What are the 3 forms of Ohm's law?

There are three forms of Ohm’s law that relate to voltage, current, and resistance. For a constant flow of current, the current is given by the ratio of voltage across terminals to the resistance of the resistor.

I = V/R

Ohm’s law can also be written as

R = V/I

Or

V = IR

  • I is the current
  • V is the voltage
  • R is the resistance

Related Questions:

  1. What is the necessary condition for a conductor to obey Ohm's Law?
  2. How do you find the resistance in Ohm's law?
  3. Is resistance constant in Ohm's law?
  4. Why is the curve representing Ohm's law linear?

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CBSE CLASS XII Related Questions

  • 1.

    Find the ratio $\dfrac{\lambda_\alpha}{\lambda_p}$ of the de Broglie wavelength $\lambda_\alpha$ associated with an alpha particle to the de Broglie wavelength $\lambda_p$ associated with a proton if both are moving with the same velocity.


      • 2.

        A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.


          • 3.
            In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

              • \(10\ \text{N/C}\) along \(+x\)
              • \(10\ \text{N/C}\) along \(-x\)
              • \(50\ \text{N/C}\) along \(+x\)
              • \(50\ \text{N/C}\) along \(-x\)

            • 4.

              A parallel beam of monochromatic light falls normally on a single slit of width $a$ and a diffraction pattern is observed on a screen placed at distance $D$ from the slit. Explain:

              (I) the formation of maxima and minima in the diffraction pattern, and

              (II) why the maxima go on becoming weaker and weaker with its increasing number $n$.


                • 5.
                  Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

                    • \(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (\hat{i} - \hat{j})\)
                    • \(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (-\hat{i} - \hat{j})\)
                    • \(\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (\hat{i} - \hat{j})\)
                    • \(\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (-\hat{i} - \hat{j})\)

                  • 6.

                    A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

                      • 5 V
                      • 3.5 V
                      • 2.5 V
                      • Zero V
                    CBSE CLASS XII Previous Year Papers

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