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Current Electricity

152536 The voltage $V$ and current $I$ graph for a conductor at two different temperatures $T_{1}$ and $T_{2}$ and shown in the figure. The relation between $T_{1}$ and $T_{2}$ is :

1

T_{1} > T_{2}

2

T_{1} < T_{2}

3

T_{1} = T_{2}

4

T_{1} = \frac{1}{T_{2}}

Explanation:

A Slope of the V-I curve at any point equal to resistance at that point.
From the curve
Slope of $T_{1} >$ Slope for $T_{2}$

Karnataka CET-2011

Current Electricity

152537 A battery of emf $E$ has an internal resistance $r$ . $A$ variable resistance $R$ is connected to the terminals of the battery. A current $i$ is drawn from the battery. $V$ is the terminal potential difference. If $R$ alone is gradually reduced to zero, which of the following best describes $i$ and $V$ ?

1

i

approaches zero,

V

approaches

E

2 i approaches

\frac{E}{r}

, V approaches zero

3 i approaches

\frac{E}{r}

, V approaches

E

4 i approaches infinity,

V

approaches

E

Explanation:

B
We know that,
$Current (I) = \frac{E}{R + r}$
When $R$ decreases to 0 ,
$I = \frac{E}{r}$
Similarly, potential difference $(V) = IR$
When $R$ decrease to $0, V = 0$

Karnataka CET-2010

Current Electricity

152538 Two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition (such that $E_{1} > E_{2}$ ). If $r_{1}$ and $r_{2}$ be the internal resistances and $R$ be the external resistance, then the terminal potential difference is :

1

\frac{E_{1} - E_{2}}{r_{1} + r_{2}} \times R

2

\frac{E_{1} + E_{2}}{r_{1} + r_{2}} \times R

3

\frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} \times R

4

\frac{E_{1} + E_{2}}{r_{1} + r_{2} + R} \times R

Explanation:

C Given, two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition such that $E_{1} > E_{2}$

$Current (I) = \frac{E_{1} - E_{2}}{r_{1} + r_{2} + R}$
Now, terminal potential difference across a circuit,
$V = IR$
$V = (\frac{E_{1} - E_{2}}{r_{1} + r_{2} + R}) R$

Karnataka CET-2015

Current Electricity

152539 Four identical cells of emf $E$ and internal resistance $r$ are to be connected in series. Suppose, if one of the cell is connected wrongly, the equivalent emf and effective internal resistance of the combination is :

1

2 E

and

4 r

2

4 E

and

4 r

3

2 E

and

2 r

4

4 E

and

2 r

Explanation:

A Total internal resistance does not charge.
$∴ R^{'} = 4 r$
Net emf $(E^{'}) = E (n - 2 m)$
Where, $n =$ total number of cells $= 4$
$m =$ wrong connection $= 1$
$E^{'} = E (4 - 2)$
$E^{'} = 2 E$

Karnataka CET-2015

Current Electricity

152536 The voltage $V$ and current $I$ graph for a conductor at two different temperatures $T_{1}$ and $T_{2}$ and shown in the figure. The relation between $T_{1}$ and $T_{2}$ is :

1

T_{1} > T_{2}

2

T_{1} < T_{2}

3

T_{1} = T_{2}

4

T_{1} = \frac{1}{T_{2}}

Explanation:

A Slope of the V-I curve at any point equal to resistance at that point.
From the curve
Slope of $T_{1} >$ Slope for $T_{2}$

Karnataka CET-2011

Current Electricity

152537 A battery of emf $E$ has an internal resistance $r$ . $A$ variable resistance $R$ is connected to the terminals of the battery. A current $i$ is drawn from the battery. $V$ is the terminal potential difference. If $R$ alone is gradually reduced to zero, which of the following best describes $i$ and $V$ ?

1

i

approaches zero,

V

approaches

E

2 i approaches

\frac{E}{r}

, V approaches zero

3 i approaches

\frac{E}{r}

, V approaches

E

4 i approaches infinity,

V

approaches

E

Explanation:

B
We know that,
$Current (I) = \frac{E}{R + r}$
When $R$ decreases to 0 ,
$I = \frac{E}{r}$
Similarly, potential difference $(V) = IR$
When $R$ decrease to $0, V = 0$

Karnataka CET-2010

Current Electricity

152538 Two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition (such that $E_{1} > E_{2}$ ). If $r_{1}$ and $r_{2}$ be the internal resistances and $R$ be the external resistance, then the terminal potential difference is :

1

\frac{E_{1} - E_{2}}{r_{1} + r_{2}} \times R

2

\frac{E_{1} + E_{2}}{r_{1} + r_{2}} \times R

3

\frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} \times R

4

\frac{E_{1} + E_{2}}{r_{1} + r_{2} + R} \times R

Explanation:

C Given, two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition such that $E_{1} > E_{2}$

$Current (I) = \frac{E_{1} - E_{2}}{r_{1} + r_{2} + R}$
Now, terminal potential difference across a circuit,
$V = IR$
$V = (\frac{E_{1} - E_{2}}{r_{1} + r_{2} + R}) R$

Karnataka CET-2015

Current Electricity

152539 Four identical cells of emf $E$ and internal resistance $r$ are to be connected in series. Suppose, if one of the cell is connected wrongly, the equivalent emf and effective internal resistance of the combination is :

1

2 E

and

4 r

2

4 E

and

4 r

3

2 E

and

2 r

4

4 E

and

2 r

Explanation:

A Total internal resistance does not charge.
$∴ R^{'} = 4 r$
Net emf $(E^{'}) = E (n - 2 m)$
Where, $n =$ total number of cells $= 4$
$m =$ wrong connection $= 1$
$E^{'} = E (4 - 2)$
$E^{'} = 2 E$

Karnataka CET-2015

Current Electricity

152536 The voltage $V$ and current $I$ graph for a conductor at two different temperatures $T_{1}$ and $T_{2}$ and shown in the figure. The relation between $T_{1}$ and $T_{2}$ is :

1

T_{1} > T_{2}

2

T_{1} < T_{2}

3

T_{1} = T_{2}

4

T_{1} = \frac{1}{T_{2}}

Explanation:

A Slope of the V-I curve at any point equal to resistance at that point.
From the curve
Slope of $T_{1} >$ Slope for $T_{2}$

Karnataka CET-2011

Current Electricity

152537 A battery of emf $E$ has an internal resistance $r$ . $A$ variable resistance $R$ is connected to the terminals of the battery. A current $i$ is drawn from the battery. $V$ is the terminal potential difference. If $R$ alone is gradually reduced to zero, which of the following best describes $i$ and $V$ ?

1

i

approaches zero,

V

approaches

E

2 i approaches

\frac{E}{r}

, V approaches zero

3 i approaches

\frac{E}{r}

, V approaches

E

4 i approaches infinity,

V

approaches

E

Explanation:

B
We know that,
$Current (I) = \frac{E}{R + r}$
When $R$ decreases to 0 ,
$I = \frac{E}{r}$
Similarly, potential difference $(V) = IR$
When $R$ decrease to $0, V = 0$

Karnataka CET-2010

Current Electricity

152538 Two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition (such that $E_{1} > E_{2}$ ). If $r_{1}$ and $r_{2}$ be the internal resistances and $R$ be the external resistance, then the terminal potential difference is :

1

\frac{E_{1} - E_{2}}{r_{1} + r_{2}} \times R

2

\frac{E_{1} + E_{2}}{r_{1} + r_{2}} \times R

3

\frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} \times R

4

\frac{E_{1} + E_{2}}{r_{1} + r_{2} + R} \times R

Explanation:

C Given, two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition such that $E_{1} > E_{2}$

$Current (I) = \frac{E_{1} - E_{2}}{r_{1} + r_{2} + R}$
Now, terminal potential difference across a circuit,
$V = IR$
$V = (\frac{E_{1} - E_{2}}{r_{1} + r_{2} + R}) R$

Karnataka CET-2015

Current Electricity

152539 Four identical cells of emf $E$ and internal resistance $r$ are to be connected in series. Suppose, if one of the cell is connected wrongly, the equivalent emf and effective internal resistance of the combination is :

1

2 E

and

4 r

2

4 E

and

4 r

3

2 E

and

2 r

4

4 E

and

2 r

Explanation:

A Total internal resistance does not charge.
$∴ R^{'} = 4 r$
Net emf $(E^{'}) = E (n - 2 m)$
Where, $n =$ total number of cells $= 4$
$m =$ wrong connection $= 1$
$E^{'} = E (4 - 2)$
$E^{'} = 2 E$

Karnataka CET-2015

Current Electricity

152536 The voltage $V$ and current $I$ graph for a conductor at two different temperatures $T_{1}$ and $T_{2}$ and shown in the figure. The relation between $T_{1}$ and $T_{2}$ is :

1

T_{1} > T_{2}

2

T_{1} < T_{2}

3

T_{1} = T_{2}

4

T_{1} = \frac{1}{T_{2}}

Explanation:

A Slope of the V-I curve at any point equal to resistance at that point.
From the curve
Slope of $T_{1} >$ Slope for $T_{2}$

Karnataka CET-2011

Current Electricity

152537 A battery of emf $E$ has an internal resistance $r$ . $A$ variable resistance $R$ is connected to the terminals of the battery. A current $i$ is drawn from the battery. $V$ is the terminal potential difference. If $R$ alone is gradually reduced to zero, which of the following best describes $i$ and $V$ ?

1

i

approaches zero,

V

approaches

E

2 i approaches

\frac{E}{r}

, V approaches zero

3 i approaches

\frac{E}{r}

, V approaches

E

4 i approaches infinity,

V

approaches

E

Explanation:

B
We know that,
$Current (I) = \frac{E}{R + r}$
When $R$ decreases to 0 ,
$I = \frac{E}{r}$
Similarly, potential difference $(V) = IR$
When $R$ decrease to $0, V = 0$

Karnataka CET-2010

Current Electricity

152538 Two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition (such that $E_{1} > E_{2}$ ). If $r_{1}$ and $r_{2}$ be the internal resistances and $R$ be the external resistance, then the terminal potential difference is :

1

\frac{E_{1} - E_{2}}{r_{1} + r_{2}} \times R

2

\frac{E_{1} + E_{2}}{r_{1} + r_{2}} \times R

3

\frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} \times R

4

\frac{E_{1} + E_{2}}{r_{1} + r_{2} + R} \times R

Explanation:

C Given, two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition such that $E_{1} > E_{2}$

$Current (I) = \frac{E_{1} - E_{2}}{r_{1} + r_{2} + R}$
Now, terminal potential difference across a circuit,
$V = IR$
$V = (\frac{E_{1} - E_{2}}{r_{1} + r_{2} + R}) R$

Karnataka CET-2015

Current Electricity

152539 Four identical cells of emf $E$ and internal resistance $r$ are to be connected in series. Suppose, if one of the cell is connected wrongly, the equivalent emf and effective internal resistance of the combination is :

1

2 E

and

4 r

2

4 E

and

4 r

3

2 E

and

2 r

4

4 E

and

2 r

Explanation:

A Total internal resistance does not charge.
$∴ R^{'} = 4 r$
Net emf $(E^{'}) = E (n - 2 m)$
Where, $n =$ total number of cells $= 4$
$m =$ wrong connection $= 1$
$E^{'} = E (4 - 2)$
$E^{'} = 2 E$

Karnataka CET-2015

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