Electricity Test

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Electricity Test - 75

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Question 1
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Directions For Questions

Three resistors $${ R }_{ 1 },{ R }_{ 2 },{ R }_{ 3 }$$ are to be combined to form an electrical circuit as shown in the figure. It is found that when $${ R }_{ 1 },{ R }_{ 2 },{ R }_{ 3 }$$ are put respectively in positions A, B, C the effecitve resistance of the circuit is $$70\Omega$$. When $${R}_{2},{R}_{3}$$ and $${R}_{1}$$ are put respectively in position A, B, C the effective resistance is $$35\Omega$$ and when $${R}_{3},{R}_{1}$$ and$${R}_{2}$$ are respectively put in the position A, B, C the effective resistance is $$42\Omega$$

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If $${ R }_{ 1 },{ R }_{ 2 },{ R }_{ 3 }$$ are all connected in parallel, the effective resistance will be:

A
$$37\Omega$$

B
$$7\Omega$$

C
$$50.7\Omega$$

D
$$8.57\Omega$$
Question 2
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Two wire of same metal have same length but their cross-sections are in the ratio 3:1. They are joined in series. The resistance of thick wire is $$10\Omega$$. The total resistance of combination will be

A
$$10\Omega$$

B
$$20\Omega$$

C
$$40\Omega$$

D
$$100\Omega$$

Solution
Question 3
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Three equal resistors connected in series across a source of emf together dissipate $$10$$ watt. If the same resistors are connected in parallel across the same emf, then the power dissipated will be:

A
$$10\ watt$$

B
$$30\ watt$$

C
$$\dfrac{10}{3}\ watt$$

D
$$90\ watt$$

Solution
Question 4
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For a circuit shown in the figure, the total current in the circuit is:

A
$$0.1\ A$$

B
$$0.2\ A$$

C
$$18\ A$$

D
$$1\ A$$

Solution

Given,
Resistance of each resistors, $$R = 1\ \Omega$$
$$R_1 = R_2 = R_3 = R$$
Voltage across the battery, $$V = 6\ V$$
Equivalent resistances of 3 resistors connected in parallel
$$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$$

$$\dfrac{1}{R_{eq}} = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} = \dfrac{3}{R} =3$$

$$R_{eq} = \dfrac{1}{3}\ \Omega$$

Current, $$I = \dfrac{V}{R} = \dfrac{6}{1}\times 3$$
$$I = 18\ A$$
Question 5
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A wire of resistance $$12\Omega$$ per metre is bent to form a complete circle of radius $$10\ cm$$. The resistance between its two diametrically opposite points $$A$$ and $$B$$ as shown in the figure is

A
$$6\Omega$$

B
$$0.6\pi \Omega$$

C
$$3\Omega$$

D
$$4\Omega$$
Question 6
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The equivalent resistance between points $$A$$ and $$B$$ is

A
$$\dfrac {91}{2}\Omega$$

B
$$\dfrac {65}{2}\Omega$$

C
$$\dfrac {45}{2}\Omega$$

D
$$\dfrac {5}{2}\Omega$$

Solution
Question 7
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Calculate the potential difference between, points A and B current flowing in $$10\Omega $$ resistor in the part of the network below:-

A
$$20 V,2 A$$

B
$$50 V, 1 A$$

C
$$40 V, 1 A$$

D
$$30 V,1 A$$

Solution
Question 8
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A wire of resistance R divided in 10 equal parts. These parts are connected in parallel. Then resistance equivalent is

A
$$0.01 R$$

B
$$0.1 R$$

C
$$10 R$$

D
$$100 R.$$

Solution

Resistance of a wire is proportional to its length. The wire is cut into 10 equal parts. So, resistance of each part =$$\dfrac{R}{10}$$

Then all 10 wires are connected in parallel. Then the equivalent resistance is

$$ \dfrac{1}{{{R}_{eq}}}=\dfrac{1}{\dfrac{R}{10}}+\dfrac{1}{\dfrac{R}{10}}+.........(10\,\,times) $$

$$ \dfrac{1}{{{R}_{eq}}}=\dfrac{100}{R} $$

$$ {{R}_{eq}}=\dfrac{R}{100} $$

$$ {{R}_{eq}}=0.01R $$
Question 9
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A metal wire of resistance R is cut into three equal pieces which are then connected side by side to form a new wire, the length of which is equal to one third of the original length. The resistance of this new wire is:-

A
$$R$$

B
$$3R$$

C
$$\cfrac{R}{9}$$

D
$$\cfrac{R}{3}$$

Solution

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Question 10
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Find the potential differences across the $$24\Omega $$

A
$$48 volts$$

B
$$2 volts$$

C
$$4 volts$$

D
$$1 volts$$

Solution

Let the equivalent resistance consisting of six resistances in parallel be $$R_{eq}.$$
$$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}+\dfrac{1}{R_5}+\dfrac{1}{R_6}$$

$$\Rightarrow \dfrac{1}{R_{eq}}=\dfrac{1}{48}+\dfrac{1}{48}+\dfrac{1}{24}+\dfrac{1}{6}+\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{1+1+2+8+12+24}{48}$$

$$\Rightarrow \dfrac{1}{R_{eq}}=\dfrac{48}{48}$$

$$\Rightarrow R_{eq}=1 \Omega$$

Let the voltage across $$R_{eq}$$ be $$V_{eq}.$$
Applying Ohm's law for $$R_{eq}$$
$$V_{eq}=IR_{eq}$$
$$V_{eq}=(2 \ A) \times (1 \ \Omega)=2 \ V$$

The voltage $$V_{eq}=2 \ V$$ also appears across $$R_1$$, $$R_2, R_3 . . . R_6.$$
$$\therefore$$ Voltage across $$24 \Omega$$ is $$V_{24 \ \Omega}=V_{eq}=2 \ V$$

Answer is option (B).