Current Electricity Test

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Current Electricity Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

A student carries out an experiment and plots the V-I graph of three samples of nichrome wire
with resistances R1, R2 and R3 respectively (Figure.12.5). Which of the following is true?

A
$$R_1 = R_2 = R_3$$

B
$$R_1 > R_2 > R_3$$

C
$$R_3 > R_2 > R_1$$

D
$$R_2 > R_3 > R_1$$

Solution

Hint: Resistance can be obtained from the slope of the V-I graph.

Step 1: Comparing the slopes of the three samples.

Slope for $$R_1$$ is maximum and slope for $$R_3$$ is the minimum.

$$(Slope)_{R_1} > (Slope)_{R_2} > (Slope)_{R_3}$$

Step 2: Comparison of the resistances of the samples from the slopes.

We know that,
Resistance of the wire $$= \dfrac{1}{Slope \ of \ the \ I-V graph}$$

Therefore, $${R_3} > {R_2} > {R_1}$$

Final Step : The correct order of resistance is $${R_3} > {R_2} > {R_1}$$
Question 2
1 / -0

Find the currents $$I_1$$ and $$I_2$$ for the circuit shown in given figure.

A
$$I_1=-6 A$$ and $$I_2=9 A$$

B
$$I_1=6 A$$ and $$I_2=9 A$$

C
$$I_1=-6 A$$ and $$I_2=-9 A$$

D
$$I_1=-9 A$$ and $$I_2=6 A$$

Solution

Consider node $$a$$, using Kirchhoff's current law
(using $$+$$ for $$I_{in}$$ $$-$$ for $$I_{out}$$)
$$ \sum I_{at\, a\, node} = 0$$

$$-3+2+I_{ab} = 0$$
$$ I_{ab} = 1 A$$

Consider node $$b$$,
$$ I_{ba}+7+I_1 = 0$$ ($$ I_{ab} = -I_{ba}$$)
$$-1-7 +I_1=0$$
$$ I_1 = -6A$$

Consider node $$d$$ ,
$$I_{dc} +4-1 = 0$$
$$I_{dc} = -3 A$$

Consider node $$c$$,
$$-I_2+6+I_{cd} = 0$$
$$-I_2+6+3 = 0$$
$$I_2 = 9A$$
$$ option\, A\, is\, correct$$
Question 3
1 / -0

A potentiometer wire of length $$100 cm$$ has a resistance of $$10$$ ohm. It is connected in series with a resistance and an accumulator of emf $$2V $$ and of negligible internal resistance. A source of emf of $$10 mV$$ is balanced against a length of $$40 cm$$ of the potentiometer wire. What is the value of external resistance?

A
$$890 \Omega$$

B
$$600 \Omega$$

C
$$650 \Omega$$

D
$$790 \Omega$$

Solution

Let, R be the resistance external resistance. The current through the circuit is
$$i=\dfrac{2}{10+R}$$

We know, $$x=\dfrac{V}{l}=\dfrac{2\times10}{(R+10)}\dfrac{1}{100}$$
where, $$x$$ is the resistance of the wire.

Also, A source of emf of 10 mV is balanced against a length of 40 cm of the potentiometer wire.
Hence, $$V_{1}=xl$$
$$10\times10^{-3}=\dfrac{2\times10}{(R+10)}\times\dfrac{40}{100}$$
$$R+10=\dfrac{8}{10\times10^{-3}}$$
$$\Rightarrow R+10=800$$
$$\therefore R=790 \Omega$$
Question 4
1 / -0

Which of the following statement is true regarding the Kirchoffs laws?

A
The junction rule is a statement of conservation of energy and the loop rule is a statement of conservation of charge

B
The junction rule as well as the loop rule are statements of conservation of charge.

C
The junction rule as well as the loop rule are statements of conservation of energy.

D
The junction rule is a statement of conservation of charge and the loop rule is a statement of conservation of energy.
Question 5
1 / -0

An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

A
1A

B
2A

C
4A

D
5A

Solution

Hint : A fuse wire should be able pass that amount of current which the appliance requires

Step 1 : Find the maximum current that the appliance requires
Given, Power $$ = 1KW= 1000W$$ & Voltage $$= 220\ V$$
$$P= V \times I$$
So, $$I=\dfrac{P}{V}$$
$$I=\dfrac{1000}{220}= 4.545 A$$
The fuse wire should be able to pass at least $$4.545\ A$$ current.
So, the rating should be $$5A$$.
Question 6
1 / -0

Electrical resistivity of a given metallic wire depends upon:

A
its length

B
its thickness

C
its shape

D
nature of the material

Solution

Hint : Resistivity is an intrinsic property of a metal.

Step 1 : Explanation of Resistivity
Resistivity of a metallic wire does not depend on its geometry but depends on the nature of the material of the wire. Hence option $$(D)$$ is correct.
Question 7
1 / -0

A strip of copper and another of germanium are cooled from room temperature to 80 K. The resistance of

A
Each of these increases

B
Each of these decreases

C
Copper strip increases and that of germanium decreases

D
Copper strip decreases and that of germanium increases

Solution

Copper is a conductor while germanium is a semiconductor.
In case of conductor resistance decreases with temperature while in case of semi-conductor resistance increases with temperature. Hence the resistance of copper strip decreases and that of germanium increases.
Question 8
1 / -0

When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, resistance between any two of diametrically opposite points will be:

A
R/8

B
4R

C
R/2

D
R/4

Solution

$$\textbf{Step 1: Draw resistance structure}$$ $$\textbf{[Ref. Fig.]}$$
Resistance of both part will be $$R/2$$ because wire is uniform.

$$\textbf{Step 2: Equivalent resistance between A and B}$$,
Each resistor has both terminals connected to both terminals of the others so both are in parallel-

$$R_{eq \ parallel} = \dfrac{R_1R_2}{R_1+R_2}$$

$$\Rightarrow R_{AB} = \dfrac{\dfrac{R}{2}\times \dfrac{R}{2}}{\dfrac{R}{2} + \dfrac{R}{2}} = \dfrac{R}{4}$$

Hence equivalent resistance between any two of diametrically opposite points will be $$\dfrac{R}{4}$$

Option $$(D)$$ correct.
Question 9
1 / -0

Combined resistance is the sum of separate resistances provided that the various conductors are connected in______ .

A
Series

B
Parallel

C
Both series and parallel

D
Data insufficient

Solution
Question 10
1 / -0

Two wires of resistance $$R_1$$ and $$R_2$$ are joined in parallel. The equivalent resistance of the combination is:

A
$$R_1R_2/(R_1+R_2)$$

B
$$R_1+R_2$$

C
$$R_1 \times R_2$$

D
$$R_1R_2$$

Solution

The equivalent resistance of N resistors connected in parallel connection is given by,
$$ \dfrac{1}{R^{}_{Equivalent}} = \dfrac{1}{R^{}_{1}} + \dfrac{1}{R^{}_{2}} + \dfrac{1}{R^{}_{3}} + . . . + \dfrac{1}{R^{}_{N}}$$
Hence,
$$ \dfrac{1}{R^{}_{12}} = \dfrac{1}{R^{}_{1}} + \dfrac{1}{R^{}_{2}} $$
or,
$$ R^{}_{12} = \dfrac{R^{}_{1}R^{}_{2}}{R^{}_{1} + R^{}_{2} } $$

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Current Electricity Test - 41

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Current Electricity Test - 41

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Question 1

$$R_1 = R_2 = R_3$$

$$R_1 > R_2 > R_3$$

$$R_3 > R_2 > R_1$$

$$R_2 > R_3 > R_1$$

Solution

Question 2

$$I_1=-6 A$$ and $$I_2=9 A$$

$$I_1=6 A$$ and $$I_2=9 A$$

$$I_1=-6 A$$ and $$I_2=-9 A$$

$$I_1=-9 A$$ and $$I_2=6 A$$

Solution

Question 3

$$890 \Omega$$

$$600 \Omega$$

$$650 \Omega$$

$$790 \Omega$$

Solution

Question 4

The junction rule is a statement of conservation of energy and the loop rule is a statement of conservation of charge

The junction rule as well as the loop rule are statements of conservation of charge.

The junction rule as well as the loop rule are statements of conservation of energy.

The junction rule is a statement of conservation of charge and the loop rule is a statement of conservation of energy.

Question 5

1A

2A

4A

5A

Solution

Question 6

its length

its thickness

its shape

nature of the material

Solution

Question 7

Each of these increases

Each of these decreases

Copper strip increases and that of germanium decreases

Copper strip decreases and that of germanium increases

Solution

Question 8

R/8

4R

R/2

R/4

Solution

Question 9

Series

Parallel

Both series and parallel

Data insufficient

Solution

Question 10

$$R_1R_2/(R_1+R_2)$$

$$R_1+R_2$$

$$R_1 \times R_2$$

$$R_1R_2$$

Solution

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