Current Electricity Test

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Current Electricity Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

Four conductors of resistance 4, 3, 9 and 6 ohm are connected in AB,BC, CD and DA arms of a Wheatstone bridge. The bridge can be balanced by connecting.

A
1) 6 ohm in series with 3 ohm conductor

B
2) 4 ohm in parallel with 6 ohm conductor

C
3) 3 ohm in series with 3 ohm conductor

D
4) 5 ohm in series with 6 ohm conductor

Solution

To make it a wheatstone bridge $$\dfrac { { R }_{ AB } }{ { R }_{ AD } } =\dfrac { { R }_{ BC } }{ { R }_{ CD } } $$

Verification of options.

A) $$\Rightarrow { R }_{ BC }=9\Omega $$

$$\Rightarrow \dfrac { 2 }{ 3 } \neq 1$$

B) $$\Rightarrow { R }_{ AD }=\dfrac { 6\times 4 }{ 10 } =2-4$$

$$\Rightarrow \dfrac { 1 }{ 0.6} \neq 1/3$$

C) $$\Rightarrow { R }_{ BC }=6 $$

$$\Rightarrow \dfrac { 2 }{ 3 } =\dfrac { 2 }{ 3 } $$

$$=$$ 2/3

$$\therefore$$ Option C is correct
Question 2
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A cell of emf 2 V and negligible resistance is connected in series with a resistance of 5 ohm, and a potentiometer wire of resistance 10 ohm. The potential drop per cm if the length of the potentiometer wire is 10 m. The emf of cell which is balanced by 750 cm long wire is :

A
$$\dfrac{1}{1500}\dfrac{V}{cm},0.5V$$

B
$$\dfrac{1}{750}\dfrac{V}{cm},1V$$

C
$$\dfrac{1}{1500}\dfrac{V}{cm},0.6V$$

D
$$\dfrac{1}{250}\dfrac{V}{cm},1V$$

Solution

Part I

$$\Rightarrow IR=E$$

$$I=\dfrac { 2 }{ 15 } $$

P.D across potentiometer $$=$$ I x 10

$$=\dfrac { 2 }{ 15 }\times 10$$

$$=4/3$$ V

$$\dfrac { \Delta V }{ l } =\dfrac { 4/3 }{ 250 } V/cm$$

$$=\dfrac {1 }{ 750 }V/cm $$

Part 2

$$\Rightarrow { E }_{ 1 }={ IR }_{ AB }$$

$${ E }_{ 1 }=\dfrac { 2 }{ 15 } .r{ l }_{ AB }$$

$${ E }_{ 1 }=1V$$
Question 3
1 / -0

In a potentiometer using two cells in series gives a balance length of $$600cm$$. When the same cells are connected opposing each other then balance length is $$100cm$$. The ratio of emf of larger to smaller cell is:

A
$$7 : 5$$

B
$$5 : 7$$

C
$$6 : 1$$

D
$$1 : 6$$

Solution

Case 1

$$\Rightarrow \left( { E }_{ 1 }+{ E }_{ 2 } \right) =I\times r\left( 600 \right) $$

Case 2

$$\Rightarrow \left( { E }_{ 1 }-{ E }_{ 2 } \right) =I\times r\times100$$

$$\Rightarrow \dfrac { { E }_{ 1 }+{ E }_{ 2 } }{ { E }_{ 1 }-{ E }_{ 2 } } =\dfrac { 600 }{ 100 } $$

$${ E }_{ 1 }+{ E }_{ 2 }={ 6E }_{ 1 }-{ 6E }_{ 2 }$$

$$7{ E }_{ 2 }=5{ E }_{ 1 }$$

$$\Rightarrow \dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =7/5$$
Question 4
1 / -0

In a potentiometer, the balance length with standard cadmium cell is 509 cm. The emf of a cell which when connected in the place of the standard cell gave a balanced length of 750 cm is (emf of the standard cell is 1.018V):

A
1.5V

B
0.5V

C
1.08V

D
1.2V

Solution

Let the Emf of cell be E

$$\Rightarrow { E }_{ c }=I \times r\times { l }_{ 1 }\left[ I_{ 1 }=509cm \right] $$

r - resistance per unit length of wire

$${ E }=I \times r\times { l }_{ 2 }\left[ I_{ 2 }=750cm \right] $$

$$\Rightarrow \dfrac { { E }_{ c } }{ E } =\dfrac { { l }_{ 1 } }{ { l }_{ 2 } } =\dfrac { 509 }{ 750 } $$

$$\Rightarrow \dfrac { 1.018 } { E } =\dfrac { 509 }{ 750 } $$

$$E=1.018\times \dfrac { 750 }{ 509 } =1.5V$$
Question 5
1 / -0

A 5$$^{0}$$C rise is temperature is observed in a conductor by passing a current, when the current is doubled, the rise in temperature will be nearly

A
10$$^{0}$$ C

B
16$$^{0}$$ C

C
20$$^{0}$$ C

D
12$$^{0}$$ C

Solution

$$P={ I }^{ 2 }R$$
doubling I implies 4 times the power
$$\Rightarrow \Delta T\alpha P$$
$$ \therefore$$ 4 times power implies 4 times $$ \Delta T$$
$$4\times { 5 }^{ 0 }C$$
$$=20^{ 0 }C$$
Question 6
1 / -0

A,B,C and D are four coils of wires of 2, 2, 2 and 3 ohm resistances respectively and are arranged to form a Wheatstone bridge.The resistance which the coil 'D' must be shunted in order that the bridge may be balanced is:

A
4$$\Omega $$

B
6$$\Omega $$

C
3$$\Omega $$

D
8$$\Omega $$

Solution

Wheatstone condition

$$ \dfrac{R_{A}}{R_{C}} = \dfrac{R_{B}}{R_{D}}$$

$$ R_{B} = R_{D}$$

$$ R_{B} = \dfrac{3 \times R_{S}}{3 + R_{S}}$$

$$ 6 + 2R_{S} = 3R_{S}$$

$$ R_{S} = 6 \Omega$$
Question 7
1 / -0

A battery has four cells in parallel, each has an e.m.f $$1.5V$$ and internal resistance $$0.8 \Omega$$. The current delivered by it to a load of $$2.8 \Omega$$ is

A
0.2A

B
0.4A

C
0.5A

D
0.6A

Solution

$${ E }_{ eff }=E=1.5V$$

$$r_{ eff }=\dfrac { 0.8 }{ 4 } =0.2V$$

$$\Rightarrow E_{ eff }=I\left[ R+{ r }_{ eff } \right] $$

$$1.5=I\left[ 2.8+0.2 \right] $$

$$I=\dfrac { 1.5 }{ 3 } =0.5A$$
Question 8
1 / -0

Two batteries of different emf and internal resistances connected in series with each other and with an external load resistor. The current is 3.0 A. When the polarity of one battery is reversed, the current becomes 1.0 A. The ratio of the emf of the two batteries is :

A
$$2.5 : 1$$

B
$$2 : 1$$

C
$$3 : 2$$

D
$$1 : 1$$
Solution
Hint: By Ohm’s Law, total current $$I = \dfrac{V}{{{R_{eq}}}}$$, where $$V$$ is the total voltage and $${R_{eq}}$$is the equivalent resistance.
Explanation for correct answer:

Step 1: Find the ratio of emfs by ohm’s law.
- When two batteries are connected in series,
$${I_1} = \dfrac{{{E_1} + {E_2}}}{{R + {r_1} + {r_2}}} = 3A$$ ----------(1)
- When the polarity of a battery is changed,
$${I_2} = \dfrac{{{E_1} - {E_2}}}{{R + {r_1} + {r_2}}} = 1A$$ ----------(2)
- From equations (1) and (2),
$$\dfrac{{{E_1} + {E_2}}}{{{E_1} - {E_2}}} = \dfrac{3}{1}$$

$$ \Rightarrow \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{2}{1}$$

So, the ratio of the emfs of the two batteries is $$2:1$$.
Question 9
1 / -0

In a potentiometer whose wir e resistance is 10$$\Omega $$ the potential fall per cm is V volts. To reduce it to V/4 Volt cm$$^{-1}$$, the resistance that must be connected in series with the potentiometer wire is

A
40$$\Omega $$

B
30$$\Omega $$

C
20$$\Omega $$

D
10$$\Omega $$

Solution

Let the length and resistance of the wire be L and $$ 10 \Omega$$

$$ \dfrac{\Delta V}{l} = \dfrac{V}{1}$$

$$ \Delta V = Vl$$
Emf of cell $$ = Vl/4$$
Later

$$ \dfrac{\Delta V}{l} = \dfrac{V}{4}$$

of four parts of potential, 1 part is across wire and other 3 parts are at R.
R must be 3 times of Resistance of wire
$$ R = 3 \times 10 = 30 \Omega$$
Question 10
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Three unequal resistors in parallel are equivalent to a resistance $$1\ \Omega$$. If two of them are in the ratio $$1 : 2$$ and if no resistance value is fractional, the largest of the three resistances in $$\Omega$$ is:

A
$$4$$

B
$$6$$

C
$$8$$

D
$$12$$

Solution

Let $$R_1, R_2,$$ and $$R_3$$ be the three resistances connected in parallel.
$$R_{p}$$ is equivalent resistance in parallel combination.
Hence,
$$\dfrac {1}{R_{p}} = \dfrac {1}{R_{1}} + \dfrac {1}{R_{2}}+\dfrac {1}{R_{3}}$$
Let, $$R_{2} = 2R_{3}$$
Therefore,
$$\dfrac {1}{R_{p}} = \dfrac {1}{R_{1}} + \dfrac {1}{2R_{3}}+\dfrac {1}{R_{3}}$$

$$\dfrac {1}{R_{p}} = \dfrac {1}{R_{1}} + \dfrac {3}{2R_{3}}$$

$$R_p = 1\ \Omega$$ ; given

$$\dfrac {1}{R_{1}} = 1- \dfrac {3}{2R_{3}}$$

$$\therefore R_{1} = \dfrac {2R_{3}}{(2R_{3}-3)}$$

If $$R_{3} = 3\Omega$$, $$R_{1} = 2\Omega$$ and $$R_{2} = 6\Omega$$
$$\therefore$$ Largest resistance $$= 6\Omega $$