D The amount of charge per unit time that flows through a unit area of a chosen cross- section is called current density. So, current density, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ Where, $\mathrm{J}=$ Current density $\mathrm{I}=$ Current $\mathrm{A}=$ Cross-section area
Kerala CEE - 2016
Current Electricity
151770
For ohmic conductor the drift velocity $v_{d}$ and the electric field applied across it are related as-
1 $v_{d} \propto \sqrt{E}$
2 $v_{d} \propto E^{2}$
3 $v_{d} \propto E$
4 $v_{d} \propto \frac{1}{E}$
Explanation:
C We know that, $\mathrm{I}=\mathrm{neAv}_{\mathrm{d}}$ $\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{I}}{\mathrm{neA}}$ $\because \quad \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$ $\therefore \quad \mathrm{v}_{\mathrm{d}}=\frac{\mathrm{V}}{\mathrm{neAR}}$ $\frac{\mathrm{V}}{\mathrm{R}}=n e A v_{\mathrm{d}}$ $\mathrm{V}=\mathrm{neAv}_{\mathrm{d}} \mathrm{R}$ $\mathrm{V}=\operatorname{neAv}_{\mathrm{d}}\left(\frac{\rho l}{\mathrm{~A}}\right)$ $\mathrm{V}=\operatorname{nev}_{\mathrm{d}} \rho l$ $\frac{\mathrm{V}}{l}=\operatorname{nev}_{\mathrm{d}} \rho$ $\mathrm{E}=\operatorname{nev}_{\mathrm{d}} \rho$ $v_{d} \propto E$ Ans: a : Given, Ans: a : Electric force (F) = e.E $\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{e} . \mathrm{E}}{\mathrm{m}}=\frac{\mathrm{eV}}{\mathrm{mL}}$ Drift velocity $\left(\mathrm{v}_{\mathrm{d}}\right)=\mathrm{a} \times \tau$ $\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{e} . \mathrm{V}}{\mathrm{mL}} \tau$ $\mathrm{v}_{\mathrm{d}} \propto \mathrm{V} \quad\left\{\mathrm{K}=\frac{\mathrm{e}}{\mathrm{mL}} \tau\right\}$ If the potential difference is doubled, then also drift velocity will be doubled.
We know that
Current Electricity
151776
Mobility of free electrons in a conductor is :
1 Directly proportional to electron density
2 Directly proportional to relaxation time
3 Inversely proportional to electron density
4 Inversely proportional to relaxation time
Explanation:
B Electron mobility, $\mu=\frac{\mathrm{e} \tau}{\mathrm{m}}$ $\mu \propto \tau \quad[\because \mathrm{e} \text { and } \mathrm{m} \text { are constants }]$ Where, $\mu=$ Mobility of free electron $\tau=\text { Relaxation time }$ Mobility of free electrons in a conductor is directly proportional to relaxation time.
Karnataka CET-2016
Current Electricity
151778
A 5.0 amp current is setup in an external circuit by a 6.0 volt storage battery for 6.0 minutes. The chemical energy of the battery is reduced by :
151782
The amount of charge $Q$ passed in time $t$ through a cross-section of a wire is $Q=5 t^{2}+3 t$ +1 . The value of current at time $t=5 \mathrm{~s}$ is
1 $9 \mathrm{~A}$
2 $49 \mathrm{~A}$
3 $53 \mathrm{~A}$
4 none of these
Explanation:
C Given that, $\mathrm{Q}=5 \mathrm{t}^{2}+3 \mathrm{t}+1, \mathrm{t}=5 \mathrm{sec}$ We know that, $I=\frac{d Q}{d t}=\frac{d}{d t}\left(5 t^{2}+3 t+1\right)=10 t+3$ The value of current, at time $t=5 \mathrm{sec}$. $\mathrm{I}=10 \times 5+3$ $\mathrm{I}=53 \mathrm{~A}$
D The amount of charge per unit time that flows through a unit area of a chosen cross- section is called current density. So, current density, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ Where, $\mathrm{J}=$ Current density $\mathrm{I}=$ Current $\mathrm{A}=$ Cross-section area
Kerala CEE - 2016
Current Electricity
151770
For ohmic conductor the drift velocity $v_{d}$ and the electric field applied across it are related as-
1 $v_{d} \propto \sqrt{E}$
2 $v_{d} \propto E^{2}$
3 $v_{d} \propto E$
4 $v_{d} \propto \frac{1}{E}$
Explanation:
C We know that, $\mathrm{I}=\mathrm{neAv}_{\mathrm{d}}$ $\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{I}}{\mathrm{neA}}$ $\because \quad \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$ $\therefore \quad \mathrm{v}_{\mathrm{d}}=\frac{\mathrm{V}}{\mathrm{neAR}}$ $\frac{\mathrm{V}}{\mathrm{R}}=n e A v_{\mathrm{d}}$ $\mathrm{V}=\mathrm{neAv}_{\mathrm{d}} \mathrm{R}$ $\mathrm{V}=\operatorname{neAv}_{\mathrm{d}}\left(\frac{\rho l}{\mathrm{~A}}\right)$ $\mathrm{V}=\operatorname{nev}_{\mathrm{d}} \rho l$ $\frac{\mathrm{V}}{l}=\operatorname{nev}_{\mathrm{d}} \rho$ $\mathrm{E}=\operatorname{nev}_{\mathrm{d}} \rho$ $v_{d} \propto E$ Ans: a : Given, Ans: a : Electric force (F) = e.E $\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{e} . \mathrm{E}}{\mathrm{m}}=\frac{\mathrm{eV}}{\mathrm{mL}}$ Drift velocity $\left(\mathrm{v}_{\mathrm{d}}\right)=\mathrm{a} \times \tau$ $\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{e} . \mathrm{V}}{\mathrm{mL}} \tau$ $\mathrm{v}_{\mathrm{d}} \propto \mathrm{V} \quad\left\{\mathrm{K}=\frac{\mathrm{e}}{\mathrm{mL}} \tau\right\}$ If the potential difference is doubled, then also drift velocity will be doubled.
We know that
Current Electricity
151776
Mobility of free electrons in a conductor is :
1 Directly proportional to electron density
2 Directly proportional to relaxation time
3 Inversely proportional to electron density
4 Inversely proportional to relaxation time
Explanation:
B Electron mobility, $\mu=\frac{\mathrm{e} \tau}{\mathrm{m}}$ $\mu \propto \tau \quad[\because \mathrm{e} \text { and } \mathrm{m} \text { are constants }]$ Where, $\mu=$ Mobility of free electron $\tau=\text { Relaxation time }$ Mobility of free electrons in a conductor is directly proportional to relaxation time.
Karnataka CET-2016
Current Electricity
151778
A 5.0 amp current is setup in an external circuit by a 6.0 volt storage battery for 6.0 minutes. The chemical energy of the battery is reduced by :
151782
The amount of charge $Q$ passed in time $t$ through a cross-section of a wire is $Q=5 t^{2}+3 t$ +1 . The value of current at time $t=5 \mathrm{~s}$ is
1 $9 \mathrm{~A}$
2 $49 \mathrm{~A}$
3 $53 \mathrm{~A}$
4 none of these
Explanation:
C Given that, $\mathrm{Q}=5 \mathrm{t}^{2}+3 \mathrm{t}+1, \mathrm{t}=5 \mathrm{sec}$ We know that, $I=\frac{d Q}{d t}=\frac{d}{d t}\left(5 t^{2}+3 t+1\right)=10 t+3$ The value of current, at time $t=5 \mathrm{sec}$. $\mathrm{I}=10 \times 5+3$ $\mathrm{I}=53 \mathrm{~A}$
D The amount of charge per unit time that flows through a unit area of a chosen cross- section is called current density. So, current density, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ Where, $\mathrm{J}=$ Current density $\mathrm{I}=$ Current $\mathrm{A}=$ Cross-section area
Kerala CEE - 2016
Current Electricity
151770
For ohmic conductor the drift velocity $v_{d}$ and the electric field applied across it are related as-
1 $v_{d} \propto \sqrt{E}$
2 $v_{d} \propto E^{2}$
3 $v_{d} \propto E$
4 $v_{d} \propto \frac{1}{E}$
Explanation:
C We know that, $\mathrm{I}=\mathrm{neAv}_{\mathrm{d}}$ $\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{I}}{\mathrm{neA}}$ $\because \quad \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$ $\therefore \quad \mathrm{v}_{\mathrm{d}}=\frac{\mathrm{V}}{\mathrm{neAR}}$ $\frac{\mathrm{V}}{\mathrm{R}}=n e A v_{\mathrm{d}}$ $\mathrm{V}=\mathrm{neAv}_{\mathrm{d}} \mathrm{R}$ $\mathrm{V}=\operatorname{neAv}_{\mathrm{d}}\left(\frac{\rho l}{\mathrm{~A}}\right)$ $\mathrm{V}=\operatorname{nev}_{\mathrm{d}} \rho l$ $\frac{\mathrm{V}}{l}=\operatorname{nev}_{\mathrm{d}} \rho$ $\mathrm{E}=\operatorname{nev}_{\mathrm{d}} \rho$ $v_{d} \propto E$ Ans: a : Given, Ans: a : Electric force (F) = e.E $\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{e} . \mathrm{E}}{\mathrm{m}}=\frac{\mathrm{eV}}{\mathrm{mL}}$ Drift velocity $\left(\mathrm{v}_{\mathrm{d}}\right)=\mathrm{a} \times \tau$ $\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{e} . \mathrm{V}}{\mathrm{mL}} \tau$ $\mathrm{v}_{\mathrm{d}} \propto \mathrm{V} \quad\left\{\mathrm{K}=\frac{\mathrm{e}}{\mathrm{mL}} \tau\right\}$ If the potential difference is doubled, then also drift velocity will be doubled.
We know that
Current Electricity
151776
Mobility of free electrons in a conductor is :
1 Directly proportional to electron density
2 Directly proportional to relaxation time
3 Inversely proportional to electron density
4 Inversely proportional to relaxation time
Explanation:
B Electron mobility, $\mu=\frac{\mathrm{e} \tau}{\mathrm{m}}$ $\mu \propto \tau \quad[\because \mathrm{e} \text { and } \mathrm{m} \text { are constants }]$ Where, $\mu=$ Mobility of free electron $\tau=\text { Relaxation time }$ Mobility of free electrons in a conductor is directly proportional to relaxation time.
Karnataka CET-2016
Current Electricity
151778
A 5.0 amp current is setup in an external circuit by a 6.0 volt storage battery for 6.0 minutes. The chemical energy of the battery is reduced by :
151782
The amount of charge $Q$ passed in time $t$ through a cross-section of a wire is $Q=5 t^{2}+3 t$ +1 . The value of current at time $t=5 \mathrm{~s}$ is
1 $9 \mathrm{~A}$
2 $49 \mathrm{~A}$
3 $53 \mathrm{~A}$
4 none of these
Explanation:
C Given that, $\mathrm{Q}=5 \mathrm{t}^{2}+3 \mathrm{t}+1, \mathrm{t}=5 \mathrm{sec}$ We know that, $I=\frac{d Q}{d t}=\frac{d}{d t}\left(5 t^{2}+3 t+1\right)=10 t+3$ The value of current, at time $t=5 \mathrm{sec}$. $\mathrm{I}=10 \times 5+3$ $\mathrm{I}=53 \mathrm{~A}$
D The amount of charge per unit time that flows through a unit area of a chosen cross- section is called current density. So, current density, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ Where, $\mathrm{J}=$ Current density $\mathrm{I}=$ Current $\mathrm{A}=$ Cross-section area
Kerala CEE - 2016
Current Electricity
151770
For ohmic conductor the drift velocity $v_{d}$ and the electric field applied across it are related as-
1 $v_{d} \propto \sqrt{E}$
2 $v_{d} \propto E^{2}$
3 $v_{d} \propto E$
4 $v_{d} \propto \frac{1}{E}$
Explanation:
C We know that, $\mathrm{I}=\mathrm{neAv}_{\mathrm{d}}$ $\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{I}}{\mathrm{neA}}$ $\because \quad \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$ $\therefore \quad \mathrm{v}_{\mathrm{d}}=\frac{\mathrm{V}}{\mathrm{neAR}}$ $\frac{\mathrm{V}}{\mathrm{R}}=n e A v_{\mathrm{d}}$ $\mathrm{V}=\mathrm{neAv}_{\mathrm{d}} \mathrm{R}$ $\mathrm{V}=\operatorname{neAv}_{\mathrm{d}}\left(\frac{\rho l}{\mathrm{~A}}\right)$ $\mathrm{V}=\operatorname{nev}_{\mathrm{d}} \rho l$ $\frac{\mathrm{V}}{l}=\operatorname{nev}_{\mathrm{d}} \rho$ $\mathrm{E}=\operatorname{nev}_{\mathrm{d}} \rho$ $v_{d} \propto E$ Ans: a : Given, Ans: a : Electric force (F) = e.E $\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{e} . \mathrm{E}}{\mathrm{m}}=\frac{\mathrm{eV}}{\mathrm{mL}}$ Drift velocity $\left(\mathrm{v}_{\mathrm{d}}\right)=\mathrm{a} \times \tau$ $\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{e} . \mathrm{V}}{\mathrm{mL}} \tau$ $\mathrm{v}_{\mathrm{d}} \propto \mathrm{V} \quad\left\{\mathrm{K}=\frac{\mathrm{e}}{\mathrm{mL}} \tau\right\}$ If the potential difference is doubled, then also drift velocity will be doubled.
We know that
Current Electricity
151776
Mobility of free electrons in a conductor is :
1 Directly proportional to electron density
2 Directly proportional to relaxation time
3 Inversely proportional to electron density
4 Inversely proportional to relaxation time
Explanation:
B Electron mobility, $\mu=\frac{\mathrm{e} \tau}{\mathrm{m}}$ $\mu \propto \tau \quad[\because \mathrm{e} \text { and } \mathrm{m} \text { are constants }]$ Where, $\mu=$ Mobility of free electron $\tau=\text { Relaxation time }$ Mobility of free electrons in a conductor is directly proportional to relaxation time.
Karnataka CET-2016
Current Electricity
151778
A 5.0 amp current is setup in an external circuit by a 6.0 volt storage battery for 6.0 minutes. The chemical energy of the battery is reduced by :
151782
The amount of charge $Q$ passed in time $t$ through a cross-section of a wire is $Q=5 t^{2}+3 t$ +1 . The value of current at time $t=5 \mathrm{~s}$ is
1 $9 \mathrm{~A}$
2 $49 \mathrm{~A}$
3 $53 \mathrm{~A}$
4 none of these
Explanation:
C Given that, $\mathrm{Q}=5 \mathrm{t}^{2}+3 \mathrm{t}+1, \mathrm{t}=5 \mathrm{sec}$ We know that, $I=\frac{d Q}{d t}=\frac{d}{d t}\left(5 t^{2}+3 t+1\right)=10 t+3$ The value of current, at time $t=5 \mathrm{sec}$. $\mathrm{I}=10 \times 5+3$ $\mathrm{I}=53 \mathrm{~A}$
D The amount of charge per unit time that flows through a unit area of a chosen cross- section is called current density. So, current density, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$ Where, $\mathrm{J}=$ Current density $\mathrm{I}=$ Current $\mathrm{A}=$ Cross-section area
Kerala CEE - 2016
Current Electricity
151770
For ohmic conductor the drift velocity $v_{d}$ and the electric field applied across it are related as-
1 $v_{d} \propto \sqrt{E}$
2 $v_{d} \propto E^{2}$
3 $v_{d} \propto E$
4 $v_{d} \propto \frac{1}{E}$
Explanation:
C We know that, $\mathrm{I}=\mathrm{neAv}_{\mathrm{d}}$ $\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{I}}{\mathrm{neA}}$ $\because \quad \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$ $\therefore \quad \mathrm{v}_{\mathrm{d}}=\frac{\mathrm{V}}{\mathrm{neAR}}$ $\frac{\mathrm{V}}{\mathrm{R}}=n e A v_{\mathrm{d}}$ $\mathrm{V}=\mathrm{neAv}_{\mathrm{d}} \mathrm{R}$ $\mathrm{V}=\operatorname{neAv}_{\mathrm{d}}\left(\frac{\rho l}{\mathrm{~A}}\right)$ $\mathrm{V}=\operatorname{nev}_{\mathrm{d}} \rho l$ $\frac{\mathrm{V}}{l}=\operatorname{nev}_{\mathrm{d}} \rho$ $\mathrm{E}=\operatorname{nev}_{\mathrm{d}} \rho$ $v_{d} \propto E$ Ans: a : Given, Ans: a : Electric force (F) = e.E $\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{e} . \mathrm{E}}{\mathrm{m}}=\frac{\mathrm{eV}}{\mathrm{mL}}$ Drift velocity $\left(\mathrm{v}_{\mathrm{d}}\right)=\mathrm{a} \times \tau$ $\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{e} . \mathrm{V}}{\mathrm{mL}} \tau$ $\mathrm{v}_{\mathrm{d}} \propto \mathrm{V} \quad\left\{\mathrm{K}=\frac{\mathrm{e}}{\mathrm{mL}} \tau\right\}$ If the potential difference is doubled, then also drift velocity will be doubled.
We know that
Current Electricity
151776
Mobility of free electrons in a conductor is :
1 Directly proportional to electron density
2 Directly proportional to relaxation time
3 Inversely proportional to electron density
4 Inversely proportional to relaxation time
Explanation:
B Electron mobility, $\mu=\frac{\mathrm{e} \tau}{\mathrm{m}}$ $\mu \propto \tau \quad[\because \mathrm{e} \text { and } \mathrm{m} \text { are constants }]$ Where, $\mu=$ Mobility of free electron $\tau=\text { Relaxation time }$ Mobility of free electrons in a conductor is directly proportional to relaxation time.
Karnataka CET-2016
Current Electricity
151778
A 5.0 amp current is setup in an external circuit by a 6.0 volt storage battery for 6.0 minutes. The chemical energy of the battery is reduced by :
151782
The amount of charge $Q$ passed in time $t$ through a cross-section of a wire is $Q=5 t^{2}+3 t$ +1 . The value of current at time $t=5 \mathrm{~s}$ is
1 $9 \mathrm{~A}$
2 $49 \mathrm{~A}$
3 $53 \mathrm{~A}$
4 none of these
Explanation:
C Given that, $\mathrm{Q}=5 \mathrm{t}^{2}+3 \mathrm{t}+1, \mathrm{t}=5 \mathrm{sec}$ We know that, $I=\frac{d Q}{d t}=\frac{d}{d t}\left(5 t^{2}+3 t+1\right)=10 t+3$ The value of current, at time $t=5 \mathrm{sec}$. $\mathrm{I}=10 \times 5+3$ $\mathrm{I}=53 \mathrm{~A}$