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PHXII03:CURRENT ELECTRICITY

356892 The external resistance of a circuit is $η$ times higher than the internal resistance of the source. The raito of the potential difference across the terminals of the source to its emf is

1

\frac{1}{η}

2

η

3

\frac{1 + η}{η}

4

\frac{η}{1 + η}

Explanation:

Let $r$ be the internal resistance $R$ is the external resistance
$i_{1} = \frac{ε}{r + R} \Rightarrow Δ V = i R$
$\frac{Δ V}{ε} = \frac{η}{1 + η} (R = η r)$

PHXII03:CURRENT ELECTRICITY

356893 A battery of $ε$ and internal resistance r is connected across a resistance $R$ . Resistance $R$ can be adjusted to any value greater than are equal to zero. A graph is plotted between the current ( $I$ ) passing through the resistance and potential difference ( $V$ ) across it.
Select the correct alternative ( $s$ ):
supporting img

1 Internal resistance of battery is

2 Maximum current which can be taken from the battery is 4

A

3

V

-

I

graph can never be a straight line as shown in figure

4 E.M.F of the battery is 20

V

Explanation:

When $I = 0$ (open circuit)
Potential difference $= ε - I r = ε = 10 V$
when potential difference $= ε - I r = 0$
$r = \frac{10}{2} = 5 Ω$

PHXII03:CURRENT ELECTRICITY

356894 When a resistor of $11 Ω$ is connected in series with an electric cell, the current flowing in it is $0.5 A$ . Instead when a resistor of $5 Ω$ is connected to the same electric cell in series, the current increases by $0.4 A$ . The internal resistance of the cell is

1

1.5 Ω

2

2 Ω

3

2.5 Ω

4

3.5 Ω

Explanation:

Let $ε$ and $r$ be emf and internal resistance of the cell.
As per question,when a resistor of $11 Ω$ is connected in series with it, then the current flowing in $11 Ω$ is
supporting img
$0.5 A = \frac{ε}{11 Ω + r}$
$ε = (0.5 A) (11 Ω) + r (0.5 A)$
$= 5.5 V + r (0.5 A) (1)$
When a resistor of $5 Ω$ is connected in series with it, then the current flowing in $5 Ω$ is
$0.9 A = \frac{ε}{5 Ω + r}$

$ε = (0.9 A) (5 Ω) + r (0.9 A)$
$= 4.5 V + r (0.9 A) (2)$
Equating eqns. (1) and (2), we get
$5.5 V + r (0.5 A) = 4.5 V + r (0.9 A)$
$5.5 V - 4.5 V = r [0.9 A - 0.5 A]$
$1 V = r (0.4 A)$ or $r = \frac{1 V}{0.4 A} = 2.5 Ω$

PHXII03:CURRENT ELECTRICITY

356895 The potential difference across the terminals of a battery is $50 V$ when $11 A$ current is drawn and $60 V$ when $1 A$ current is drawn. The emf and the internal resistance of the battery are

1

62 V, 2 Ω

2

63 V, 1 Ω

3

61 V, 1 Ω

4

64 V, 2 Ω

Explanation:

Let $ε$ and $r$ be the emf and the internal resistance of the battery.
The potential difference $V$ across the terminals of the battery when a current $I$ is drawn from it is
$V = ε - I r$
As per question
$50 V = ε - (11 A) r (1)$
$and 60 V = ε - (1 A) r (2)$
Subtracting eqn. (1) from eqn. (2), we get
$10 V = (10 A) r$
$r = \frac{10 V}{10 A} = 1 Ω$
Substituting this value of $r$ in eqn. (1), we get
$50 V = ε - (11 A) (1 Ω)$
$50 V = ε - 11 V$
$ε = 50 V + 11 V = 61 V$

PHXII03:CURRENT ELECTRICITY

356896 A cell of internal resistance $r$ is connected across an external resistance $n r$ . Then the ratio of the terminal voltage to the emf of the cell is

1

\frac{1}{n}

2

\frac{1}{n + 1}

3

\frac{n}{n + 1}

4

\frac{n - 1}{n}

Explanation:

Internal resistnace $= r$ ,
External resistance $= n r$
Let terminal voltage $= V$
then $V = E - I r \Rightarrow V = E - \frac{E r}{(n + 1) r}$ $\Rightarrow V = \frac{n E}{n + 1} \Rightarrow \frac{V}{E} = \frac{n}{n + 1}$

MHTCET - 2021

PHXII03:CURRENT ELECTRICITY

356892 The external resistance of a circuit is $η$ times higher than the internal resistance of the source. The raito of the potential difference across the terminals of the source to its emf is

1

\frac{1}{η}

2

η

3

\frac{1 + η}{η}

4

\frac{η}{1 + η}

Explanation:

Let $r$ be the internal resistance $R$ is the external resistance
$i_{1} = \frac{ε}{r + R} \Rightarrow Δ V = i R$
$\frac{Δ V}{ε} = \frac{η}{1 + η} (R = η r)$

PHXII03:CURRENT ELECTRICITY

356893 A battery of $ε$ and internal resistance r is connected across a resistance $R$ . Resistance $R$ can be adjusted to any value greater than are equal to zero. A graph is plotted between the current ( $I$ ) passing through the resistance and potential difference ( $V$ ) across it.
Select the correct alternative ( $s$ ):
supporting img

1 Internal resistance of battery is

2 Maximum current which can be taken from the battery is 4

A

3

V

-

I

graph can never be a straight line as shown in figure

4 E.M.F of the battery is 20

V

Explanation:

When $I = 0$ (open circuit)
Potential difference $= ε - I r = ε = 10 V$
when potential difference $= ε - I r = 0$
$r = \frac{10}{2} = 5 Ω$

PHXII03:CURRENT ELECTRICITY

356894 When a resistor of $11 Ω$ is connected in series with an electric cell, the current flowing in it is $0.5 A$ . Instead when a resistor of $5 Ω$ is connected to the same electric cell in series, the current increases by $0.4 A$ . The internal resistance of the cell is

1

1.5 Ω

2

2 Ω

3

2.5 Ω

4

3.5 Ω

Explanation:

Let $ε$ and $r$ be emf and internal resistance of the cell.
As per question,when a resistor of $11 Ω$ is connected in series with it, then the current flowing in $11 Ω$ is
supporting img
$0.5 A = \frac{ε}{11 Ω + r}$
$ε = (0.5 A) (11 Ω) + r (0.5 A)$
$= 5.5 V + r (0.5 A) (1)$
When a resistor of $5 Ω$ is connected in series with it, then the current flowing in $5 Ω$ is
$0.9 A = \frac{ε}{5 Ω + r}$

$ε = (0.9 A) (5 Ω) + r (0.9 A)$
$= 4.5 V + r (0.9 A) (2)$
Equating eqns. (1) and (2), we get
$5.5 V + r (0.5 A) = 4.5 V + r (0.9 A)$
$5.5 V - 4.5 V = r [0.9 A - 0.5 A]$
$1 V = r (0.4 A)$ or $r = \frac{1 V}{0.4 A} = 2.5 Ω$

PHXII03:CURRENT ELECTRICITY

356895 The potential difference across the terminals of a battery is $50 V$ when $11 A$ current is drawn and $60 V$ when $1 A$ current is drawn. The emf and the internal resistance of the battery are

1

62 V, 2 Ω

2

63 V, 1 Ω

3

61 V, 1 Ω

4

64 V, 2 Ω

Explanation:

Let $ε$ and $r$ be the emf and the internal resistance of the battery.
The potential difference $V$ across the terminals of the battery when a current $I$ is drawn from it is
$V = ε - I r$
As per question
$50 V = ε - (11 A) r (1)$
$and 60 V = ε - (1 A) r (2)$
Subtracting eqn. (1) from eqn. (2), we get
$10 V = (10 A) r$
$r = \frac{10 V}{10 A} = 1 Ω$
Substituting this value of $r$ in eqn. (1), we get
$50 V = ε - (11 A) (1 Ω)$
$50 V = ε - 11 V$
$ε = 50 V + 11 V = 61 V$

PHXII03:CURRENT ELECTRICITY

356896 A cell of internal resistance $r$ is connected across an external resistance $n r$ . Then the ratio of the terminal voltage to the emf of the cell is

1

\frac{1}{n}

2

\frac{1}{n + 1}

3

\frac{n}{n + 1}

4

\frac{n - 1}{n}

Explanation:

Internal resistnace $= r$ ,
External resistance $= n r$
Let terminal voltage $= V$
then $V = E - I r \Rightarrow V = E - \frac{E r}{(n + 1) r}$ $\Rightarrow V = \frac{n E}{n + 1} \Rightarrow \frac{V}{E} = \frac{n}{n + 1}$

MHTCET - 2021

PHXII03:CURRENT ELECTRICITY

356892 The external resistance of a circuit is $η$ times higher than the internal resistance of the source. The raito of the potential difference across the terminals of the source to its emf is

1

\frac{1}{η}

2

η

3

\frac{1 + η}{η}

4

\frac{η}{1 + η}

Explanation:

Let $r$ be the internal resistance $R$ is the external resistance
$i_{1} = \frac{ε}{r + R} \Rightarrow Δ V = i R$
$\frac{Δ V}{ε} = \frac{η}{1 + η} (R = η r)$

PHXII03:CURRENT ELECTRICITY

356893 A battery of $ε$ and internal resistance r is connected across a resistance $R$ . Resistance $R$ can be adjusted to any value greater than are equal to zero. A graph is plotted between the current ( $I$ ) passing through the resistance and potential difference ( $V$ ) across it.
Select the correct alternative ( $s$ ):
supporting img

1 Internal resistance of battery is

2 Maximum current which can be taken from the battery is 4

A

3

V

-

I

graph can never be a straight line as shown in figure

4 E.M.F of the battery is 20

V

Explanation:

When $I = 0$ (open circuit)
Potential difference $= ε - I r = ε = 10 V$
when potential difference $= ε - I r = 0$
$r = \frac{10}{2} = 5 Ω$

PHXII03:CURRENT ELECTRICITY

356894 When a resistor of $11 Ω$ is connected in series with an electric cell, the current flowing in it is $0.5 A$ . Instead when a resistor of $5 Ω$ is connected to the same electric cell in series, the current increases by $0.4 A$ . The internal resistance of the cell is

1

1.5 Ω

2

2 Ω

3

2.5 Ω

4

3.5 Ω

Explanation:

Let $ε$ and $r$ be emf and internal resistance of the cell.
As per question,when a resistor of $11 Ω$ is connected in series with it, then the current flowing in $11 Ω$ is
supporting img
$0.5 A = \frac{ε}{11 Ω + r}$
$ε = (0.5 A) (11 Ω) + r (0.5 A)$
$= 5.5 V + r (0.5 A) (1)$
When a resistor of $5 Ω$ is connected in series with it, then the current flowing in $5 Ω$ is
$0.9 A = \frac{ε}{5 Ω + r}$

$ε = (0.9 A) (5 Ω) + r (0.9 A)$
$= 4.5 V + r (0.9 A) (2)$
Equating eqns. (1) and (2), we get
$5.5 V + r (0.5 A) = 4.5 V + r (0.9 A)$
$5.5 V - 4.5 V = r [0.9 A - 0.5 A]$
$1 V = r (0.4 A)$ or $r = \frac{1 V}{0.4 A} = 2.5 Ω$

PHXII03:CURRENT ELECTRICITY

356895 The potential difference across the terminals of a battery is $50 V$ when $11 A$ current is drawn and $60 V$ when $1 A$ current is drawn. The emf and the internal resistance of the battery are

1

62 V, 2 Ω

2

63 V, 1 Ω

3

61 V, 1 Ω

4

64 V, 2 Ω

Explanation:

Let $ε$ and $r$ be the emf and the internal resistance of the battery.
The potential difference $V$ across the terminals of the battery when a current $I$ is drawn from it is
$V = ε - I r$
As per question
$50 V = ε - (11 A) r (1)$
$and 60 V = ε - (1 A) r (2)$
Subtracting eqn. (1) from eqn. (2), we get
$10 V = (10 A) r$
$r = \frac{10 V}{10 A} = 1 Ω$
Substituting this value of $r$ in eqn. (1), we get
$50 V = ε - (11 A) (1 Ω)$
$50 V = ε - 11 V$
$ε = 50 V + 11 V = 61 V$

PHXII03:CURRENT ELECTRICITY

356896 A cell of internal resistance $r$ is connected across an external resistance $n r$ . Then the ratio of the terminal voltage to the emf of the cell is

1

\frac{1}{n}

2

\frac{1}{n + 1}

3

\frac{n}{n + 1}

4

\frac{n - 1}{n}

Explanation:

Internal resistnace $= r$ ,
External resistance $= n r$
Let terminal voltage $= V$
then $V = E - I r \Rightarrow V = E - \frac{E r}{(n + 1) r}$ $\Rightarrow V = \frac{n E}{n + 1} \Rightarrow \frac{V}{E} = \frac{n}{n + 1}$

MHTCET - 2021

PHXII03:CURRENT ELECTRICITY

356892 The external resistance of a circuit is $η$ times higher than the internal resistance of the source. The raito of the potential difference across the terminals of the source to its emf is

1

\frac{1}{η}

2

η

3

\frac{1 + η}{η}

4

\frac{η}{1 + η}

Explanation:

Let $r$ be the internal resistance $R$ is the external resistance
$i_{1} = \frac{ε}{r + R} \Rightarrow Δ V = i R$
$\frac{Δ V}{ε} = \frac{η}{1 + η} (R = η r)$

PHXII03:CURRENT ELECTRICITY

356893 A battery of $ε$ and internal resistance r is connected across a resistance $R$ . Resistance $R$ can be adjusted to any value greater than are equal to zero. A graph is plotted between the current ( $I$ ) passing through the resistance and potential difference ( $V$ ) across it.
Select the correct alternative ( $s$ ):
supporting img

1 Internal resistance of battery is

2 Maximum current which can be taken from the battery is 4

A

3

V

-

I

graph can never be a straight line as shown in figure

4 E.M.F of the battery is 20

V

Explanation:

When $I = 0$ (open circuit)
Potential difference $= ε - I r = ε = 10 V$
when potential difference $= ε - I r = 0$
$r = \frac{10}{2} = 5 Ω$

PHXII03:CURRENT ELECTRICITY

356894 When a resistor of $11 Ω$ is connected in series with an electric cell, the current flowing in it is $0.5 A$ . Instead when a resistor of $5 Ω$ is connected to the same electric cell in series, the current increases by $0.4 A$ . The internal resistance of the cell is

1

1.5 Ω

2

2 Ω

3

2.5 Ω

4

3.5 Ω

Explanation:

Let $ε$ and $r$ be emf and internal resistance of the cell.
As per question,when a resistor of $11 Ω$ is connected in series with it, then the current flowing in $11 Ω$ is
supporting img
$0.5 A = \frac{ε}{11 Ω + r}$
$ε = (0.5 A) (11 Ω) + r (0.5 A)$
$= 5.5 V + r (0.5 A) (1)$
When a resistor of $5 Ω$ is connected in series with it, then the current flowing in $5 Ω$ is
$0.9 A = \frac{ε}{5 Ω + r}$

$ε = (0.9 A) (5 Ω) + r (0.9 A)$
$= 4.5 V + r (0.9 A) (2)$
Equating eqns. (1) and (2), we get
$5.5 V + r (0.5 A) = 4.5 V + r (0.9 A)$
$5.5 V - 4.5 V = r [0.9 A - 0.5 A]$
$1 V = r (0.4 A)$ or $r = \frac{1 V}{0.4 A} = 2.5 Ω$

PHXII03:CURRENT ELECTRICITY

356895 The potential difference across the terminals of a battery is $50 V$ when $11 A$ current is drawn and $60 V$ when $1 A$ current is drawn. The emf and the internal resistance of the battery are

1

62 V, 2 Ω

2

63 V, 1 Ω

3

61 V, 1 Ω

4

64 V, 2 Ω

Explanation:

Let $ε$ and $r$ be the emf and the internal resistance of the battery.
The potential difference $V$ across the terminals of the battery when a current $I$ is drawn from it is
$V = ε - I r$
As per question
$50 V = ε - (11 A) r (1)$
$and 60 V = ε - (1 A) r (2)$
Subtracting eqn. (1) from eqn. (2), we get
$10 V = (10 A) r$
$r = \frac{10 V}{10 A} = 1 Ω$
Substituting this value of $r$ in eqn. (1), we get
$50 V = ε - (11 A) (1 Ω)$
$50 V = ε - 11 V$
$ε = 50 V + 11 V = 61 V$

PHXII03:CURRENT ELECTRICITY

356896 A cell of internal resistance $r$ is connected across an external resistance $n r$ . Then the ratio of the terminal voltage to the emf of the cell is

1

\frac{1}{n}

2

\frac{1}{n + 1}

3

\frac{n}{n + 1}

4

\frac{n - 1}{n}

Explanation:

Internal resistnace $= r$ ,
External resistance $= n r$
Let terminal voltage $= V$
then $V = E - I r \Rightarrow V = E - \frac{E r}{(n + 1) r}$ $\Rightarrow V = \frac{n E}{n + 1} \Rightarrow \frac{V}{E} = \frac{n}{n + 1}$

MHTCET - 2021

PHXII03:CURRENT ELECTRICITY

356892 The external resistance of a circuit is $η$ times higher than the internal resistance of the source. The raito of the potential difference across the terminals of the source to its emf is

1

\frac{1}{η}

2

η

3

\frac{1 + η}{η}

4

\frac{η}{1 + η}

Explanation:

Let $r$ be the internal resistance $R$ is the external resistance
$i_{1} = \frac{ε}{r + R} \Rightarrow Δ V = i R$
$\frac{Δ V}{ε} = \frac{η}{1 + η} (R = η r)$

PHXII03:CURRENT ELECTRICITY

356893 A battery of $ε$ and internal resistance r is connected across a resistance $R$ . Resistance $R$ can be adjusted to any value greater than are equal to zero. A graph is plotted between the current ( $I$ ) passing through the resistance and potential difference ( $V$ ) across it.
Select the correct alternative ( $s$ ):
supporting img

1 Internal resistance of battery is

2 Maximum current which can be taken from the battery is 4

A

3

V

-

I

graph can never be a straight line as shown in figure

4 E.M.F of the battery is 20

V

Explanation:

When $I = 0$ (open circuit)
Potential difference $= ε - I r = ε = 10 V$
when potential difference $= ε - I r = 0$
$r = \frac{10}{2} = 5 Ω$

PHXII03:CURRENT ELECTRICITY

356894 When a resistor of $11 Ω$ is connected in series with an electric cell, the current flowing in it is $0.5 A$ . Instead when a resistor of $5 Ω$ is connected to the same electric cell in series, the current increases by $0.4 A$ . The internal resistance of the cell is

1

1.5 Ω

2

2 Ω

3

2.5 Ω

4

3.5 Ω

Explanation:

Let $ε$ and $r$ be emf and internal resistance of the cell.
As per question,when a resistor of $11 Ω$ is connected in series with it, then the current flowing in $11 Ω$ is
supporting img
$0.5 A = \frac{ε}{11 Ω + r}$
$ε = (0.5 A) (11 Ω) + r (0.5 A)$
$= 5.5 V + r (0.5 A) (1)$
When a resistor of $5 Ω$ is connected in series with it, then the current flowing in $5 Ω$ is
$0.9 A = \frac{ε}{5 Ω + r}$

$ε = (0.9 A) (5 Ω) + r (0.9 A)$
$= 4.5 V + r (0.9 A) (2)$
Equating eqns. (1) and (2), we get
$5.5 V + r (0.5 A) = 4.5 V + r (0.9 A)$
$5.5 V - 4.5 V = r [0.9 A - 0.5 A]$
$1 V = r (0.4 A)$ or $r = \frac{1 V}{0.4 A} = 2.5 Ω$

PHXII03:CURRENT ELECTRICITY

356895 The potential difference across the terminals of a battery is $50 V$ when $11 A$ current is drawn and $60 V$ when $1 A$ current is drawn. The emf and the internal resistance of the battery are

1

62 V, 2 Ω

2

63 V, 1 Ω

3

61 V, 1 Ω

4

64 V, 2 Ω

Explanation:

Let $ε$ and $r$ be the emf and the internal resistance of the battery.
The potential difference $V$ across the terminals of the battery when a current $I$ is drawn from it is
$V = ε - I r$
As per question
$50 V = ε - (11 A) r (1)$
$and 60 V = ε - (1 A) r (2)$
Subtracting eqn. (1) from eqn. (2), we get
$10 V = (10 A) r$
$r = \frac{10 V}{10 A} = 1 Ω$
Substituting this value of $r$ in eqn. (1), we get
$50 V = ε - (11 A) (1 Ω)$
$50 V = ε - 11 V$
$ε = 50 V + 11 V = 61 V$

PHXII03:CURRENT ELECTRICITY

356896 A cell of internal resistance $r$ is connected across an external resistance $n r$ . Then the ratio of the terminal voltage to the emf of the cell is

1

\frac{1}{n}

2

\frac{1}{n + 1}

3

\frac{n}{n + 1}

4

\frac{n - 1}{n}

Explanation:

Internal resistnace $= r$ ,
External resistance $= n r$
Let terminal voltage $= V$
then $V = E - I r \Rightarrow V = E - \frac{E r}{(n + 1) r}$ $\Rightarrow V = \frac{n E}{n + 1} \Rightarrow \frac{V}{E} = \frac{n}{n + 1}$

MHTCET - 2021

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