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Current Electricity

152611 The temperature of the cold junction of a thermocouple is $0^{0} C$ and the temperature of the hot junction is $T^{\circ} C$ . The relation for the thermo emf is given by, $E = AT - \frac{1}{2} {BT}^{2}$ (When $A = 16$ and $B = 0.08$ ). The temperature of inversion will be :

1

500^{\circ} C

2

460^{\circ} C

3

600 C

4

400^{\circ} C

Explanation:

D The given relation for thermo emf is,
$E = AT - \frac{1}{2} {BT}^{2}$
At $A = 16$ and $B = 0.08$ , equation (i) becomes -
$E = 16 T - \frac{1}{2} \times 0.08 T^{2}$
$E = 16 T - 0.04 T^{2}$
As the inversion temperature $(T_{i})$ of a thermocouple is the temperature of hot junction at which thermo emf becomes zero.
$i.e. At, T = T_{i}, E = 0$
Now, from equation (ii),
$0 = 16 T_{i} - 0.04 T_{i}^{2}$
$0.04 T_{i}^{2} = 16 T_{i}$
$T_{i} = \frac{16}{0.04} = \frac{1600}{4} = 400^{\circ} C$

AIIMS-2001

Current Electricity

152606 Assertion: In a simple battery circuit, the point of the lowest potential is negative terminal of the battery.
Reason: The current flows towards the point of the higher potential, as it does in such a circuit from the negative to the positive terminal.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-2002

Current Electricity

152612 A battery of e.m.f. $20 V$ and internal resistance $6 Ω$ is connected to a resistor as shown in figure. If the current in the circuit is 1 amp, the resistance of the resistor will be :

1

14 Ω

2

16 Ω

3

28 Ω

4

7 Ω

Explanation:

A Given,

E.m.f of battery, $E = 20 V$
Internal resistance, $r = 6 Ω$
Current I $= 1 amp$
Resistance $(R) =$ ? From the figure,
$Current, I = \frac{E}{r + R}$
$1 = \frac{20}{6 + R}$
$6 + R = 20$
$R = 14 Ω$

AIIMS-1997

Current Electricity

152613 When a battery of emf $8 V$ with internal resistance $0.5 Ω$ is charged by a $120 V$ DC supply using a series resistance of $15.5 Ω$ , then the terminal voltage of the battery is-

1

11.5 V

2

20 V

3

21.5 V

4

12.3 V

Explanation:

A Given,
Emf of battery, $E_{1} = 8 V$
Internal resistance of battery, $r = 0.5 Ω$
DC Supply voltage, $V = 120 V$
External resistance, $R = 15.5 Ω$
Terminal voltage of battery $=$ ?
After charging the battery by $120 V$
DC supply, the net emf of the circuit, $E_{net} = 120 - 8 = 112 V$
Current through circuit, $I = \frac{E_{n e t}}{r + R} = \frac{12}{0.5 + 15.5}$
$I = \frac{112}{16} = 7 A$
Now, Terminal voltage of battery $= E_{1} + Ir$
$= 8 + 7 \times (0.5)$
$= 8 + 3.5$
$= 11.5 V$

BCECE-2015

Current Electricity

152611 The temperature of the cold junction of a thermocouple is $0^{0} C$ and the temperature of the hot junction is $T^{\circ} C$ . The relation for the thermo emf is given by, $E = AT - \frac{1}{2} {BT}^{2}$ (When $A = 16$ and $B = 0.08$ ). The temperature of inversion will be :

1

500^{\circ} C

2

460^{\circ} C

3

600 C

4

400^{\circ} C

Explanation:

D The given relation for thermo emf is,
$E = AT - \frac{1}{2} {BT}^{2}$
At $A = 16$ and $B = 0.08$ , equation (i) becomes -
$E = 16 T - \frac{1}{2} \times 0.08 T^{2}$
$E = 16 T - 0.04 T^{2}$
As the inversion temperature $(T_{i})$ of a thermocouple is the temperature of hot junction at which thermo emf becomes zero.
$i.e. At, T = T_{i}, E = 0$
Now, from equation (ii),
$0 = 16 T_{i} - 0.04 T_{i}^{2}$
$0.04 T_{i}^{2} = 16 T_{i}$
$T_{i} = \frac{16}{0.04} = \frac{1600}{4} = 400^{\circ} C$

AIIMS-2001

Current Electricity

152606 Assertion: In a simple battery circuit, the point of the lowest potential is negative terminal of the battery.
Reason: The current flows towards the point of the higher potential, as it does in such a circuit from the negative to the positive terminal.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-2002

Current Electricity

152612 A battery of e.m.f. $20 V$ and internal resistance $6 Ω$ is connected to a resistor as shown in figure. If the current in the circuit is 1 amp, the resistance of the resistor will be :

1

14 Ω

2

16 Ω

3

28 Ω

4

7 Ω

Explanation:

A Given,

E.m.f of battery, $E = 20 V$
Internal resistance, $r = 6 Ω$
Current I $= 1 amp$
Resistance $(R) =$ ? From the figure,
$Current, I = \frac{E}{r + R}$
$1 = \frac{20}{6 + R}$
$6 + R = 20$
$R = 14 Ω$

AIIMS-1997

Current Electricity

152613 When a battery of emf $8 V$ with internal resistance $0.5 Ω$ is charged by a $120 V$ DC supply using a series resistance of $15.5 Ω$ , then the terminal voltage of the battery is-

1

11.5 V

2

20 V

3

21.5 V

4

12.3 V

Explanation:

A Given,
Emf of battery, $E_{1} = 8 V$
Internal resistance of battery, $r = 0.5 Ω$
DC Supply voltage, $V = 120 V$
External resistance, $R = 15.5 Ω$
Terminal voltage of battery $=$ ?
After charging the battery by $120 V$
DC supply, the net emf of the circuit, $E_{net} = 120 - 8 = 112 V$
Current through circuit, $I = \frac{E_{n e t}}{r + R} = \frac{12}{0.5 + 15.5}$
$I = \frac{112}{16} = 7 A$
Now, Terminal voltage of battery $= E_{1} + Ir$
$= 8 + 7 \times (0.5)$
$= 8 + 3.5$
$= 11.5 V$

BCECE-2015

Current Electricity

152611 The temperature of the cold junction of a thermocouple is $0^{0} C$ and the temperature of the hot junction is $T^{\circ} C$ . The relation for the thermo emf is given by, $E = AT - \frac{1}{2} {BT}^{2}$ (When $A = 16$ and $B = 0.08$ ). The temperature of inversion will be :

1

500^{\circ} C

2

460^{\circ} C

3

600 C

4

400^{\circ} C

Explanation:

D The given relation for thermo emf is,
$E = AT - \frac{1}{2} {BT}^{2}$
At $A = 16$ and $B = 0.08$ , equation (i) becomes -
$E = 16 T - \frac{1}{2} \times 0.08 T^{2}$
$E = 16 T - 0.04 T^{2}$
As the inversion temperature $(T_{i})$ of a thermocouple is the temperature of hot junction at which thermo emf becomes zero.
$i.e. At, T = T_{i}, E = 0$
Now, from equation (ii),
$0 = 16 T_{i} - 0.04 T_{i}^{2}$
$0.04 T_{i}^{2} = 16 T_{i}$
$T_{i} = \frac{16}{0.04} = \frac{1600}{4} = 400^{\circ} C$

AIIMS-2001

Current Electricity

152606 Assertion: In a simple battery circuit, the point of the lowest potential is negative terminal of the battery.
Reason: The current flows towards the point of the higher potential, as it does in such a circuit from the negative to the positive terminal.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-2002

Current Electricity

152612 A battery of e.m.f. $20 V$ and internal resistance $6 Ω$ is connected to a resistor as shown in figure. If the current in the circuit is 1 amp, the resistance of the resistor will be :

1

14 Ω

2

16 Ω

3

28 Ω

4

7 Ω

Explanation:

A Given,

E.m.f of battery, $E = 20 V$
Internal resistance, $r = 6 Ω$
Current I $= 1 amp$
Resistance $(R) =$ ? From the figure,
$Current, I = \frac{E}{r + R}$
$1 = \frac{20}{6 + R}$
$6 + R = 20$
$R = 14 Ω$

AIIMS-1997

Current Electricity

152613 When a battery of emf $8 V$ with internal resistance $0.5 Ω$ is charged by a $120 V$ DC supply using a series resistance of $15.5 Ω$ , then the terminal voltage of the battery is-

1

11.5 V

2

20 V

3

21.5 V

4

12.3 V

Explanation:

A Given,
Emf of battery, $E_{1} = 8 V$
Internal resistance of battery, $r = 0.5 Ω$
DC Supply voltage, $V = 120 V$
External resistance, $R = 15.5 Ω$
Terminal voltage of battery $=$ ?
After charging the battery by $120 V$
DC supply, the net emf of the circuit, $E_{net} = 120 - 8 = 112 V$
Current through circuit, $I = \frac{E_{n e t}}{r + R} = \frac{12}{0.5 + 15.5}$
$I = \frac{112}{16} = 7 A$
Now, Terminal voltage of battery $= E_{1} + Ir$
$= 8 + 7 \times (0.5)$
$= 8 + 3.5$
$= 11.5 V$

BCECE-2015

Current Electricity

152611 The temperature of the cold junction of a thermocouple is $0^{0} C$ and the temperature of the hot junction is $T^{\circ} C$ . The relation for the thermo emf is given by, $E = AT - \frac{1}{2} {BT}^{2}$ (When $A = 16$ and $B = 0.08$ ). The temperature of inversion will be :

1

500^{\circ} C

2

460^{\circ} C

3

600 C

4

400^{\circ} C

Explanation:

D The given relation for thermo emf is,
$E = AT - \frac{1}{2} {BT}^{2}$
At $A = 16$ and $B = 0.08$ , equation (i) becomes -
$E = 16 T - \frac{1}{2} \times 0.08 T^{2}$
$E = 16 T - 0.04 T^{2}$
As the inversion temperature $(T_{i})$ of a thermocouple is the temperature of hot junction at which thermo emf becomes zero.
$i.e. At, T = T_{i}, E = 0$
Now, from equation (ii),
$0 = 16 T_{i} - 0.04 T_{i}^{2}$
$0.04 T_{i}^{2} = 16 T_{i}$
$T_{i} = \frac{16}{0.04} = \frac{1600}{4} = 400^{\circ} C$

AIIMS-2001

Current Electricity

152606 Assertion: In a simple battery circuit, the point of the lowest potential is negative terminal of the battery.
Reason: The current flows towards the point of the higher potential, as it does in such a circuit from the negative to the positive terminal.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-2002

Current Electricity

152612 A battery of e.m.f. $20 V$ and internal resistance $6 Ω$ is connected to a resistor as shown in figure. If the current in the circuit is 1 amp, the resistance of the resistor will be :

1

14 Ω

2

16 Ω

3

28 Ω

4

7 Ω

Explanation:

A Given,

E.m.f of battery, $E = 20 V$
Internal resistance, $r = 6 Ω$
Current I $= 1 amp$
Resistance $(R) =$ ? From the figure,
$Current, I = \frac{E}{r + R}$
$1 = \frac{20}{6 + R}$
$6 + R = 20$
$R = 14 Ω$

AIIMS-1997

Current Electricity

152613 When a battery of emf $8 V$ with internal resistance $0.5 Ω$ is charged by a $120 V$ DC supply using a series resistance of $15.5 Ω$ , then the terminal voltage of the battery is-

1

11.5 V

2

20 V

3

21.5 V

4

12.3 V

Explanation:

A Given,
Emf of battery, $E_{1} = 8 V$
Internal resistance of battery, $r = 0.5 Ω$
DC Supply voltage, $V = 120 V$
External resistance, $R = 15.5 Ω$
Terminal voltage of battery $=$ ?
After charging the battery by $120 V$
DC supply, the net emf of the circuit, $E_{net} = 120 - 8 = 112 V$
Current through circuit, $I = \frac{E_{n e t}}{r + R} = \frac{12}{0.5 + 15.5}$
$I = \frac{112}{16} = 7 A$
Now, Terminal voltage of battery $= E_{1} + Ir$
$= 8 + 7 \times (0.5)$
$= 8 + 3.5$
$= 11.5 V$

BCECE-2015