QBManager

Current Electricity

152519 A battery of emf $8 V$ and internal resistance 0.5 $Ω$ is being charged by a $120 V$ d.c. supply using a series resistor of $15.5 Ω$ . The terminal voltage of $8 V$ battery during charging is

1

11.5 V

2

112 V

3

14 V

4

6 V

Explanation:

A Given,
emf of the storage battery $(E) = 8 V$
Internal resistance of the battery $(r) = 0.5 Ω$
DC supply voltage $(V) = 120 V$
Resistance of the resistor $(R) = 15.5 Ω$
$R$ is connected to the storage battery in series.
Effective voltage in the circuit $(V^{'}) = V - E$
$= 120 - 8$
$= 112 V$
According to Ohm's law,
$I = \frac{V^{'}}{R + r} = \frac{112}{15.5 + 0.5} = \frac{112}{10} = 7 A$
Voltage across resistance
$V = IR$
$= 7 \times 15.5 = 108.5 V$
DC supply voltage $=$ Terminal voltage of battery + Voltage drop across $R$
Terminal voltage of battery $= 120 - 108.5$
$= 11.5 V$

AP EAMCET-28.04.2017

Current Electricity

152520 A circuit is shown in the figure. The e.m.f. of the battery is

1

4 V

2

6 V

3

12 V

4

8 V

Explanation:

C
$2 Ω$ and $1 Ω$ resistance are in series.

$2 Ω$ resistor through which $1 A$ current is flowing, we get-
$V_{CO} = 1 \times 2 = 2 V$
$∴ I_{C} = \frac{V_{CO}}{3} = \frac{2}{3} A$
Applying Kirchhoff's current law,
$I_{B} = 1 + I_{C} = 1 + \frac{2}{3} = \frac{5}{3} A$
$∵ V_{BC} = I_{B} \times 2 = \frac{5}{3} \times 2 = \frac{10}{3} V$
$∴ V_{BO} = V_{BC} + V_{CO} = \frac{10}{3} + 2 = \frac{16}{3} V$
$I_{D} = \frac{V_{BO}}{6} = \frac{16 / 3}{6} = \frac{8}{9} A$
Again applying Kirchhoff's current law,
$I_{A} = I_{B} + I_{D}$
$= \frac{5}{3} + \frac{8}{9} = \frac{23}{9} A$
$V_{AB} = I_{A} \times 2 = \frac{23}{9} \times 2 = \frac{46}{9} V$
Therefore,
$V_{AO} = V_{AB} + V_{BO} = \frac{46}{9} + \frac{16}{3} = \frac{94}{9} V$
Voltage drop across the battery is the same as the voltage drop between $V_{A}$ and $V_{B}$ .
The emf of the battery
$E = V_{AO} = \frac{94}{9} = 10.44 V \approx 12 V$

AMU-2017

Current Electricity

152521 Two identical cells are first connected in series and then in parallel. The ratio of power consumed by them is

1

1 : 1

2

1 : 2

3

1 : 3

4

1 : 4

Explanation:

A Let, the emf of a cell $= E$
Internal Resistance $= r$
then in series,
$P_{1} = {(\frac{2 E}{r + r})}^{2} \times 2 r = 2 E^{2} r$
In parallel,
$P_{2} = {(\frac{E}{r / 2})}^{2} \times r / 2 = 2 E^{2} r$
$∴ \frac{P_{1}}{P_{2}} = \frac{2 E^{2} r}{2 E^{2} r}$
$P_{1} : P_{2} = 1 : 1$

AMU-2017

Current Electricity

152523 In the circuit shown below emf of each battery is $5 V$ ; and has an internal resistance of $1.0 Ω$ . The current in the circuit $(I_{0})$ and the reading in an ideal voltmeter $(V)$ are:

1

I_{0} = 1 A, V = 0

2

I_{0} = 5 A, V = 0

3

I_{0} = 1 A, V = 1

4

I_{0} = 5 A, V = 1

Explanation:

B Given, Internal resistance (r) $= 1 Ω$ ,
And emf $(E) = 5 V$
Let, $I$ is the current flow in the circuit
Net emf of the circuit $= 8 \times 5 V = 40 V$
and Net resistance in the circuit $= 8 \times 1 Ω = 8 Ω$
then, current in the circuit
$I = \frac{40}{8} = 5 A$
And voltmeter reading
$V = E - Ir$
$V = 5 - 5 \times 1$
$V = 0$

TS EAMCET(Medical)-2017

Current Electricity

152524 Each of the six ideal batteries of emf $20 V$ is connected to an external resistance of $4 Ω$ as shown in the figure. The current through the resistance is

1

6 A

2

3 A

3

4 A

4

5 A

Explanation:

D Given, emf of each batteries $(E) = 20 V$
External Resistance $(R) = 4 Ω$

Voltage drop across $AB$ and $CD$ are same i.e. $20 V$ Hence, current $(I) = \frac{V}{R} = \frac{20}{4} = 5 A$

TS EAMCET (Engg.)-2017

Current Electricity

152519 A battery of emf $8 V$ and internal resistance 0.5 $Ω$ is being charged by a $120 V$ d.c. supply using a series resistor of $15.5 Ω$ . The terminal voltage of $8 V$ battery during charging is

1

11.5 V

2

112 V

3

14 V

4

6 V

Explanation:

A Given,
emf of the storage battery $(E) = 8 V$
Internal resistance of the battery $(r) = 0.5 Ω$
DC supply voltage $(V) = 120 V$
Resistance of the resistor $(R) = 15.5 Ω$
$R$ is connected to the storage battery in series.
Effective voltage in the circuit $(V^{'}) = V - E$
$= 120 - 8$
$= 112 V$
According to Ohm's law,
$I = \frac{V^{'}}{R + r} = \frac{112}{15.5 + 0.5} = \frac{112}{10} = 7 A$
Voltage across resistance
$V = IR$
$= 7 \times 15.5 = 108.5 V$
DC supply voltage $=$ Terminal voltage of battery + Voltage drop across $R$
Terminal voltage of battery $= 120 - 108.5$
$= 11.5 V$

AP EAMCET-28.04.2017

Current Electricity

152520 A circuit is shown in the figure. The e.m.f. of the battery is

1

4 V

2

6 V

3

12 V

4

8 V

Explanation:

C
$2 Ω$ and $1 Ω$ resistance are in series.

$2 Ω$ resistor through which $1 A$ current is flowing, we get-
$V_{CO} = 1 \times 2 = 2 V$
$∴ I_{C} = \frac{V_{CO}}{3} = \frac{2}{3} A$
Applying Kirchhoff's current law,
$I_{B} = 1 + I_{C} = 1 + \frac{2}{3} = \frac{5}{3} A$
$∵ V_{BC} = I_{B} \times 2 = \frac{5}{3} \times 2 = \frac{10}{3} V$
$∴ V_{BO} = V_{BC} + V_{CO} = \frac{10}{3} + 2 = \frac{16}{3} V$
$I_{D} = \frac{V_{BO}}{6} = \frac{16 / 3}{6} = \frac{8}{9} A$
Again applying Kirchhoff's current law,
$I_{A} = I_{B} + I_{D}$
$= \frac{5}{3} + \frac{8}{9} = \frac{23}{9} A$
$V_{AB} = I_{A} \times 2 = \frac{23}{9} \times 2 = \frac{46}{9} V$
Therefore,
$V_{AO} = V_{AB} + V_{BO} = \frac{46}{9} + \frac{16}{3} = \frac{94}{9} V$
Voltage drop across the battery is the same as the voltage drop between $V_{A}$ and $V_{B}$ .
The emf of the battery
$E = V_{AO} = \frac{94}{9} = 10.44 V \approx 12 V$

AMU-2017

Current Electricity

152521 Two identical cells are first connected in series and then in parallel. The ratio of power consumed by them is

1

1 : 1

2

1 : 2

3

1 : 3

4

1 : 4

Explanation:

A Let, the emf of a cell $= E$
Internal Resistance $= r$
then in series,
$P_{1} = {(\frac{2 E}{r + r})}^{2} \times 2 r = 2 E^{2} r$
In parallel,
$P_{2} = {(\frac{E}{r / 2})}^{2} \times r / 2 = 2 E^{2} r$
$∴ \frac{P_{1}}{P_{2}} = \frac{2 E^{2} r}{2 E^{2} r}$
$P_{1} : P_{2} = 1 : 1$

AMU-2017

Current Electricity

152523 In the circuit shown below emf of each battery is $5 V$ ; and has an internal resistance of $1.0 Ω$ . The current in the circuit $(I_{0})$ and the reading in an ideal voltmeter $(V)$ are:

1

I_{0} = 1 A, V = 0

2

I_{0} = 5 A, V = 0

3

I_{0} = 1 A, V = 1

4

I_{0} = 5 A, V = 1

Explanation:

B Given, Internal resistance (r) $= 1 Ω$ ,
And emf $(E) = 5 V$
Let, $I$ is the current flow in the circuit
Net emf of the circuit $= 8 \times 5 V = 40 V$
and Net resistance in the circuit $= 8 \times 1 Ω = 8 Ω$
then, current in the circuit
$I = \frac{40}{8} = 5 A$
And voltmeter reading
$V = E - Ir$
$V = 5 - 5 \times 1$
$V = 0$

TS EAMCET(Medical)-2017

Current Electricity

152524 Each of the six ideal batteries of emf $20 V$ is connected to an external resistance of $4 Ω$ as shown in the figure. The current through the resistance is

1

6 A

2

3 A

3

4 A

4

5 A

Explanation:

D Given, emf of each batteries $(E) = 20 V$
External Resistance $(R) = 4 Ω$

Voltage drop across $AB$ and $CD$ are same i.e. $20 V$ Hence, current $(I) = \frac{V}{R} = \frac{20}{4} = 5 A$

TS EAMCET (Engg.)-2017

Current Electricity

152519 A battery of emf $8 V$ and internal resistance 0.5 $Ω$ is being charged by a $120 V$ d.c. supply using a series resistor of $15.5 Ω$ . The terminal voltage of $8 V$ battery during charging is

1

11.5 V

2

112 V

3

14 V

4

6 V

Explanation:

A Given,
emf of the storage battery $(E) = 8 V$
Internal resistance of the battery $(r) = 0.5 Ω$
DC supply voltage $(V) = 120 V$
Resistance of the resistor $(R) = 15.5 Ω$
$R$ is connected to the storage battery in series.
Effective voltage in the circuit $(V^{'}) = V - E$
$= 120 - 8$
$= 112 V$
According to Ohm's law,
$I = \frac{V^{'}}{R + r} = \frac{112}{15.5 + 0.5} = \frac{112}{10} = 7 A$
Voltage across resistance
$V = IR$
$= 7 \times 15.5 = 108.5 V$
DC supply voltage $=$ Terminal voltage of battery + Voltage drop across $R$
Terminal voltage of battery $= 120 - 108.5$
$= 11.5 V$

AP EAMCET-28.04.2017

Current Electricity

152520 A circuit is shown in the figure. The e.m.f. of the battery is

1

4 V

2

6 V

3

12 V

4

8 V

Explanation:

C
$2 Ω$ and $1 Ω$ resistance are in series.

$2 Ω$ resistor through which $1 A$ current is flowing, we get-
$V_{CO} = 1 \times 2 = 2 V$
$∴ I_{C} = \frac{V_{CO}}{3} = \frac{2}{3} A$
Applying Kirchhoff's current law,
$I_{B} = 1 + I_{C} = 1 + \frac{2}{3} = \frac{5}{3} A$
$∵ V_{BC} = I_{B} \times 2 = \frac{5}{3} \times 2 = \frac{10}{3} V$
$∴ V_{BO} = V_{BC} + V_{CO} = \frac{10}{3} + 2 = \frac{16}{3} V$
$I_{D} = \frac{V_{BO}}{6} = \frac{16 / 3}{6} = \frac{8}{9} A$
Again applying Kirchhoff's current law,
$I_{A} = I_{B} + I_{D}$
$= \frac{5}{3} + \frac{8}{9} = \frac{23}{9} A$
$V_{AB} = I_{A} \times 2 = \frac{23}{9} \times 2 = \frac{46}{9} V$
Therefore,
$V_{AO} = V_{AB} + V_{BO} = \frac{46}{9} + \frac{16}{3} = \frac{94}{9} V$
Voltage drop across the battery is the same as the voltage drop between $V_{A}$ and $V_{B}$ .
The emf of the battery
$E = V_{AO} = \frac{94}{9} = 10.44 V \approx 12 V$

AMU-2017

Current Electricity

152521 Two identical cells are first connected in series and then in parallel. The ratio of power consumed by them is

1

1 : 1

2

1 : 2

3

1 : 3

4

1 : 4

Explanation:

A Let, the emf of a cell $= E$
Internal Resistance $= r$
then in series,
$P_{1} = {(\frac{2 E}{r + r})}^{2} \times 2 r = 2 E^{2} r$
In parallel,
$P_{2} = {(\frac{E}{r / 2})}^{2} \times r / 2 = 2 E^{2} r$
$∴ \frac{P_{1}}{P_{2}} = \frac{2 E^{2} r}{2 E^{2} r}$
$P_{1} : P_{2} = 1 : 1$

AMU-2017

Current Electricity

152523 In the circuit shown below emf of each battery is $5 V$ ; and has an internal resistance of $1.0 Ω$ . The current in the circuit $(I_{0})$ and the reading in an ideal voltmeter $(V)$ are:

1

I_{0} = 1 A, V = 0

2

I_{0} = 5 A, V = 0

3

I_{0} = 1 A, V = 1

4

I_{0} = 5 A, V = 1

Explanation:

B Given, Internal resistance (r) $= 1 Ω$ ,
And emf $(E) = 5 V$
Let, $I$ is the current flow in the circuit
Net emf of the circuit $= 8 \times 5 V = 40 V$
and Net resistance in the circuit $= 8 \times 1 Ω = 8 Ω$
then, current in the circuit
$I = \frac{40}{8} = 5 A$
And voltmeter reading
$V = E - Ir$
$V = 5 - 5 \times 1$
$V = 0$

TS EAMCET(Medical)-2017

Current Electricity

152524 Each of the six ideal batteries of emf $20 V$ is connected to an external resistance of $4 Ω$ as shown in the figure. The current through the resistance is

1

6 A

2

3 A

3

4 A

4

5 A

Explanation:

D Given, emf of each batteries $(E) = 20 V$
External Resistance $(R) = 4 Ω$

Voltage drop across $AB$ and $CD$ are same i.e. $20 V$ Hence, current $(I) = \frac{V}{R} = \frac{20}{4} = 5 A$

TS EAMCET (Engg.)-2017

Current Electricity

152519 A battery of emf $8 V$ and internal resistance 0.5 $Ω$ is being charged by a $120 V$ d.c. supply using a series resistor of $15.5 Ω$ . The terminal voltage of $8 V$ battery during charging is

1

11.5 V

2

112 V

3

14 V

4

6 V

Explanation:

A Given,
emf of the storage battery $(E) = 8 V$
Internal resistance of the battery $(r) = 0.5 Ω$
DC supply voltage $(V) = 120 V$
Resistance of the resistor $(R) = 15.5 Ω$
$R$ is connected to the storage battery in series.
Effective voltage in the circuit $(V^{'}) = V - E$
$= 120 - 8$
$= 112 V$
According to Ohm's law,
$I = \frac{V^{'}}{R + r} = \frac{112}{15.5 + 0.5} = \frac{112}{10} = 7 A$
Voltage across resistance
$V = IR$
$= 7 \times 15.5 = 108.5 V$
DC supply voltage $=$ Terminal voltage of battery + Voltage drop across $R$
Terminal voltage of battery $= 120 - 108.5$
$= 11.5 V$

AP EAMCET-28.04.2017

Current Electricity

152520 A circuit is shown in the figure. The e.m.f. of the battery is

1

4 V

2

6 V

3

12 V

4

8 V

Explanation:

C
$2 Ω$ and $1 Ω$ resistance are in series.

$2 Ω$ resistor through which $1 A$ current is flowing, we get-
$V_{CO} = 1 \times 2 = 2 V$
$∴ I_{C} = \frac{V_{CO}}{3} = \frac{2}{3} A$
Applying Kirchhoff's current law,
$I_{B} = 1 + I_{C} = 1 + \frac{2}{3} = \frac{5}{3} A$
$∵ V_{BC} = I_{B} \times 2 = \frac{5}{3} \times 2 = \frac{10}{3} V$
$∴ V_{BO} = V_{BC} + V_{CO} = \frac{10}{3} + 2 = \frac{16}{3} V$
$I_{D} = \frac{V_{BO}}{6} = \frac{16 / 3}{6} = \frac{8}{9} A$
Again applying Kirchhoff's current law,
$I_{A} = I_{B} + I_{D}$
$= \frac{5}{3} + \frac{8}{9} = \frac{23}{9} A$
$V_{AB} = I_{A} \times 2 = \frac{23}{9} \times 2 = \frac{46}{9} V$
Therefore,
$V_{AO} = V_{AB} + V_{BO} = \frac{46}{9} + \frac{16}{3} = \frac{94}{9} V$
Voltage drop across the battery is the same as the voltage drop between $V_{A}$ and $V_{B}$ .
The emf of the battery
$E = V_{AO} = \frac{94}{9} = 10.44 V \approx 12 V$

AMU-2017

Current Electricity

152521 Two identical cells are first connected in series and then in parallel. The ratio of power consumed by them is

1

1 : 1

2

1 : 2

3

1 : 3

4

1 : 4

Explanation:

A Let, the emf of a cell $= E$
Internal Resistance $= r$
then in series,
$P_{1} = {(\frac{2 E}{r + r})}^{2} \times 2 r = 2 E^{2} r$
In parallel,
$P_{2} = {(\frac{E}{r / 2})}^{2} \times r / 2 = 2 E^{2} r$
$∴ \frac{P_{1}}{P_{2}} = \frac{2 E^{2} r}{2 E^{2} r}$
$P_{1} : P_{2} = 1 : 1$

AMU-2017

Current Electricity

152523 In the circuit shown below emf of each battery is $5 V$ ; and has an internal resistance of $1.0 Ω$ . The current in the circuit $(I_{0})$ and the reading in an ideal voltmeter $(V)$ are:

1

I_{0} = 1 A, V = 0

2

I_{0} = 5 A, V = 0

3

I_{0} = 1 A, V = 1

4

I_{0} = 5 A, V = 1

Explanation:

B Given, Internal resistance (r) $= 1 Ω$ ,
And emf $(E) = 5 V$
Let, $I$ is the current flow in the circuit
Net emf of the circuit $= 8 \times 5 V = 40 V$
and Net resistance in the circuit $= 8 \times 1 Ω = 8 Ω$
then, current in the circuit
$I = \frac{40}{8} = 5 A$
And voltmeter reading
$V = E - Ir$
$V = 5 - 5 \times 1$
$V = 0$

TS EAMCET(Medical)-2017

Current Electricity

152524 Each of the six ideal batteries of emf $20 V$ is connected to an external resistance of $4 Ω$ as shown in the figure. The current through the resistance is

1

6 A

2

3 A

3

4 A

4

5 A

Explanation:

D Given, emf of each batteries $(E) = 20 V$
External Resistance $(R) = 4 Ω$

Voltage drop across $AB$ and $CD$ are same i.e. $20 V$ Hence, current $(I) = \frac{V}{R} = \frac{20}{4} = 5 A$

TS EAMCET (Engg.)-2017

Current Electricity

152519 A battery of emf $8 V$ and internal resistance 0.5 $Ω$ is being charged by a $120 V$ d.c. supply using a series resistor of $15.5 Ω$ . The terminal voltage of $8 V$ battery during charging is

1

11.5 V

2

112 V

3

14 V

4

6 V

Explanation:

A Given,
emf of the storage battery $(E) = 8 V$
Internal resistance of the battery $(r) = 0.5 Ω$
DC supply voltage $(V) = 120 V$
Resistance of the resistor $(R) = 15.5 Ω$
$R$ is connected to the storage battery in series.
Effective voltage in the circuit $(V^{'}) = V - E$
$= 120 - 8$
$= 112 V$
According to Ohm's law,
$I = \frac{V^{'}}{R + r} = \frac{112}{15.5 + 0.5} = \frac{112}{10} = 7 A$
Voltage across resistance
$V = IR$
$= 7 \times 15.5 = 108.5 V$
DC supply voltage $=$ Terminal voltage of battery + Voltage drop across $R$
Terminal voltage of battery $= 120 - 108.5$
$= 11.5 V$

AP EAMCET-28.04.2017

Current Electricity

152520 A circuit is shown in the figure. The e.m.f. of the battery is

1

4 V

2

6 V

3

12 V

4

8 V

Explanation:

C
$2 Ω$ and $1 Ω$ resistance are in series.

$2 Ω$ resistor through which $1 A$ current is flowing, we get-
$V_{CO} = 1 \times 2 = 2 V$
$∴ I_{C} = \frac{V_{CO}}{3} = \frac{2}{3} A$
Applying Kirchhoff's current law,
$I_{B} = 1 + I_{C} = 1 + \frac{2}{3} = \frac{5}{3} A$
$∵ V_{BC} = I_{B} \times 2 = \frac{5}{3} \times 2 = \frac{10}{3} V$
$∴ V_{BO} = V_{BC} + V_{CO} = \frac{10}{3} + 2 = \frac{16}{3} V$
$I_{D} = \frac{V_{BO}}{6} = \frac{16 / 3}{6} = \frac{8}{9} A$
Again applying Kirchhoff's current law,
$I_{A} = I_{B} + I_{D}$
$= \frac{5}{3} + \frac{8}{9} = \frac{23}{9} A$
$V_{AB} = I_{A} \times 2 = \frac{23}{9} \times 2 = \frac{46}{9} V$
Therefore,
$V_{AO} = V_{AB} + V_{BO} = \frac{46}{9} + \frac{16}{3} = \frac{94}{9} V$
Voltage drop across the battery is the same as the voltage drop between $V_{A}$ and $V_{B}$ .
The emf of the battery
$E = V_{AO} = \frac{94}{9} = 10.44 V \approx 12 V$

AMU-2017

Current Electricity

152521 Two identical cells are first connected in series and then in parallel. The ratio of power consumed by them is

1

1 : 1

2

1 : 2

3

1 : 3

4

1 : 4

Explanation:

A Let, the emf of a cell $= E$
Internal Resistance $= r$
then in series,
$P_{1} = {(\frac{2 E}{r + r})}^{2} \times 2 r = 2 E^{2} r$
In parallel,
$P_{2} = {(\frac{E}{r / 2})}^{2} \times r / 2 = 2 E^{2} r$
$∴ \frac{P_{1}}{P_{2}} = \frac{2 E^{2} r}{2 E^{2} r}$
$P_{1} : P_{2} = 1 : 1$

AMU-2017

Current Electricity

152523 In the circuit shown below emf of each battery is $5 V$ ; and has an internal resistance of $1.0 Ω$ . The current in the circuit $(I_{0})$ and the reading in an ideal voltmeter $(V)$ are:

1

I_{0} = 1 A, V = 0

2

I_{0} = 5 A, V = 0

3

I_{0} = 1 A, V = 1

4

I_{0} = 5 A, V = 1

Explanation:

B Given, Internal resistance (r) $= 1 Ω$ ,
And emf $(E) = 5 V$
Let, $I$ is the current flow in the circuit
Net emf of the circuit $= 8 \times 5 V = 40 V$
and Net resistance in the circuit $= 8 \times 1 Ω = 8 Ω$
then, current in the circuit
$I = \frac{40}{8} = 5 A$
And voltmeter reading
$V = E - Ir$
$V = 5 - 5 \times 1$
$V = 0$

TS EAMCET(Medical)-2017

Current Electricity

152524 Each of the six ideal batteries of emf $20 V$ is connected to an external resistance of $4 Ω$ as shown in the figure. The current through the resistance is

1

6 A

2

3 A

3

4 A

4

5 A

Explanation:

D Given, emf of each batteries $(E) = 20 V$
External Resistance $(R) = 4 Ω$

Voltage drop across $AB$ and $CD$ are same i.e. $20 V$ Hence, current $(I) = \frac{V}{R} = \frac{20}{4} = 5 A$

TS EAMCET (Engg.)-2017

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