Current Electricity Test

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Current Electricity Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

Two non-ideal batteries are connected in series. Consider the following statements:
(A) The equivalent emf is larger than either of the two emfs.
(B) The equivalent internal resistance is smaller than either of the two internal resistances

A
Each of A and B is correct

B
A is correct but B is wrong

C
B is correct but A is wrong

D
Each of A and B is wrong

Solution

When two non-ideal batteries are connected in series, their equivalent emf is
$$E_{eq} = E_1 + E_2$$
Hence, the equivalent emf is larger than either of the two emfs.
Also, their equivalent internal resistance is
$$r_{eq} = r_1 + r_2$$
Hence, the equivalent resistance is larger than either of the two internal resistances.
Therefore, statement A is correct but B is wrong.
Question 2
1 / -0

Consider the following two statements:
(A) Kirchhoff's Junction Law follows from conservation of charge.
(B) Kirchhoff's Loop Law follows from conservative nature of electric field.

A
Both A and B are correct

B
A is correct but B is wrong

C
A is correct but A is wrong

D
Both A and B are wrong

Solution

Kirchhoff's Junction law states that total current(charge) entering the junction is equal to total current(charge) leaving the junction hence, total charge across junction remains conserved.
Also, Kirchhoff's loop law states that the directed sum of the electrical potential differences (voltage) around any closed network is zero or the sum of the emfs in any closed loop is equivalent to the sum of the potential drops in that loop.
Question 3
1 / -0

A uniform wire of resistance $$50\Omega$$ is cut into 5 equal parts. These parts are now connected in parallel. The equivalent resistance of the combination is

A
$$2\Omega$$

B
$$10\Omega$$

C
$$250\Omega$$

D
$$6250\Omega$$

Solution

Here, a uniform wire of resistance $$50 \Omega$$ is cut into $$5$$ equal parts. Hence, resistance of each part is
$$r = \dfrac{R}{n} $$
$$ \Rightarrow r = \dfrac{50}{5} $$
$$ \Rightarrow r= 10\ \Omega$$
Now, the parts are connected in parallel, the equivalent resistance of the combination is
$$\dfrac{1}{R_{eq}} = \dfrac{1}{10} + \dfrac{1}{10} + \dfrac{1}{10} + \dfrac{1}{10} + \dfrac{1}{10}$$
$$\dfrac{1}{R_{eq}} = \dfrac{5}{10} = \dfrac{1}{2}$$
$$R_{eq} = 2\Omega$$
Question 4
1 / -0

The two cells are connected in series, in a potentiometer experiment, in such a way so as to support each other and to oppose each other. The balancing lengths in two conditions are obtained as 150 cm and 50 cm respectively. The ratio of emf's of two cells will be

A
1:2

B
2:1

C
1:4

D
4:1

Solution

Let, $$E_1$$ and $$E_2$$ be the emf's of two cells, $$l_1$$ and $$l_2$$ be
the balancing lengths obtained for two cells respectively.

When the cells support each other,

$$E_1 + E_2 = l_1 + l_2$$..........................$$(\because E \propto l)$$

When the cells oppose each other,

$$E_1 - E_2 = l_1 - l_2$$

But, $$l_1 + l_2 = 150 cm$$...............................$$(1)$$

and $$l_1 - l_2 = 50 cm$$....................................$$(2)$$

From $$(1)$$ and $$(2)$$, we get,

$$2l_1 = 200 cm$$

$$l_1 = 100 cm$$

Using this in $$(1)$$, we get,

$$l_2 = 50 cm$$
Thereore,

$$\dfrac{l_1}{l_2} = \dfrac{100}{50} = \dfrac{2}{1}$$
Question 5
1 / -0

A student connects four cells, each of internal resistance $$\dfrac{1}{4}\Omega$$, in series. One of the cells is incorrectly connected because its terminals are reversed. The value of external resistance is $$1\Omega$$. If the emf of each cell is 1.5 volt then current in the circuit will be

A
$$\dfrac {4}{3}A$$

B
$$0$$

C
$$\dfrac {3}{4}A$$

D
$$1.5A$$

Solution

The cells are connected in series hence, the equivalent of internal resistances of cells is

$$r = \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4}$$

$$r = \dfrac{4}{4} = 1 \Omega$$

One of the cells is incorrectly connected because its terminals are reversed.
The cell has equal emf hence in series total emf is

$$E = 1.5 - 1.5 + 1.5 +1.5 = 3 V$$

Let, the external resistance is, $$R = 1\Omega$$

The current in circuit is

$$I = \dfrac{E}{R + r}$$

$$I = \dfrac{3}{1 + 1}$$

$$I = \dfrac{3}{2} = 1.5 A$$
Question 6
1 / -0

The resistance of an iron wire is $$10\Omega$$ and its temperature coefficient of resistance is $$5\times 10^{-3}/^oC$$. A current of 30 mA is flowing in it at $$20^oC$$. Keeping potential difference across its ends constant, if its temperature is increased to $$120^oC$$ then the current flowing in the wire will be (in mA)

A
$$20$$

B
$$15$$

C
$$10$$

D
$$40$$

Solution

Given:
Temperature coefficient of resistance, $$\alpha = 5 \times 10^{-3}/^\circ C$$

Initial temperature, $$T_0 = 20^\circ$$

Initial resistance at $$T_0$$, $$R_0 =10 \Omega$$

Initial current at $$T_0$$, $$I_0 =30 mA$$

Increased temperature, $$T_1 = 120^\circ$$

Increased resistance is,

$$R_1 = R_0 [1 + \alpha(T_1 - T_0)]$$

$$R_1 = 10 [1 + 5 \times 10^{-3}(120 - 20)]$$

$$R_1 = 10 [1 + 5 \times 10^{-1}]$$

$$R_1 = 10 [1 + 0.5]$$

$$R_1 = 10 [1.5] = 15 \Omega$$

Since, potential difference across the ends of wire is constant,

$$V = I_0R_0 = I_1R_1$$

$$\Rightarrow I_1 = \dfrac{I_0R_0}{R_1}$$

$$\therefore I_1 = \dfrac{(30)(10)}{15}$$

$$\therefore I_1 = \dfrac{300}{15} = 20\ mA$$
Question 7
1 / -0

In an experiment of Wheatstone bridge, if the positions of cells and galvanometer are interchanged, then the balance points will

A
change

B
remain unchanged

C
depend on the internal resistance of the cell and resistance of the galvanometer

D
none of these

Solution

When Wheatstone bridge is balanced, $$\dfrac{R_1}{R_2}=\dfrac{R_3}{R_4} or \dfrac{R_1}{R_3}=\dfrac{R_2}{R_4}$$

If the galvanometer is replaced with a cell in balanced Wheatstone bridge, the condition for balanced bridge will be unchanged. i.e $$\dfrac{R_1}{R_2}=\dfrac{R_3}{R_4} $$. In balanced condition no current pass through the galvanometer. Thus, the null points will remian same.
Question 8
1 / -0

Which of the following statements is/are correct for potentiometer circuit?

A
Sensitivity doesn't depend on the length of the potentiometer wire.

B
Sensitivity is inversely proportional to potential difference applied across the potentiometer wire.

C
Accuracy of potentiometer can be increased, only by increasing length of the wire.

D
Range is independent of potential difference applied across the potentiometer wire.

Solution

$$\text{Sensitivity} \ $$$$\propto \dfrac{1}{\text{length of potentiometer wire}}$$

$$\propto \text{potential difference across potentiometer wire}$$

On increasing length of potentiometer wire, the least count decreases, hence the accuracy increases. Hence the correct option is option C.
Question 9
1 / -0

If I current is flowing in a potentiometer wire of length $$L$$ and resistance $$R$$, then potential gradient will be

A
$$\dfrac {IR}{L}$$

B
$$IRL$$

C
$$\dfrac {RL}{I}$$

D
$$\dfrac {IL}{R}$$

Solution

The potential is the fall of potential per unit length of potentiometer wire, hence
$$k = \dfrac{V}{L}$$

$$k = \dfrac{IR}{L}$$
Question 10
1 / -0

In the following circuit the resistance of wire AB is $$10\Omega$$ and its length is 1m. Rest of the quantities are given in the diagram. The potential gradient on the wire will be

A
$$0.08\dfrac {V}{m}$$

B
$$0.008\dfrac {V}{m}$$

C
$$0.8\dfrac {V}{m}$$

D
$$none$$

Solution

Potential gradient depends only on the primary circuit $$\Rightarrow$$

voltage drop per unit length of the wire$$(v)$$

current in primary circuit = $${\dfrac {2}{15+10}}={\dfrac {2}{25}}A$$

voltage drop per unit length of the wire AB$$\Rightarrow v={\dfrac {IR}{L}}={\dfrac {{\dfrac {2}{25}}{\times}{10}}{1}}$$

$$v={\dfrac {20}{25}}={0.8}{\dfrac {V}{m}}$$

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Re-Attempt Weekly Quiz Competition

Current Electricity Test - 46

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Current Electricity Test - 46

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SHARING IS CARING

Question 1

Each of A and B is correct

A is correct but B is wrong

B is correct but A is wrong

Each of A and B is wrong

Solution

Question 2

Both A and B are correct

A is correct but B is wrong

A is correct but A is wrong

Both A and B are wrong

Solution

Question 3

$$2\Omega$$

$$10\Omega$$

$$250\Omega$$

$$6250\Omega$$

Solution

Question 4

1:2

2:1

1:4

4:1

Solution

Question 5

$$\dfrac {4}{3}A$$

$$0$$

$$\dfrac {3}{4}A$$

$$1.5A$$

Solution

Question 6

$$20$$

$$15$$

$$10$$

$$40$$

Solution

Question 7

change

remain unchanged

depend on the internal resistance of the cell and resistance of the galvanometer

none of these

Solution

Question 8

Sensitivity doesn't depend on the length of the potentiometer wire.

Sensitivity is inversely proportional to potential difference applied across the potentiometer wire.

Accuracy of potentiometer can be increased, only by increasing length of the wire.

Range is independent of potential difference applied across the potentiometer wire.

Solution

Question 9

$$\dfrac {IR}{L}$$

$$IRL$$

$$\dfrac {RL}{I}$$

$$\dfrac {IL}{R}$$

Solution

Question 10

$$0.08\dfrac {V}{m}$$

$$0.008\dfrac {V}{m}$$

$$0.8\dfrac {V}{m}$$

$$none$$

Solution

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