Electricity Test

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Electricity Test - 71

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Weekly Quiz Competition

Question 1
1 / -0

If you are provided three resistances $$2 \Omega, 3 \Omega$$ and $$6 \Omega$$. How will you connect them so as to obtain the equivalent resistance of $$4\Omega$$

A

B

C

D
None of these
Question 2
1 / -0

A nichrome wire $$50\ cm$$ long and one square millimeter cross-section carries a current of $$4A$$ when connected to a $$2V$$ battery. The resistivity of nichrome wire in ohm meter is

A
$$1\times 10^{-6}$$

B
$$4\times 10^{-7}$$

C
$$3\times 10^{-7}$$

D
$$2\times 10^{-7}$$

Solution

Given,
$$V=2\ volt\\l=50\ cm=50\times 10^{-2}\ cm\\A=1\ mm^2\ =10^{-6}\ m^2\\i=4\ A\\ \rho=?$$
According to Ohm's law,
$$R=\dfrac Vi$$............(1)
Resistance or any resistor can also be written as
$$R=\rho \dfrac{l}{A}$$............(2)

From equation 1 and 2,
$$R=\dfrac {V}{i}=\rho \dfrac {l}{A}$$

$$\Rightarrow \dfrac {2}{4}=\rho \dfrac {50\times 10^{-2}}{( 10^{-6})}$$

$$\Rightarrow \rho =1\times 10^{-6}\Omega m$$
Question 3
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The resistors of resistances $$2\ \Omega, 4\ \Omega$$ and $$8\ \Omega$$ are connected in parallel, then the equivalent resistance of the combination will be:

A
$$\dfrac{8}{7}\ \Omega$$

B
$$\dfrac{7}{8}\ \Omega$$

C
$$\dfrac{7}{4}\ \Omega$$

D
$$\dfrac{4}{9}\ \Omega$$

Solution

Equivalent of the parallelly connected resistance is given as,
$$\dfrac {1}{R_{eq}}=\dfrac {1}{2}+\dfrac {1}{4}+\dfrac {1}{8}=\dfrac {4+2+1}{8}$$
$$\Rightarrow R_{eq}=\dfrac {8}{7}\Omega$$.
Question 4
1 / -0

There are $$n$$ similar conductor each of resistance $$R$$. The resultant resistance comes out to be $$x$$ when connected in parallel. If they are connected in series, the resistance comes out to be

A
$$x/n^{2}$$

B
$$n^{2}x$$

C
$$x/n$$

D
$$nx$$

Solution

In parallel, combination of n resistance of value R each , Equivalent resistance $$x=\dfrac{R}{n}\implies R=nx$$
In series, $$R_{eq}= R+R+R.... n$$ times $$=nR=n(nx)=n^2x$$
Question 5
1 / -0

Two bulbs, one of 50 watt and another of 25 watt are connected in series to the mains. The ratio of the currents through them is

A
$$2 : 1$$

B
$$1 : 2$$

C
$$1 : 1$$

D
Without voltage, cannot be calculated

Solution

IN series combination, same amount of current flows through elements
The bulbs are in series, hence they will have the same current through them.
Question 6
1 / -0

An electric iron draws 5 amp, a TV set draws 3 amp and refrigerator drawn 2 amp from a 220 volt main line. The three appliances are connected in parallel. If all the three are operating at the same time, The fuse may be of

A
$$20 $$ amp

B
$$5$$ amp

C
$$15$$ amp

D
$$10$$ amp

Solution

Given,
Current in iron $$i_1=5\ A$$
Current in TV $$i_2=3\ A$$
Current in refrigerator $$i_3=2\ A$$
Total load current $$i=i_1+i_2+i_3= 5+3+2 = 10$$
The current capacity of a fuse wire should be slightly greater than the total rated load current.
Question 7
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Two bulbs of $$100 W$$ and $$200 W$$ working at $$220$$ volt are joined in series with 220 volt supply. Total power consumed wll be approximately.

A
$$65$$ watt

B
$$33$$ watt

C
$$300$$ watt

D
$$100$$ watt

Solution

The two bulbs of 100W and 200W are working at 220V are joined in series with 220V supply
Thus ,$$P_{s}=\dfrac{P_{1}P_{2}}{P_{1}+P_{2}}=\dfrac{100\times 200}{100+200}=\dfrac{200}{3}\approx 65watt$$
Question 8
1 / -0

If a high power heater is connected to electric mains then the bulbs in the house become dim, becuase there is a

A
Current drop

B
Potential drop

C
No current drop

D
No Potential drop

Solution

Brightness depend on power of bulb and
power $$\propto i^2$$
If a high power heater is connected to electric mains then the bulbs in the house bcome dim, becuase there is a change in voltage across bulb, which result in current drop.
Question 9
1 / -0

A $$60 \ Watt$$ bulb operates on $$220 \ V$$ supply. The current flowing through the bulb is

A
$$\dfrac{11}{3} \ amp$$

B
$$\dfrac{3}{11} \ amp$$

C
$$3 \ amp$$

D
$$6 \ amp$$

Solution

The power rated on the bulb $$P = 60 \ W$$ ,it operates on $$V = 220 \ V$$ supply .

Power is given by $$P=Vi$$

Hence,
$$\Rightarrow i = \dfrac{P}{V} = \dfrac{60}{220} = \dfrac{3}{11} \ amp $$
Question 10
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Electric radiator which operates at 225 volts has resistance of 50 ohms. Power of the radiator is approximately

A
$$1000$$ W

B
$$450$$ W

C
$$750$$ W

D
$$1500$$ W

Solution

It is known that power = $$P=\dfrac{V^2}{R}$$
$$P=\dfrac{225^2}{50}=1012.5\approx 1000\,W$$
so Power of the radiator is approximately 1000 W