357465
For a metallic wire, the ratio \(\dfrac{V}{i}(V=\) applied potential difference and \(i=\) current flowing) is
1 independent of temperature
2 increases as the temperature rises
3 decreases as the temperature rises
4 increases or decreases as temperature rises depending upon the metal
Explanation:
The resistance of a metallic wire at temperature \(t^{\circ} \mathrm{C}\) is given by \(R_{t}=R_{0}(1+\alpha t)\) where, \(\alpha\) is the coefficient of expansion. The resistance of wire increases on increasing the temperature. Also, from ohm's law, the ratio of \(V / i\) is equal to \(R\). Hence, on increasing the temperature, the resistance of metallic wire is increased, so the ratio \(V / i\) also increases.
PHXII03:CURRENT ELECTRICITY
357466
Statement A : Ohm’s law is not valid if current depends on voltage non-linearly. Statement B : Ohm’s law only holds if the temperature is maintained as constant.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357467
For a wire \(\dfrac{R}{l}=\dfrac{1}{2}\) and length of wire is \(l = 5\;cm\). If potential difference of \(1\,V\) is applied across it, then current through wire will be
1 \(40\,A\)
2 \(4\,A\)
3 \(25\,A\)
4 \(2.5\,A\)
Explanation:
Given, for a wire, \(\dfrac{R}{l}=\dfrac{1}{2}\) Length of wire, \(l = 5\;cm = 5 \times {10^{ - 2}}\;m\) \(\therefore R=\dfrac{l}{2}=2.5 \times 10^{-2} \Omega\) Potential difference, \(V=1 V\) or IR \(=1\) \(I=\dfrac{1}{R}=\dfrac{1}{2.5 \times 10^{-2}}=\dfrac{100}{2.5}=40 \mathrm{~A}\) So, correct option is (1)
AIIMS - 2019
PHXII03:CURRENT ELECTRICITY
357468
\(I\)-\(V\) characteristic of a copper wire of length \(L\) and area of cross-section \(A\) is shown in figure. The slope of the curve becomes
1 Less if the area of the wire is increased
2 Less if the length of the wire is increased
3 More if experiment is performed at higher temperature
4 More if a wire of steel of same dimension is used
Explanation:
Here, slope of the curve,\(\frac{{\Delta I}}{{\Delta V}} = \frac{1}{R} = \frac{A}{{\rho l}}\) \(\therefore \) With increase in length (\(l\)), slope decreases. \(\therefore \) (2) is the only correct option.
KCET - 2020
PHXII03:CURRENT ELECTRICITY
357469
\({\vec{E}}\) is the electric field inside a conductor whose material has conductivity \({\sigma}\) and resistivity \({\rho}\). The current density inside the conductor is \({\vec{j}}\). The correct form of \(Ohm's\) law is
1 \({\vec{E}=\sigma \vec{j}}\)
2 \({\vec{j}=\rho \vec{E}}\)
3 \({\vec{E}=\rho \vec{j}}\)
4 \({\vec{E} \cdot \vec{j}=\rho}\)
Explanation:
Ohm's law \({V=i R}\), and \({R=\rho \dfrac{l}{A}}\) \({\therefore V=i \rho \dfrac{l}{A}}\) \({\Rightarrow \dfrac{V}{l}=\dfrac{i}{A} \rho}\) \({\therefore \vec{E}=\rho \vec{j}}\).
357465
For a metallic wire, the ratio \(\dfrac{V}{i}(V=\) applied potential difference and \(i=\) current flowing) is
1 independent of temperature
2 increases as the temperature rises
3 decreases as the temperature rises
4 increases or decreases as temperature rises depending upon the metal
Explanation:
The resistance of a metallic wire at temperature \(t^{\circ} \mathrm{C}\) is given by \(R_{t}=R_{0}(1+\alpha t)\) where, \(\alpha\) is the coefficient of expansion. The resistance of wire increases on increasing the temperature. Also, from ohm's law, the ratio of \(V / i\) is equal to \(R\). Hence, on increasing the temperature, the resistance of metallic wire is increased, so the ratio \(V / i\) also increases.
PHXII03:CURRENT ELECTRICITY
357466
Statement A : Ohm’s law is not valid if current depends on voltage non-linearly. Statement B : Ohm’s law only holds if the temperature is maintained as constant.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357467
For a wire \(\dfrac{R}{l}=\dfrac{1}{2}\) and length of wire is \(l = 5\;cm\). If potential difference of \(1\,V\) is applied across it, then current through wire will be
1 \(40\,A\)
2 \(4\,A\)
3 \(25\,A\)
4 \(2.5\,A\)
Explanation:
Given, for a wire, \(\dfrac{R}{l}=\dfrac{1}{2}\) Length of wire, \(l = 5\;cm = 5 \times {10^{ - 2}}\;m\) \(\therefore R=\dfrac{l}{2}=2.5 \times 10^{-2} \Omega\) Potential difference, \(V=1 V\) or IR \(=1\) \(I=\dfrac{1}{R}=\dfrac{1}{2.5 \times 10^{-2}}=\dfrac{100}{2.5}=40 \mathrm{~A}\) So, correct option is (1)
AIIMS - 2019
PHXII03:CURRENT ELECTRICITY
357468
\(I\)-\(V\) characteristic of a copper wire of length \(L\) and area of cross-section \(A\) is shown in figure. The slope of the curve becomes
1 Less if the area of the wire is increased
2 Less if the length of the wire is increased
3 More if experiment is performed at higher temperature
4 More if a wire of steel of same dimension is used
Explanation:
Here, slope of the curve,\(\frac{{\Delta I}}{{\Delta V}} = \frac{1}{R} = \frac{A}{{\rho l}}\) \(\therefore \) With increase in length (\(l\)), slope decreases. \(\therefore \) (2) is the only correct option.
KCET - 2020
PHXII03:CURRENT ELECTRICITY
357469
\({\vec{E}}\) is the electric field inside a conductor whose material has conductivity \({\sigma}\) and resistivity \({\rho}\). The current density inside the conductor is \({\vec{j}}\). The correct form of \(Ohm's\) law is
1 \({\vec{E}=\sigma \vec{j}}\)
2 \({\vec{j}=\rho \vec{E}}\)
3 \({\vec{E}=\rho \vec{j}}\)
4 \({\vec{E} \cdot \vec{j}=\rho}\)
Explanation:
Ohm's law \({V=i R}\), and \({R=\rho \dfrac{l}{A}}\) \({\therefore V=i \rho \dfrac{l}{A}}\) \({\Rightarrow \dfrac{V}{l}=\dfrac{i}{A} \rho}\) \({\therefore \vec{E}=\rho \vec{j}}\).
357465
For a metallic wire, the ratio \(\dfrac{V}{i}(V=\) applied potential difference and \(i=\) current flowing) is
1 independent of temperature
2 increases as the temperature rises
3 decreases as the temperature rises
4 increases or decreases as temperature rises depending upon the metal
Explanation:
The resistance of a metallic wire at temperature \(t^{\circ} \mathrm{C}\) is given by \(R_{t}=R_{0}(1+\alpha t)\) where, \(\alpha\) is the coefficient of expansion. The resistance of wire increases on increasing the temperature. Also, from ohm's law, the ratio of \(V / i\) is equal to \(R\). Hence, on increasing the temperature, the resistance of metallic wire is increased, so the ratio \(V / i\) also increases.
PHXII03:CURRENT ELECTRICITY
357466
Statement A : Ohm’s law is not valid if current depends on voltage non-linearly. Statement B : Ohm’s law only holds if the temperature is maintained as constant.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357467
For a wire \(\dfrac{R}{l}=\dfrac{1}{2}\) and length of wire is \(l = 5\;cm\). If potential difference of \(1\,V\) is applied across it, then current through wire will be
1 \(40\,A\)
2 \(4\,A\)
3 \(25\,A\)
4 \(2.5\,A\)
Explanation:
Given, for a wire, \(\dfrac{R}{l}=\dfrac{1}{2}\) Length of wire, \(l = 5\;cm = 5 \times {10^{ - 2}}\;m\) \(\therefore R=\dfrac{l}{2}=2.5 \times 10^{-2} \Omega\) Potential difference, \(V=1 V\) or IR \(=1\) \(I=\dfrac{1}{R}=\dfrac{1}{2.5 \times 10^{-2}}=\dfrac{100}{2.5}=40 \mathrm{~A}\) So, correct option is (1)
AIIMS - 2019
PHXII03:CURRENT ELECTRICITY
357468
\(I\)-\(V\) characteristic of a copper wire of length \(L\) and area of cross-section \(A\) is shown in figure. The slope of the curve becomes
1 Less if the area of the wire is increased
2 Less if the length of the wire is increased
3 More if experiment is performed at higher temperature
4 More if a wire of steel of same dimension is used
Explanation:
Here, slope of the curve,\(\frac{{\Delta I}}{{\Delta V}} = \frac{1}{R} = \frac{A}{{\rho l}}\) \(\therefore \) With increase in length (\(l\)), slope decreases. \(\therefore \) (2) is the only correct option.
KCET - 2020
PHXII03:CURRENT ELECTRICITY
357469
\({\vec{E}}\) is the electric field inside a conductor whose material has conductivity \({\sigma}\) and resistivity \({\rho}\). The current density inside the conductor is \({\vec{j}}\). The correct form of \(Ohm's\) law is
1 \({\vec{E}=\sigma \vec{j}}\)
2 \({\vec{j}=\rho \vec{E}}\)
3 \({\vec{E}=\rho \vec{j}}\)
4 \({\vec{E} \cdot \vec{j}=\rho}\)
Explanation:
Ohm's law \({V=i R}\), and \({R=\rho \dfrac{l}{A}}\) \({\therefore V=i \rho \dfrac{l}{A}}\) \({\Rightarrow \dfrac{V}{l}=\dfrac{i}{A} \rho}\) \({\therefore \vec{E}=\rho \vec{j}}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
357465
For a metallic wire, the ratio \(\dfrac{V}{i}(V=\) applied potential difference and \(i=\) current flowing) is
1 independent of temperature
2 increases as the temperature rises
3 decreases as the temperature rises
4 increases or decreases as temperature rises depending upon the metal
Explanation:
The resistance of a metallic wire at temperature \(t^{\circ} \mathrm{C}\) is given by \(R_{t}=R_{0}(1+\alpha t)\) where, \(\alpha\) is the coefficient of expansion. The resistance of wire increases on increasing the temperature. Also, from ohm's law, the ratio of \(V / i\) is equal to \(R\). Hence, on increasing the temperature, the resistance of metallic wire is increased, so the ratio \(V / i\) also increases.
PHXII03:CURRENT ELECTRICITY
357466
Statement A : Ohm’s law is not valid if current depends on voltage non-linearly. Statement B : Ohm’s law only holds if the temperature is maintained as constant.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357467
For a wire \(\dfrac{R}{l}=\dfrac{1}{2}\) and length of wire is \(l = 5\;cm\). If potential difference of \(1\,V\) is applied across it, then current through wire will be
1 \(40\,A\)
2 \(4\,A\)
3 \(25\,A\)
4 \(2.5\,A\)
Explanation:
Given, for a wire, \(\dfrac{R}{l}=\dfrac{1}{2}\) Length of wire, \(l = 5\;cm = 5 \times {10^{ - 2}}\;m\) \(\therefore R=\dfrac{l}{2}=2.5 \times 10^{-2} \Omega\) Potential difference, \(V=1 V\) or IR \(=1\) \(I=\dfrac{1}{R}=\dfrac{1}{2.5 \times 10^{-2}}=\dfrac{100}{2.5}=40 \mathrm{~A}\) So, correct option is (1)
AIIMS - 2019
PHXII03:CURRENT ELECTRICITY
357468
\(I\)-\(V\) characteristic of a copper wire of length \(L\) and area of cross-section \(A\) is shown in figure. The slope of the curve becomes
1 Less if the area of the wire is increased
2 Less if the length of the wire is increased
3 More if experiment is performed at higher temperature
4 More if a wire of steel of same dimension is used
Explanation:
Here, slope of the curve,\(\frac{{\Delta I}}{{\Delta V}} = \frac{1}{R} = \frac{A}{{\rho l}}\) \(\therefore \) With increase in length (\(l\)), slope decreases. \(\therefore \) (2) is the only correct option.
KCET - 2020
PHXII03:CURRENT ELECTRICITY
357469
\({\vec{E}}\) is the electric field inside a conductor whose material has conductivity \({\sigma}\) and resistivity \({\rho}\). The current density inside the conductor is \({\vec{j}}\). The correct form of \(Ohm's\) law is
1 \({\vec{E}=\sigma \vec{j}}\)
2 \({\vec{j}=\rho \vec{E}}\)
3 \({\vec{E}=\rho \vec{j}}\)
4 \({\vec{E} \cdot \vec{j}=\rho}\)
Explanation:
Ohm's law \({V=i R}\), and \({R=\rho \dfrac{l}{A}}\) \({\therefore V=i \rho \dfrac{l}{A}}\) \({\Rightarrow \dfrac{V}{l}=\dfrac{i}{A} \rho}\) \({\therefore \vec{E}=\rho \vec{j}}\).
357465
For a metallic wire, the ratio \(\dfrac{V}{i}(V=\) applied potential difference and \(i=\) current flowing) is
1 independent of temperature
2 increases as the temperature rises
3 decreases as the temperature rises
4 increases or decreases as temperature rises depending upon the metal
Explanation:
The resistance of a metallic wire at temperature \(t^{\circ} \mathrm{C}\) is given by \(R_{t}=R_{0}(1+\alpha t)\) where, \(\alpha\) is the coefficient of expansion. The resistance of wire increases on increasing the temperature. Also, from ohm's law, the ratio of \(V / i\) is equal to \(R\). Hence, on increasing the temperature, the resistance of metallic wire is increased, so the ratio \(V / i\) also increases.
PHXII03:CURRENT ELECTRICITY
357466
Statement A : Ohm’s law is not valid if current depends on voltage non-linearly. Statement B : Ohm’s law only holds if the temperature is maintained as constant.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357467
For a wire \(\dfrac{R}{l}=\dfrac{1}{2}\) and length of wire is \(l = 5\;cm\). If potential difference of \(1\,V\) is applied across it, then current through wire will be
1 \(40\,A\)
2 \(4\,A\)
3 \(25\,A\)
4 \(2.5\,A\)
Explanation:
Given, for a wire, \(\dfrac{R}{l}=\dfrac{1}{2}\) Length of wire, \(l = 5\;cm = 5 \times {10^{ - 2}}\;m\) \(\therefore R=\dfrac{l}{2}=2.5 \times 10^{-2} \Omega\) Potential difference, \(V=1 V\) or IR \(=1\) \(I=\dfrac{1}{R}=\dfrac{1}{2.5 \times 10^{-2}}=\dfrac{100}{2.5}=40 \mathrm{~A}\) So, correct option is (1)
AIIMS - 2019
PHXII03:CURRENT ELECTRICITY
357468
\(I\)-\(V\) characteristic of a copper wire of length \(L\) and area of cross-section \(A\) is shown in figure. The slope of the curve becomes
1 Less if the area of the wire is increased
2 Less if the length of the wire is increased
3 More if experiment is performed at higher temperature
4 More if a wire of steel of same dimension is used
Explanation:
Here, slope of the curve,\(\frac{{\Delta I}}{{\Delta V}} = \frac{1}{R} = \frac{A}{{\rho l}}\) \(\therefore \) With increase in length (\(l\)), slope decreases. \(\therefore \) (2) is the only correct option.
KCET - 2020
PHXII03:CURRENT ELECTRICITY
357469
\({\vec{E}}\) is the electric field inside a conductor whose material has conductivity \({\sigma}\) and resistivity \({\rho}\). The current density inside the conductor is \({\vec{j}}\). The correct form of \(Ohm's\) law is
1 \({\vec{E}=\sigma \vec{j}}\)
2 \({\vec{j}=\rho \vec{E}}\)
3 \({\vec{E}=\rho \vec{j}}\)
4 \({\vec{E} \cdot \vec{j}=\rho}\)
Explanation:
Ohm's law \({V=i R}\), and \({R=\rho \dfrac{l}{A}}\) \({\therefore V=i \rho \dfrac{l}{A}}\) \({\Rightarrow \dfrac{V}{l}=\dfrac{i}{A} \rho}\) \({\therefore \vec{E}=\rho \vec{j}}\).
