How do you find the resistance in Ohm's law?

According to Ohm’s law, the voltage across a conductor is directly proportional to the current that flows through it, provided the temperature and physical conditions remain constant.

Mathematically, Ohm’s law can be denoted as:

V = IR
In order to find the resistance, the formula can be rewritten as:

\(R = {V \over I}\)

Related Questions:

Read More:

Ohm’s Law – Important Topics
Current Electricity	Difference Between Primary Cell and Secondary Cell	Types of Resistors
Wheatstone Bridge	Relation between kva and kw	Combination of Resistors
Potentiometer	Cells, emf, Internal Resistance	Unit of Specific Resistance
Static Electricity	Difference between AC and DC	Resistor Colour Codes
Ampere	Cells in Series and in Parallel	Permittivity And Permeability
Electric Dipole	Resistivity Temperature Dependence	Unit of Current
Meter Bridge	Difference between rtd and thermocouple	Electromotive Force
Kirchhoff’s Rules	Difference between air conditioning and refrigeration	Internal resistance formula
Types of Current	Relation between resistance and length	Derivation of Drift Velocity
Current Density	Power Of Alternating Current	Difference between kva and kw
Drift Velocity	Differences Between Acceleration and Velocity	Electric Current Formula
Resistivity formula	Relation between power and resistance	Current density formula
Carbon Resistor	Difference between earth and neutral	Resistance Formula
Uses of Resistor	Difference between emf and voltage	Circuit Diagram
Power transformers	Current Electricity NCERT Solutions	Voltage divider formula

CBSE CLASS XII Related Questions

1.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

\( \frac{q}{4 \pi \epsilon_0 l^2} \) pointing along MA
\( \frac{q}{\pi \epsilon_0 l^2} \) pointing along AM
\( \frac{q}{2 \pi \epsilon_0 l^2} \) pointing along AM
Zero

View Solution

2.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

View Solution

3.
Two parallel plate capacitors X and Y are connected in series to a 6 V battery. They have the same plate area and same plate separation but capacitor X has air between its plates, whereas capacitor Y contains a material of dielectric constant 4. Calculate the capacitances of X and Y, if the equivalent capacitance of the combination of X and Y is \( 4 \, \mu\text{F} \). Calculate the potential difference across the plates of X and Y.

View Solution

4.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

\((P_1+P_2)\)
\(\dfrac{P_1+P_2}{2}\)
\(\dfrac{P_1P_2}{2(P_1+P_2)}\)
\(\dfrac{P_1P_2}{4(P_1+P_2)}\)

View Solution

5.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

View Solution

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° in 0.2 s. The value of emf induced in the loop will be:

5 V
3.5 V
2.5 V
Zero V

View Solution

How do you find the resistance in Ohm's law?

CBSE CLASS XII Related Questions

1.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

2.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

4.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

5.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

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° in 0.2 s. The value of emf induced in the loop will be:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

How do you find the resistance in Ohm's law?

CBSE CLASS XII Related Questions

1.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

2.A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

4.Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

5. Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

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° in 0.2 s. The value of emf induced in the loop will be:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

2.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

4.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

5.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

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° in 0.2 s. The value of emf induced in the loop will be: