In a graph between current i and voltage v, find the portion corresponding to negative resistance.

Considering that the temperature and physical conditions are constant, Ohm’s law claims that the flowing current via any conductor is directly proportional to the potential difference across the ends. Ohm’s law can be described mathematically as:

V = IR

Here,

V = Voltage
I = Current
R = Resistance

Solution of the Above Equation

Thus, Negative resistance = – (V/I) = [V/(– I)]

Therefore, the part is CD where resistance is negative.

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In a graph between current i and voltage v, find the portion corresponding to negative resistance.

Solution of the Above Equation

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1.
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

2.
A magnet held vertically, with its north pole down, is dropped along the axis of a closed solenoid placed vertically on a table. If the observer looks down from the top,

3.
A current of \(4.0\,\text{A}\) flows through a wire of length \(1\,\text{m}\) and cross-sectional area \(1.0\,\text{mm}^2\), when a potential difference of \(2\,\text{V}\) is applied across its ends. Calculate the resistivity of the material of the wire.

4.
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:

5.
An electron is revolving around a proton in a circular orbit of radius $r$. Using classical physics, prove that the ratio of value of the angular momentum and the value of magnetic moment of electron revolving in the orbit is a constant.

Similar Physics Concepts

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In a graph between current i and voltage v, find the portion corresponding to negative resistance.

Solution of the Above Equation

Related Questions

CBSE CLASS XII Related Questions

1.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

2.A magnet held vertically, with its north pole down, is dropped along the axis of a closed solenoid placed vertically on a table. If the observer looks down from the top,

3.A current of \(4.0\,\text{A}\) flows through a wire of length \(1\,\text{m}\) and cross-sectional area \(1.0\,\text{mm}^2\), when a potential difference of \(2\,\text{V}\) is applied across its ends. Calculate the resistivity of the material of the wire.

4.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:

5.An electron is revolving around a proton in a circular orbit of radius $r$. Using classical physics, prove that the ratio of value of the angular momentum and the value of magnetic moment of electron revolving in the orbit is a constant.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
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

2.
A magnet held vertically, with its north pole down, is dropped along the axis of a closed solenoid placed vertically on a table. If the observer looks down from the top,

3.
A current of \(4.0\,\text{A}\) flows through a wire of length \(1\,\text{m}\) and cross-sectional area \(1.0\,\text{mm}^2\), when a potential difference of \(2\,\text{V}\) is applied across its ends. Calculate the resistivity of the material of the wire.

4.
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:

5.
An electron is revolving around a proton in a circular orbit of radius $r$. Using classical physics, prove that the ratio of value of the angular momentum and the value of magnetic moment of electron revolving in the orbit is a constant.