Current Electricity Test

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Current Electricity Test - 40

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Question 1
1 / -0

The resistance of wire is 5 ohm at 50C and 6 ohm at 100C. The resistance of the wire at 0C willbe

A
2 ohm

B
1 ohm

C
4 ohm

D
3 ohm

Solution

Resistance of the wire must vary linearly with temperature.
Hence

$$R = aT + b$$ (let T be in degree scale)

Using $$R(50) = 5.$$

$$R(100) = 6$$

We get b = 4,

and $$a = \dfrac{1}{50}$$

Hence,

$$R = \dfrac{T}{50} + 4$$

Putting R(0) for resistance at $$0^{\circ}$$ we get R(0) $$= 4$$ ohm
Question 2
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A metallic resistor is connected across a battery. If the number of collisions of the free electrons
with the lattice is somehow decreased in the resistor (for example by cooling it), the current
will:

A
Increase

B
decrease

C
remain constant

D
become zero

Solution
Question 3
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The length of a wire of a potentiometer is 100 cm, and the e.m.f. of its stand and cell is $$\mathrm{E}$$ volt. lt is employed to measure the e.m.f. of a battery whose internal resistance is 0.5 $$\Omega$$. lf the balance point is obtained at $$l=30$$ cm from the positive end, the e.m.f. of the battery is

A
$$\displaystyle \frac{30\mathrm{E}}{100.5}$$

B
$$\displaystyle \frac{30\mathrm{E}}{100-0.5}$$

C
$$\displaystyle \frac{30(\mathrm{E}-0.5\mathrm{i})}{100}$$, where i is the current in the potentiometer wire.

D
$$\displaystyle \frac{30\mathrm{E}}{100}$$

Solution

Since galvanometer is used to find the null point when the battery of unknown EMF is connected to the potentiometer wire at a particular point(null position), no current flows through the unknown battery. Hence, there is no drop across the internal resistance of the battery.
Therefore, the potential across the given segment of wire is the absolute EMF of the unknown battery $$=\dfrac{30}{100}E$$
Hence, correct answer is option D.
Question 4
1 / -0

The Kirchhoffs first law $$(\sum i=0)$$ and second law $$(\sum \sum iR=\sum E)$$ where the symbols have their usual meanings, are respectively based on

A
conservation of charge, conservation of energy

B
conservation of charge, conservation of momentum

C
conservation of energy, conservation of charge

D
conservation of momentum, conservation of charge

Solution

Law of conservation of charge states that charge can neither be created nor destroyed. This means that the amount of charge entering a node should be equal to the amount of charge leaving the node. This directly implies that the sum of the amount of current entering or leaving a node is zero i.e. KIrchoff's first law.
Law of conservation of energy states that energy can neither be created nor destroyed. Kirchoff's second law can be interpreted as follows:
In a loop, the electron gets excited by the voltage difference and the energy gained by the electron is lost within the loop thus implying that the voltage rise in the loop is equal to the voltage drop in the loop. Thus, Kirchoff's second law is derived from conservation of energy.
Question 5
1 / -0

A piece of copper and another of germanium are cooled from room temperature to 77 K, the resistance of :

A
each of them increases

B
each of them decreases

C
copper decreases and germanium increases

D
copper increases and germanium decreases

Solution

For conductor , the resistance $$R\propto \Delta T$$ and for semiconductor, the resistance $$R\propto \dfrac{1}{\Delta T}$$
As copper is a conductor and germanium is a semiconductor so according to above relation the resistance of copper decreases and germanium increases.
Question 6
1 / -0

A potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths $$l_{1}$$cm and $$l_{2}$$ cm respectively. The magnitudes, of the resistors R and X, in ohms, are then, equal respectively to:

A
$$k(l_{2}-l_{1})$$ and $$kl_{2}$$

B
$$kl_{1}$$ and $$k(l_{2}-l_{1})$$

C
$$k(l_{2}-l_{1})$$ and $$kl_{1}$$

D
$$kl_{1}$$ and $$kl_{2}$$

Solution
Question 7
1 / -0

The following diagram represents an electrical circuit containing two uniform resistance wires A and B connected to a single ideal cell. Both wires have same length, but thickness of wire B is twice that of wire A. The conducting wires connecting A and B to the ideal cell are resistance less. The dependence of electric potential on position along the length of two wires is given in option.

A

B

C

D

E

Solution

The resistance is different for both the wires. As the length of the both the wires is same the electric field in wires $$A$$ and $$B$$ is same. Therefore potential $$V$$ in wires shall be linear function of $$x$$.
At $$x = 0$$ (i.e. at left end) $$V = 0$$. And at $$x =L$$(i.e. at right end) potential of both wires must be same.
If we plot a graph taking $$x$$ along $$X-$$axis and $$V$$ along $$Y-$$axis for both the wires, it will be a straight line.
Let $$m_{a}$$ and $$m_{b}$$ be the slopes of the graphs for wires $$A$$ and $$B$$ respectively.
At $$x = 0$$ (i.e. at left end) $$V = 0$$, we can write
$$V_{a} = m_{a}x $$ and $$V_{b} = m_{b}x $$ .........(1)
where, $$V_{a}$$ and $$V_{b}$$ potentials for wires A and B respectively.
Now, at $$x =L$$(i.e. at right end) potential of both wires is same.
$$\therefore m_{a} = m_{b}$$
Hence, $$V_{a} = V_{b} = mx$$
where, $$m = m_{a} = m_{b}$$ a positive constant.
Question 8
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You have a large supply of light bulbs and a battery. You start with one light bulb connected to the battery and notice its brightness. You then add one light bulb at a time, each new bulb being added in series to the previous bulbs then

A
Brightness of the bulbs will increase

B
Current through the bulbs will increase

C
Power transferred from the battery will decrease

D
Lifetime of the battery will decrease
Question 9
1 / -0

Analyse the given statements and choose the

correct option.

Statement 1 : Bending a wire does not affect the electrical

resistance.

Statement 2 : Resistance of a wire is proportional to resistivity

of the material.

A
Both statement 1 and statement 2 are CORRECT and statement 2 is the CORRECT explanation of thestatement 1.

B
Both statement 1 and statement 2 are CORRECT, but statement 2 is NOT THE CORRECT explanation of the statement 1.

C
Statement 1 is CORRECT, but statement 2 is INCORRECT.

D
Statement 1 is INCORRECT, but statement 2 is CORRECT.0
Question 10
1 / -0

In the electric circuit shown in the figure, the effective resistance of two 8 $$\Omega$$ resistors in the combination is :

A
$$4 \Omega$$

B
$$16 \Omega$$

C
$$2 \Omega$$

D
$$ 12 \Omega$$

Solution

In this case, the 8 ohms resistances are connected in parallel.
So the effective resistance ($$R_{eff}$$) is given as:
$$\dfrac{1}{R_{eff}}=\dfrac { 1 }{ 8 } +\dfrac { 1 }{ 8 } =\dfrac { 1 }{ 4 }$$.
Hence, the effective resistance of two 8 ohms resistors in the combination is given as $$ 4 $$ ohm.