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PHXII03:CURRENT ELECTRICITY

356905 For a cell,the graph between the potential difference $(V)$ across terminals of the cell and the current $I$ drawn
supporting img
from the cell is shown in the figure. Then,
(a) $e m f$ of the cell is $2.5 V$
(b) $e m f$ of the cell is $2.0 V$
(c) the internal resistance of cell is $0.5 Ω$
(d) The internal resistance of the cell is $0.4 Ω$
Choose the correct statements.

1

a, c

2

a, d

3

b, c

4

b, d

Explanation:

Use $V = E - I r$
$V = - I r + E$
Internal resistance of cell $(r) =$ slope of $V - I$ graph
$= \frac{d V}{d I} = \frac{2 - 0}{5 - 0} = \frac{2}{5} = 0.4 Ω$
T.P.D. at $I = 0$ is $2 V$
So $e m f = 2 V$ Statement (b).
So correct option is (4)

PHXII03:CURRENT ELECTRICITY

356906 For a cell, the terminal potential difference is $2.2 V$ when circuit is open and reduces to $1.8 V$ when cell is connected to a resistance $R = 5 Ω$ , the internal resistance $(r)$ of cell is:

1

\frac{10}{9} Ω

2

\frac{9}{10} Ω

3

\frac{11}{9} Ω

4

\frac{5}{9} Ω

Explanation:

In an open circuit, emf of cell $E = 2.2 V$ In a closed circuit, terminal potential difference $E = 1.8 V$
External resistance, $R = 5 Ω$
Thus, internal resistance of cell is
$r = (\frac{E}{V} - 1) R = (\frac{2.2}{1.8} - 1) 5 = \frac{10}{9} Ω$ .

PHXII03:CURRENT ELECTRICITY

356907 A wire of resistance $R$ is connected across a cell of $e m f$ $ε$ and internal resistance $r$ . The current through the circuit is $I$ . In time $t$ , the work done by battery to establish the current $I$ is

1

I R t

2

I^{2} R t

3

ε I t

4

\frac{ε^{2} t}{R}

Explanation:

$P = \frac{W}{t} \Rightarrow I^{2} R = \frac{W}{t}$
$\Rightarrow W = I^{2} R t$
Correct option is (2).

KECT - 2023

PHXII03:CURRENT ELECTRICITY

356908 A storage battery of e.m.f 8.0 $V$ and internal resistance $0.5 Ω$ is being charged by a 120 $V$ $d$ . $c$ . supply using a series resistor of $15.5 Ω$ . What is the terminal voltage of the battery during charging ?

1 11.5

V

2 10.6

V

3 12.1

V

4 15.1

V

Explanation:

$I = \frac{E}{R + r} = \frac{120 - 8}{15.5 + 5} = 7 A$
Terminal voltage $= 8.0 + 0.5 \times 7 = 11.5 V$

NCERT Exemplar

PHXII03:CURRENT ELECTRICITY

356909 Two resistors of resistances $2 Ω a n d 6 Ω$ are connected in parallel. This combination is then connected to a battery of emf 2 $V$ and internal resistance $0.5 Ω$ . What is the current flowing through the battery ?

1

\frac{4}{17} A

2

4 A

3

1 A

4

\frac{4}{3} A

Explanation:

Resistors 2 $W$ and 6 $W$ are connected in parallel. Their equivalent resistance $R_{P}$ is calculated as:
$\frac{1}{R_{P}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$
$R_{P} = \frac{3}{2} Ω = 1.5 Ω$
Current flowing through the battery is
$I = \frac{ε}{R_{P} + r} = \frac{2 V}{1.5 Ω + 0.5 Ω} = 1 A$
supporting img

KCET - 2014

PHXII03:CURRENT ELECTRICITY

356905 For a cell,the graph between the potential difference $(V)$ across terminals of the cell and the current $I$ drawn
supporting img
from the cell is shown in the figure. Then,
(a) $e m f$ of the cell is $2.5 V$
(b) $e m f$ of the cell is $2.0 V$
(c) the internal resistance of cell is $0.5 Ω$
(d) The internal resistance of the cell is $0.4 Ω$
Choose the correct statements.

1

a, c

2

a, d

3

b, c

4

b, d

Explanation:

Use $V = E - I r$
$V = - I r + E$
Internal resistance of cell $(r) =$ slope of $V - I$ graph
$= \frac{d V}{d I} = \frac{2 - 0}{5 - 0} = \frac{2}{5} = 0.4 Ω$
T.P.D. at $I = 0$ is $2 V$
So $e m f = 2 V$ Statement (b).
So correct option is (4)

PHXII03:CURRENT ELECTRICITY

356906 For a cell, the terminal potential difference is $2.2 V$ when circuit is open and reduces to $1.8 V$ when cell is connected to a resistance $R = 5 Ω$ , the internal resistance $(r)$ of cell is:

1

\frac{10}{9} Ω

2

\frac{9}{10} Ω

3

\frac{11}{9} Ω

4

\frac{5}{9} Ω

Explanation:

In an open circuit, emf of cell $E = 2.2 V$ In a closed circuit, terminal potential difference $E = 1.8 V$
External resistance, $R = 5 Ω$
Thus, internal resistance of cell is
$r = (\frac{E}{V} - 1) R = (\frac{2.2}{1.8} - 1) 5 = \frac{10}{9} Ω$ .

PHXII03:CURRENT ELECTRICITY

356907 A wire of resistance $R$ is connected across a cell of $e m f$ $ε$ and internal resistance $r$ . The current through the circuit is $I$ . In time $t$ , the work done by battery to establish the current $I$ is

1

I R t

2

I^{2} R t

3

ε I t

4

\frac{ε^{2} t}{R}

Explanation:

$P = \frac{W}{t} \Rightarrow I^{2} R = \frac{W}{t}$
$\Rightarrow W = I^{2} R t$
Correct option is (2).

KECT - 2023

PHXII03:CURRENT ELECTRICITY

356908 A storage battery of e.m.f 8.0 $V$ and internal resistance $0.5 Ω$ is being charged by a 120 $V$ $d$ . $c$ . supply using a series resistor of $15.5 Ω$ . What is the terminal voltage of the battery during charging ?

1 11.5

V

2 10.6

V

3 12.1

V

4 15.1

V

Explanation:

$I = \frac{E}{R + r} = \frac{120 - 8}{15.5 + 5} = 7 A$
Terminal voltage $= 8.0 + 0.5 \times 7 = 11.5 V$

NCERT Exemplar

PHXII03:CURRENT ELECTRICITY

356909 Two resistors of resistances $2 Ω a n d 6 Ω$ are connected in parallel. This combination is then connected to a battery of emf 2 $V$ and internal resistance $0.5 Ω$ . What is the current flowing through the battery ?

1

\frac{4}{17} A

2

4 A

3

1 A

4

\frac{4}{3} A

Explanation:

Resistors 2 $W$ and 6 $W$ are connected in parallel. Their equivalent resistance $R_{P}$ is calculated as:
$\frac{1}{R_{P}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$
$R_{P} = \frac{3}{2} Ω = 1.5 Ω$
Current flowing through the battery is
$I = \frac{ε}{R_{P} + r} = \frac{2 V}{1.5 Ω + 0.5 Ω} = 1 A$
supporting img

KCET - 2014

PHXII03:CURRENT ELECTRICITY

356905 For a cell,the graph between the potential difference $(V)$ across terminals of the cell and the current $I$ drawn
supporting img
from the cell is shown in the figure. Then,
(a) $e m f$ of the cell is $2.5 V$
(b) $e m f$ of the cell is $2.0 V$
(c) the internal resistance of cell is $0.5 Ω$
(d) The internal resistance of the cell is $0.4 Ω$
Choose the correct statements.

1

a, c

2

a, d

3

b, c

4

b, d

Explanation:

Use $V = E - I r$
$V = - I r + E$
Internal resistance of cell $(r) =$ slope of $V - I$ graph
$= \frac{d V}{d I} = \frac{2 - 0}{5 - 0} = \frac{2}{5} = 0.4 Ω$
T.P.D. at $I = 0$ is $2 V$
So $e m f = 2 V$ Statement (b).
So correct option is (4)

PHXII03:CURRENT ELECTRICITY

356906 For a cell, the terminal potential difference is $2.2 V$ when circuit is open and reduces to $1.8 V$ when cell is connected to a resistance $R = 5 Ω$ , the internal resistance $(r)$ of cell is:

1

\frac{10}{9} Ω

2

\frac{9}{10} Ω

3

\frac{11}{9} Ω

4

\frac{5}{9} Ω

Explanation:

In an open circuit, emf of cell $E = 2.2 V$ In a closed circuit, terminal potential difference $E = 1.8 V$
External resistance, $R = 5 Ω$
Thus, internal resistance of cell is
$r = (\frac{E}{V} - 1) R = (\frac{2.2}{1.8} - 1) 5 = \frac{10}{9} Ω$ .

PHXII03:CURRENT ELECTRICITY

356907 A wire of resistance $R$ is connected across a cell of $e m f$ $ε$ and internal resistance $r$ . The current through the circuit is $I$ . In time $t$ , the work done by battery to establish the current $I$ is

1

I R t

2

I^{2} R t

3

ε I t

4

\frac{ε^{2} t}{R}

Explanation:

$P = \frac{W}{t} \Rightarrow I^{2} R = \frac{W}{t}$
$\Rightarrow W = I^{2} R t$
Correct option is (2).

KECT - 2023

PHXII03:CURRENT ELECTRICITY

356908 A storage battery of e.m.f 8.0 $V$ and internal resistance $0.5 Ω$ is being charged by a 120 $V$ $d$ . $c$ . supply using a series resistor of $15.5 Ω$ . What is the terminal voltage of the battery during charging ?

1 11.5

V

2 10.6

V

3 12.1

V

4 15.1

V

Explanation:

$I = \frac{E}{R + r} = \frac{120 - 8}{15.5 + 5} = 7 A$
Terminal voltage $= 8.0 + 0.5 \times 7 = 11.5 V$

NCERT Exemplar

PHXII03:CURRENT ELECTRICITY

356909 Two resistors of resistances $2 Ω a n d 6 Ω$ are connected in parallel. This combination is then connected to a battery of emf 2 $V$ and internal resistance $0.5 Ω$ . What is the current flowing through the battery ?

1

\frac{4}{17} A

2

4 A

3

1 A

4

\frac{4}{3} A

Explanation:

Resistors 2 $W$ and 6 $W$ are connected in parallel. Their equivalent resistance $R_{P}$ is calculated as:
$\frac{1}{R_{P}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$
$R_{P} = \frac{3}{2} Ω = 1.5 Ω$
Current flowing through the battery is
$I = \frac{ε}{R_{P} + r} = \frac{2 V}{1.5 Ω + 0.5 Ω} = 1 A$
supporting img

KCET - 2014

PHXII03:CURRENT ELECTRICITY

356905 For a cell,the graph between the potential difference $(V)$ across terminals of the cell and the current $I$ drawn
supporting img
from the cell is shown in the figure. Then,
(a) $e m f$ of the cell is $2.5 V$
(b) $e m f$ of the cell is $2.0 V$
(c) the internal resistance of cell is $0.5 Ω$
(d) The internal resistance of the cell is $0.4 Ω$
Choose the correct statements.

1

a, c

2

a, d

3

b, c

4

b, d

Explanation:

Use $V = E - I r$
$V = - I r + E$
Internal resistance of cell $(r) =$ slope of $V - I$ graph
$= \frac{d V}{d I} = \frac{2 - 0}{5 - 0} = \frac{2}{5} = 0.4 Ω$
T.P.D. at $I = 0$ is $2 V$
So $e m f = 2 V$ Statement (b).
So correct option is (4)

PHXII03:CURRENT ELECTRICITY

356906 For a cell, the terminal potential difference is $2.2 V$ when circuit is open and reduces to $1.8 V$ when cell is connected to a resistance $R = 5 Ω$ , the internal resistance $(r)$ of cell is:

1

\frac{10}{9} Ω

2

\frac{9}{10} Ω

3

\frac{11}{9} Ω

4

\frac{5}{9} Ω

Explanation:

In an open circuit, emf of cell $E = 2.2 V$ In a closed circuit, terminal potential difference $E = 1.8 V$
External resistance, $R = 5 Ω$
Thus, internal resistance of cell is
$r = (\frac{E}{V} - 1) R = (\frac{2.2}{1.8} - 1) 5 = \frac{10}{9} Ω$ .

PHXII03:CURRENT ELECTRICITY

356907 A wire of resistance $R$ is connected across a cell of $e m f$ $ε$ and internal resistance $r$ . The current through the circuit is $I$ . In time $t$ , the work done by battery to establish the current $I$ is

1

I R t

2

I^{2} R t

3

ε I t

4

\frac{ε^{2} t}{R}

Explanation:

$P = \frac{W}{t} \Rightarrow I^{2} R = \frac{W}{t}$
$\Rightarrow W = I^{2} R t$
Correct option is (2).

KECT - 2023

PHXII03:CURRENT ELECTRICITY

356908 A storage battery of e.m.f 8.0 $V$ and internal resistance $0.5 Ω$ is being charged by a 120 $V$ $d$ . $c$ . supply using a series resistor of $15.5 Ω$ . What is the terminal voltage of the battery during charging ?

1 11.5

V

2 10.6

V

3 12.1

V

4 15.1

V

Explanation:

$I = \frac{E}{R + r} = \frac{120 - 8}{15.5 + 5} = 7 A$
Terminal voltage $= 8.0 + 0.5 \times 7 = 11.5 V$

NCERT Exemplar

PHXII03:CURRENT ELECTRICITY

356909 Two resistors of resistances $2 Ω a n d 6 Ω$ are connected in parallel. This combination is then connected to a battery of emf 2 $V$ and internal resistance $0.5 Ω$ . What is the current flowing through the battery ?

1

\frac{4}{17} A

2

4 A

3

1 A

4

\frac{4}{3} A

Explanation:

Resistors 2 $W$ and 6 $W$ are connected in parallel. Their equivalent resistance $R_{P}$ is calculated as:
$\frac{1}{R_{P}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$
$R_{P} = \frac{3}{2} Ω = 1.5 Ω$
Current flowing through the battery is
$I = \frac{ε}{R_{P} + r} = \frac{2 V}{1.5 Ω + 0.5 Ω} = 1 A$
supporting img

KCET - 2014

PHXII03:CURRENT ELECTRICITY

356905 For a cell,the graph between the potential difference $(V)$ across terminals of the cell and the current $I$ drawn
supporting img
from the cell is shown in the figure. Then,
(a) $e m f$ of the cell is $2.5 V$
(b) $e m f$ of the cell is $2.0 V$
(c) the internal resistance of cell is $0.5 Ω$
(d) The internal resistance of the cell is $0.4 Ω$
Choose the correct statements.

1

a, c

2

a, d

3

b, c

4

b, d

Explanation:

Use $V = E - I r$
$V = - I r + E$
Internal resistance of cell $(r) =$ slope of $V - I$ graph
$= \frac{d V}{d I} = \frac{2 - 0}{5 - 0} = \frac{2}{5} = 0.4 Ω$
T.P.D. at $I = 0$ is $2 V$
So $e m f = 2 V$ Statement (b).
So correct option is (4)

PHXII03:CURRENT ELECTRICITY

356906 For a cell, the terminal potential difference is $2.2 V$ when circuit is open and reduces to $1.8 V$ when cell is connected to a resistance $R = 5 Ω$ , the internal resistance $(r)$ of cell is:

1

\frac{10}{9} Ω

2

\frac{9}{10} Ω

3

\frac{11}{9} Ω

4

\frac{5}{9} Ω

Explanation:

In an open circuit, emf of cell $E = 2.2 V$ In a closed circuit, terminal potential difference $E = 1.8 V$
External resistance, $R = 5 Ω$
Thus, internal resistance of cell is
$r = (\frac{E}{V} - 1) R = (\frac{2.2}{1.8} - 1) 5 = \frac{10}{9} Ω$ .

PHXII03:CURRENT ELECTRICITY

356907 A wire of resistance $R$ is connected across a cell of $e m f$ $ε$ and internal resistance $r$ . The current through the circuit is $I$ . In time $t$ , the work done by battery to establish the current $I$ is

1

I R t

2

I^{2} R t

3

ε I t

4

\frac{ε^{2} t}{R}

Explanation:

$P = \frac{W}{t} \Rightarrow I^{2} R = \frac{W}{t}$
$\Rightarrow W = I^{2} R t$
Correct option is (2).

KECT - 2023

PHXII03:CURRENT ELECTRICITY

356908 A storage battery of e.m.f 8.0 $V$ and internal resistance $0.5 Ω$ is being charged by a 120 $V$ $d$ . $c$ . supply using a series resistor of $15.5 Ω$ . What is the terminal voltage of the battery during charging ?

1 11.5

V

2 10.6

V

3 12.1

V

4 15.1

V

Explanation:

$I = \frac{E}{R + r} = \frac{120 - 8}{15.5 + 5} = 7 A$
Terminal voltage $= 8.0 + 0.5 \times 7 = 11.5 V$

NCERT Exemplar

PHXII03:CURRENT ELECTRICITY

356909 Two resistors of resistances $2 Ω a n d 6 Ω$ are connected in parallel. This combination is then connected to a battery of emf 2 $V$ and internal resistance $0.5 Ω$ . What is the current flowing through the battery ?

1

\frac{4}{17} A

2

4 A

3

1 A

4

\frac{4}{3} A

Explanation:

Resistors 2 $W$ and 6 $W$ are connected in parallel. Their equivalent resistance $R_{P}$ is calculated as:
$\frac{1}{R_{P}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$
$R_{P} = \frac{3}{2} Ω = 1.5 Ω$
Current flowing through the battery is
$I = \frac{ε}{R_{P} + r} = \frac{2 V}{1.5 Ω + 0.5 Ω} = 1 A$
supporting img

KCET - 2014

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