Current Electricity Test

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Current Electricity Test - 43

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Question 1
1 / -0

If a man has five resistors each of value $$\dfrac { 1 }{ 5 } \Omega $$ then is the maximum resistance he can obtain by connecting them, is :

A
$$1\Omega $$

B
$$5\Omega $$

C
$$\dfrac { 1 }{ 2 } \Omega $$

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

Solution

Given,
$$R=\dfrac15\ Omega$$ total number of resistance, n=5
For maximum effective resistance, all resistors should connect in series.
For series combination
$$R_{eq}=R_1+R_2+ ..........+R_n\\R_{eq}=nR$$ (All resistor have same resistance)

On putting value of $$n$$ and $$R$$
$$R_{eq}=5\times \dfrac15=1\ \Omega$$

Hence maximum resistance is 1 ohm.
Question 2
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The length of a wire of a potentiometer is 100 cm, and the emf of its standard cell is $$E$$ volt. It is employed to measure the emf of a battery whose internal resistance is $$0.5\Omega$$. If the balance point is obtained at $$l=30$$ cm from the positive end, the emf of the battery is

A
$$\dfrac {30E}{100}$$

B
$$\dfrac {30E}{100.5}$$

C
$$\dfrac {30R}{(100-0.5)}$$

D
$$\dfrac {30(E-0.5i)}{100}$$

Solution

Let $$V$$ be the potential across balance point and one end of wire.
Hence acccording to the principle of potentiometer
$$V\propto l$$.
Also if a cell of emf $$E$$ is employed in the circuit between the ends of potentiometer wire of length $$L$$ then
$$E\propto L$$.
Therefore,
$$\dfrac {V}{E}=\dfrac {l}{L}$$

$$V=\dfrac {l}{L}E=\dfrac {30}{100}E=\dfrac {30E}{100}$$
Question 3
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The arm PQ can revolve with uniform speed continuously about P round the circular uniform potentiometer track XYZ. The voltage between RS will vary with respect to time :

A
Sinusoidally

B
Linearly

C
Rectangularly

D
Like saw tooth

Solution

As the voltage between $$RS$$ $$(V) \propto $$ Resistance between $$P$$ and $$Z$$

Arm $$PQ$$ is revolving uniformly so resistance between $$P$$ and $$Z$$ is changing uniformly

Let resistance $$XZ = R$$
So we can write resistance between $$P$$ and $$Z$$ as function of time : $$r = R - kt$$
where $$k$$ is some constant and $$r$$ is resistance between $$P$$ and $$Z$$ which is a equation of decreasing straight line
As arm comes to point $$Z$$ resistance $$r = 0$$ and again it starts from $$R$$ as it comes to point $$X$$.
Question 4
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There is a fixed potential difference between the two ends of a potentiometer. Two cells are connected in series in such a way that in one arrangement they help each other where as in the second arrangement they oppose each other. The balance point for these two combinations is obtained at 120 cm and 60 cm length respectively. The ratio of the emf of the cell is:

A
2 :1

B
3 :1

C
1 : 1

D
4: 1

Solution

There is a fixed potential difference between the two ends of a potentiometer. Two cells are connected in series in such a way that in one arrangement they help each other where as in the second arrangement they oppose each other. The balance point for these two combinations is obtained at 120 cm and 60 cm length respectively. The ratio of the emf of the cell is
$$E=\dfrac{x_1+x_2}{x_2}=\dfrac{120+60}{60}=3:1$$
Question 5
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A resistance of $$0.01\ \Omega$$ is connected in parallel with a resistance of $$1\ k\Omega$$ The resistance of the combination will be:

A
$$1000\ \Omega$$

B
$$10\ \Omega$$

C
$$1\ \Omega$$

D
$$0.01\ \Omega$$

Solution

initial velocity, $$u = 10\ m/s$$
Resistances,
$$R_1 = 0.01\ \Omega$$
$$R_2 = 1K\ \Omega = 1000\ \Omega$$
for 2 resistances, connected in parallel equivalent resistance, $$R_{eq}$$

$$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{1}{0.01} + \dfrac{1}{1000}$$

$$\dfrac{1}{R_{eq}} = \dfrac{100000 + 1}{1000}$$

$$R_{eq} = \dfrac{1000}{100001} = 0.009999$$

$$R_{eq} \sim 0.01\ \Omega$$
Question 6
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Three equal resistors connected in series across a source of e.m.f. together dissipate 10 watt of power. What would be the power dissipated if the same resistors are connected in parallel across the same source of e.m.f.?

A
90 watts

B
80 watts

C
70 watts

D
75 watts

Solution

When connected in series,
$$[R_{eq}= R + R + R]$$
$$[R_{eq}= 3 \times R]$$
Power dissipation will be,
$$P=\dfrac{V^2}{R_{eq}}$$
i.e. $$ P=\dfrac{V^2}{(3R)}$$

Given,$$[P = 10]$$
Now, when they are connected in parallel
$$\dfrac{1}{R_{eq}} = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R}$$
$$ R_{eq} = R/3$$

$$[P = V^2/R_{eq}]$$

$$ P=3V^2/R$$

From above, $$[P = 90]$$
Question 7
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Three resistors each of $$10 \Omega$$ are connected in series to a battery of potential difference 150 v. The current flowing through it is ..... A.

A
45

B
5

C
15

D
20

Solution

$$V=IR$$
Current is given by,
$$V=I\times{R_{eq}}$$
$$I=\frac{V}{R_{eq}}$$

The three resistance are in series. So, the equivalent resistance will be:
$$R_{eq}=R_1+R_2+R_3$$
$$R_{eq}=10+10+10$$
$$=30\ \Omega$$

Substitute the values in the given formula:
$$I=150/30$$
$$I=5$$ $$A$$
Question 8
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The colour code of a resistor is brown, black and brown. Then the value of resistance is .....

A
$$10 \Omega$$

B
$$100 m \Omega$$

C
$$0.1 k\Omega$$

D
100 + 5%

Solution

Answer is C.

Carbon-composition and carbon film resistors are too small to have the resistance value printed on their housings. Therefore, bands of color are used to represent the resistance value.
The first and second band represents the numerical value of the resistor, and the color of the third band specify the power-of-ten multiplier. The color bands are always read from left to right starting with the side that has a band closer to the edge.
Hence, for a brown, black, brown color code of a resistor. The resistance value is 100 ohms, that is, $$0.1k\Omega $$.
Question 9
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The given figure shows a network of currents. The current $$i$$ is

A
23 A

B
22 A

C
20 A

D
25 A

Solution

Here, consider the above part of circuit as shown in figure and applying Kirchhoff's current law we obtain

$$I_1 + 5 -I_2 = 0$$

$$I_1 =I_2 - 5$$..................................................(1)

$$I_2 + 15 - I_3 = 0$$

$$I_2 = I_3 -15 $$..........................................(2)

$$I_3 + 3 -I_4 = 0$$

$$I_3 = I_4 - 3 $$..........................................(3)

and $$I_4 - i - I_1 = 0$$

$$I_4 = I_1 + i$$...............................................(4)

now using values from equations(2), (3) and (4) into (1)we get

$$I_1 = I_1+i-23$$

$$\therefore i=23$$ A
Question 10
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Which of the following graphs represents a ohmic conductor?

A

B

C

D

Solution

$$\bullet$$ An ohmic conductor must satisfy ohm's law i.e. $$V=iR$$,
Where $$V$$ is potential difference across it, $$I$$ is current flowing through it and $$R$$ is the resistance of the conductor, which is constant.
$$\bullet$$ Therefore $$V\text{-}I$$ graph of a ohmic conductor will be a straight line.
$$\bullet$$ Hence, Option $$(C)$$ is correct