arrangement not used for domestic circuits?
  3. What is the necessary condition for a conductor to obey Ohm's Law?
  4. Define: 1 volt, Potential difference
  5. Draw a circuit diagram to verify ohm’s law.
  6. Is resistance constant in Ohm's law?
  7. The resistance R= V/I, where V=100 ± 5.0V and I = 10 ± 0.2A. What is the total error in R?
  8. What are the applications of ohm's law used in daily life?
  9. Why is the curve representing Ohm's law linear?
  10. What are ohmic and non-ohmic devices? Give examples.
  11. State And Explain Joules Law Of Heating.
  12. Obtain the equation J = σE of Ohm's law on the basis of drift velocity
  13. A small bulb has a resistance of 2 ohms when it is cold. It draws 0.4-ampere current from a source of4V and then starts glowing. Calculate the resistance when it is glowing
  14. Find the current through a resistance 2 ohm if the voltage across the resistance is 6 V.
  15. Why Do We Use Ohm's Law?

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CBSE CLASS XII Related Questions

  • 1.
    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.


      • 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.
          Capacitors are manufactured with certain standard capacitances and working voltages. However, these standard values may not be the ones that are actually needed in a particular application. Two or more capacitors can be grouped in series or in parallel to achieve desired capacitance and voltage. When connected in series, the total capacitance decreases while the voltage rating increases, whereas in parallel connections, the total capacitance increases and maintains the same voltage rating. A capacitor stores energy in the electric field between its plates and stored energy is proportional to the square of the voltage and capacitance $U = \frac{1}{2}CV^2$, where symbols have their usual meanings.
          Two capacitors, one of $3 \ \mu$F and the other of $6 \ \mu$F, are connected in series in the circuit as shown in the figure, for a long time. }


            • 4.
              Consider a cylindrical conductor of length \( l \) and area of cross-section \( A \). Current \( I \) is maintained in the conductor and electrons drift with velocity \( \vec{v}_d \, (|\vec{v}_d| = \frac{eE}{m} \tau) \), where symbols have their usual meanings. Show that the conductivity of the material of the conductor is given by \[ \sigma = \frac{n e^2 \tau}{m}. \]


                • 5.
                  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.


                    • 6.
                      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.

                        CBSE CLASS XII Previous Year Papers

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