Current Electricity Test

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Current Electricity Test - 59

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Question 1
1 / -0

In a potentiometer experiment of a cell of emf 1.25 V gives balancing length of 30 cm. If the cell is replaced by another cell, balancing length is found to be 40 cm. What is the emf of second cell?

A
$$\simeq 1.57 V$$

B
$$\simeq 1.67 V$$

C
$$\simeq 1.47 V$$

D
$$\simeq 1.37 V$$

Solution

Given : $$E_1 = 1.25$$ V $$l_1 = 30$$ cm $$l_2 = 40$$ cm
Emf of the cell $$E\propto l$$
$$\implies$$ $$\dfrac{E_2}{E_1} = \dfrac{l_2}{l_1}$$
$$\therefore$$ $$\dfrac{E_2}{1.25} = \dfrac{40}{30}$$
$$\implies$$ $$E_2 = 1.67$$ V
Question 2
1 / -0

Four identical cells of emf $$E$$ and internal resistance $$r$$ are to be connected in series. Suppose if one of the cell is connected wrongly, the equivalent emf and effective internal resistance of the combination is

A
$$4E$$ and $$4r$$

B
$$4E$$ and $$2r$$

C
$$2E$$ and $$4r$$

D
$$2E$$ and $$2r$$

Solution

Equivalent resistance in series connection $$r_{eq} = r_1+r_2+r_3+r_4$$
$$\therefore$$ $$r_{eq} = r+r+r+r = 4r$$
Equivalent emf in series connection $$E_{eq} = E_1+E_2+E_3-E_4$$ (as one is connected wrong)
$$\therefore$$ $$E_{eq} = E+E+E-E = 2E$$
Question 3
1 / -0

Resistance of $$2\ \Omega$$ and $$3\ \Omega$$ are connected in series. If the potential difference across the $$2\ \Omega$$ resistor is $$3\ V$$, the potential difference across $$3\Omega$$ is:

A
$$4.5\ V$$

B
$$9\ V$$

C
$$3\ V$$

D
$$2\ V$$

Solution

Given,
Resistors of resistances, $$R_1=2\ \Omega$$ and $$R_2=3\ \Omega$$ are connected in series.
Potential difference across $$2\ \Omega$$ resistor, $$V_1=3\ V$$
Potential difference across $$2\ \Omega$$ resistor, $$V_2$$

According to Ohm's law,
$$V=IR$$
$$\implies I=\dfrac VR$$

$$\implies I=\dfrac {V_1}{R_1}$$

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

$$I=1.5\ A$$

We know, current in a series circuit remains constant.
$$V_2=IR_2$$
$$V_2=1.5\times 3$$
$$V_2=4.5\ V$$
Question 4
1 / -0

Two resistances are connected in series as shown in the diagram. What is the current through $$R$$?

A
$$1 A$$

B
$$2 A$$

C
$$3 A$$

D
$$4 A$$

Solution

The voltage across $$5\Omega$$ resistor is $$V=10V$$

Current,
$$I=\dfrac{V}{R}$$; according to ohm' law

$$\Rightarrow I=\dfrac{10}{5}$$

$$\Rightarrow I=2A$$

Since resistance $$5\Omega$$ and $$R$$ are in series and hence current through both resistance is the same.
Hence current through $$R$$ is also $$2A$$.
Question 5
1 / -0

Which of the following is true for carbon resistors?

A
Resistance value does not get changed by change in temperature

B
Current carrying capacity and hence power wattage are limited to about $$50$$ Watts.

C
It has low-temperature coefficient value.

D
Stability and reliability are very good.

Solution

$$A\rightarrow$$ Resistance value of carbon resistor with change in temperature it has negative thermal coefficient.
$$B\rightarrow$$ Carbon resistors are not current carrying capacitor.
$$C\rightarrow$$ Carbon has low temperature coefficient value i.e. higher the temperature, lower the resistivity.
$$D\rightarrow$$ Because of low stability of resistance value this type of resistors are not suitable.
Question 6
1 / -0

Which of the following is not the advantage of Carbon resistors?

A
It is small in size and is light in weight.

B
It has low-temperature coefficient value.

C
Using it high-value resistor can easily be made.

D
It has high power rating with a low tolerance value.
Solution
Advantages of carbon resistors are:
It is small in size and is light in weight.
It has low temperature coefficient that means small change in temperature leds to high resistivity change.
It provides greater resistance, so using this high value resistor can easy be made.
They have low to medium type power rating of range $$0.125w$$ to $$5w$$.
They have high tolerance values such as $$\pm 5\%,\pm 10\%,$$ and $$\pm 20\%$$.
Hence statement $$(D)$$ is wrong.
Question 7
1 / -0

Copper and Germanium are cooled from room temperature to $$100 K$$. Then the resistance of

A
Germanium decreases, Copper increases

B
Germanium decreases, Copper decreases

C
Germanium increases, Copper decreases

D
Germanium increases, Copper increases

Solution

As we know that resistance of a conductor increases with increase in temperature whereas that of semiconductor decreases with increase in temperature.
Copper is a very good conductor and germanium is a semiconductor. Thus resistance of copper decreases but that of germanium increases when they are cooled from room temperature to $$100K$$.
Question 8
1 / -0

A resistor of $$6 k\Omega $$ with tolerance $$10$$% and another of $$4k\Omega $$ with tolerance $$10$$% are connected in series. The tolerance of combination is about

A
$$5$$%

B
$$10$$%

C
$$12$$%

D
$$15$$%

Solution

Let the resistances are $$R_1= 6\ k\Omega$$ and $$R_2=4\ k\Omega$$.
Tolerence in the resistances : $$\Delta R_1 = 600\Omega$$ $$\Delta R_2 = 400\Omega$$
Total resistance of the combination $$R_{eq} = R_1+R_2 =6+4 =10k\Omega$$
Total tolerance in the equivalent resistance: $$\Delta R_{eq} =\Delta R_1+\Delta R_2 =600+400 = 1k\Omega $$
Hence percentage tolerance of combination $$ = \dfrac{1k\Omega}{10k\Omega} \times 100 =10$$ %
Question 9
1 / -0

The________ indicates the multiplier telling you the power of ten to which the two significant digits in color coding of the resistor must be multiplied (or how many zeros to add).

A
2nd band

B
3rd band

C
4th band

D
last band

Solution

The 3rd band indicates the multiplier telling the power of ten to which the two significant digits in color coding of the resistor must be multiplied.

Answer-(B)
Question 10
1 / -0

Red,orange,black are the color bands on resistors.Find its resistance?

A
$$23\Omega \pm20$$ %

B
$$25\Omega \pm10$$ %

C
$$32\Omega \pm20$$ %

D
$$53\Omega \pm15$$ %

Solution

For first two significant digits of resistance , we see the first two colours .
And, third color signifies the multiplier.

Here, red color means digit $$2$$ and orange color means $$3$$.
And for multiplier, black colour means multiplier of $$1$$.

Hence resistance is $$(23\times 1)=23\Omega$$

Since , nothing is said about fourth color, we can check answer from the calculated resistance.

Answer-(A)

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Current Electricity Test - 59

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Current Electricity Test - 59

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Question 1

$$\simeq 1.57 V$$

$$\simeq 1.67 V$$

$$\simeq 1.47 V$$

$$\simeq 1.37 V$$

Solution

Question 2

$$4E$$ and $$4r$$

$$4E$$ and $$2r$$

$$2E$$ and $$4r$$

$$2E$$ and $$2r$$

Solution

Question 3

$$4.5\ V$$

$$9\ V$$

$$3\ V$$

$$2\ V$$

Solution

Question 4

$$1 A$$

$$2 A$$

$$3 A$$

$$4 A$$

Solution

Question 5

Resistance value does not get changed by change in temperature

Current carrying capacity and hence power wattage are limited to about $$50$$ Watts.

It has low-temperature coefficient value.

Stability and reliability are very good.

Solution

Question 6

It is small in size and is light in weight.

It has low-temperature coefficient value.

Using it high-value resistor can easily be made.

It has high power rating with a low tolerance value.

Solution

Question 7

Germanium decreases, Copper increases

Germanium decreases, Copper decreases

Germanium increases, Copper decreases

Germanium increases, Copper increases

Solution

Question 8

$$5$$%

$$10$$%

$$12$$%

$$15$$%

Solution

Question 9

2nd band

3rd band

4th band

last band

Solution

Question 10

$$23\Omega \pm20$$ %

$$25\Omega \pm10$$ %

$$32\Omega \pm20$$ %

$$53\Omega \pm15$$ %

Solution

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