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CBSE Questions for Class 10 Physics Electricity Quiz 12 - MCQExams.com
CBSE
Class 10 Physics
Electricity
Quiz 12
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
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$$x/n^{2}$$
0%
$$n^{2}x$$
0%
$$x/n$$
0%
$$nx$$
Explanation
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 $$=nR=n(nx)=n^2x$$
If three resistors of resistance $$2\Omega, 4\Omega$$ and $$5 \Omega$$ are connected in parallel then the total resistance of the combination will be
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$$\dfrac{20}{19}\Omega$$
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$$\dfrac{19}{20}\Omega$$
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$$\dfrac{19}{10}\Omega$$
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$$\dfrac{10}{19}\Omega$$
Explanation
For parallel resistance connection we know that,
$$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{19}{20}\Rightarrow R_{eq}=\dfrac{20}{19}\Omega$$
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
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$$2 : 1$$
0%
$$1 : 2$$
0%
$$1 : 1$$
0%
Without voltage, cannot be calculated
Explanation
IN series combination, same amount of current flows through elements
The bulbs are in series, hence they will have the same current through them.
Which of the following statement is false
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Heat produced in a conductor is proportional to its resistance
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Heat produced in a conductor is proportional to the square of the current
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Heat produced in a conductor is proportional to charge
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Heat produced in a conductor is proportional to the time for which current is passed
Explanation
According to Joule's heating law -
$$H = i^{2} Rt$$
We also know that current $$i = \dfrac{q}{t}$$.
Hence $$H = \dfrac{q^{2}R}{t}$$
$$\therefore H \propto q^{2}$$
An electric heater kept in vacuum is heated continuously by passing electric current. Its temperature
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Will go on rising with time
0%
Will stop after sometime as it will lose heat to the surroundings by conduction
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Will rise after sometime and there after will start falling
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Will become constant after sometime because of loss of heat due to radiation
Explanation
Unless the heat generation rate is matched by the rate of radiation, the temperature will continue to rise.
After that, the temperature will become constant.
$$10$$ wires (same length, same area, same material) are connected in parallel and each has $$1\Omega$$ resistance, then the equivalent resistance will be
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0%
$$10\Omega$$
0%
$$1\Omega$$
0%
$$0.1\Omega$$
0%
$$0.001\Omega$$
Explanation
When $$n$$ no. of resistances are connected parallelly then
$$ \dfrac{1}{R_{eq} } = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + ................$$
$$ \dfrac{1}{R_{eq} } = \frac{n}{R}$$
$$\Rightarrow$$ $$R_{eq} = \dfrac{R}{n}$$
$$\Rightarrow$$ $$R_{eq}=\dfrac {R}{n}=\dfrac {1}{10}=0.1\Omega$$
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:
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$$10\ watt$$
0%
$$30\ watt$$
0%
$$\dfrac{10}{3}\ watt$$
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$$90\ watt$$
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
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$$1\times 10^{-6}$$
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$$4\times 10^{-7}$$
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$$3\times 10^{-7}$$
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$$2\times 10^{-7}$$
Explanation
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,
$$R=\dfrac {V}{i}=\rho \dfrac {l}{A}$$
$$\Rightarrow \dfrac {2}{4}=\rho \dfrac {50\times 10^{-2}}{( 10^{-6})}$$
$$\Rightarrow \rho =1\times 10^{-6}\Omega m$$
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 $$4\Omega$$
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0%
0%
0%
None of these
The resistors of resistances $$2\ \Omega, 4\ \Omega$$ and $$8\ \Omega$$ are connected in parallel, then the equivalent resistance of the combination will be:
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$$\dfrac{8}{7}\ \Omega$$
0%
$$\dfrac{7}{8}\ \Omega$$
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$$\dfrac{7}{4}\ \Omega$$
0%
$$\dfrac{4}{9}\ \Omega$$
Explanation
Equivalent of the parallelly connected resistance is given as,
$$\dfrac {1}{R_{eq}}=\dfrac {1}{2}+\dfrac {1}{4}+\dfrac {1}{8}=\dfrac {4+2+1}{8}$$
$$\Rightarrow R_{eq}=\dfrac {8}{7}\Omega$$.
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.
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$$65$$ watt
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$$33$$ watt
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$$300$$ watt
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$$100$$ watt
Explanation
The two bulbs of 100W and 200W are working at 220V are joined in series with 220V supply
Thus ,$$P_{s}=\dfrac{P_{1}P_{2}}{P_{1}+P_{2}}=\dfrac{100\times 200}{100+200}=\dfrac{200}{3}\approx 65watt$$
If a high power heater is connected to electric mains then the bulbs in the house become dim, becuase there is a
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Current drop
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Potential drop
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No current drop
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No Potential drop
Explanation
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.
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
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$$20 $$ amp
0%
$$5$$ amp
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$$15$$ amp
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$$10$$ amp
Explanation
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.
The electric passing through a metallic wire produces heat because of
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Collisions of conduction electrons with each other
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Collisions of the atoms of the metal with each other
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The energy released in the ionization of the atoms of the metal
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Collisions of the conduction electrons with the atoms of the metallic wires
Explanation
When
an electric
current
passes through a metallic wire
, electrons move inside the
wire
and they collide with each other during their motion which
produces heat
in the
wire
.
A $$60 \ Watt$$ bulb operates on $$220 \ V$$ supply. The current flowing through the bulb is
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$$\dfrac{11}{3} \ amp$$
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$$\dfrac{3}{11} \ amp$$
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$$3 \ amp$$
0%
$$6 \ amp$$
Explanation
The power rated on the bulb $$P = 60 \ W$$ ,it operates on $$V = 220 \ V$$ supply .
Power is given by $$P=Vi$$
Hence,
$$\Rightarrow i = \dfrac{P}{V} = \dfrac{60}{220} = \dfrac{3}{11} \ amp $$
Electric radiator which operates at 225 volts has resistance of 50 ohms. Power of the radiator is approximately
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$$1000$$ W
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$$450$$ W
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$$750$$ W
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$$1500$$ W
Explanation
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
A steel wire has a resistance twice that of an aluminium wire. Both of them are connected with a constant voltage supply. More heat will be dissipated in:
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Steel wire when both are connected in series
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Steel wire when both are connected in parallel
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Aluminium wire when both are connected in series
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Aluminium wire when both are connected in parallel
Explanation
Let the resistance of steel wire $$=R_{steel}$$
Resistance of aluminium wire $$=R_{aluminium}$$
Given,
$$ R_{steel} = 2R_{aluminium}$$
We know,
Heat dissipated, $$H=I^2R$$
Also, $$H=\dfrac{V^2}{R}$$
In series, the current $$I$$ will be the same in both the wires.
$$H=I^2R$$
$$H\propto R$$\
$$\therefore$$ More heat will dissipate in the steel wire when both are connected in series.
In parallel, the voltage $$V$$ will be the same across both the wires.
$$H=\dfrac{V^2}{R}$$
$$H\propto \dfrac 1R$$
$$\therefore$$ More heat will dissipate in the aluminium wire when both are connected in parallel.
An electric bulb marked 40 W and 200 V, is used in a circuit of supply voltage 100 v, Now its power is
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$$100 W$$
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$$40 W$$
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$$20 W$$
0%
$$10 W$$
Explanation
The power genrated is given by
$$P=\dfrac{V^{2}}{R}\Rightarrow \dfrac{P_{2}}{P_{1}}=\dfrac{V^{2}_{1}}{V^{2}_{1}}(\because R is constant )\Rightarrow \dfrac{P_{2}}{P_{1}}=\left ( \dfrac{100}{200} \right )^{2}=\dfrac{1}{4}\Rightarrow P_{2}=\frac{P_{1}}{4}=\dfrac{40}{4}=10W$$
The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated $$220 \ V$$ and $$100 \ W$$ is connected to $$(220\times 0.8) \ V$$ source, then the power would be :
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$$100\times 0.8watt$$
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$$100\times (0.8)^{2}watt$$
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Between $$100\times 0.8watt$$ and 100 watt
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Between $$100\times (0.8)^{2}watt$$ and $$100\times 0.8watt$$ watt
Explanation
Since Power $$ P = \dfrac{V^2}{R}$$
For the first case, when $$V=220 \ V$$
$$P_{1}=\dfrac{(220)^2}{R_1} $$
For the second case, when $$V=220 \times 0.8 \ V$$
$$ P_{2}=\dfrac{(220\times 0.8)^{2}}{R_{2}}$$
Now, $$\dfrac{P_{2}}{P_{1}}=\dfrac{\left ( 220\times 0.8 \right )^{2}}{(220)^{2}}\times \dfrac{R_{1}}{R_{2}}$$
$$\Rightarrow \dfrac{P_{2}}{P_{1}}=(0.8)^{2}\times \dfrac{R_{1}}{R_{2}}$$
Since $$V_2<V_1$$, voltage has been decreased and is directly proportional to the resistance, from Ohm's law.
Hence, $$R_{2}< R_{1}$$
$$\Rightarrow \dfrac{R_1}{R_2} > 1$$
$$\Rightarrow \dfrac{P_2}{P_1} > (0.8)^2$$
$$\Rightarrow P_2 > 100 \times (0.8)^2 \ W$$
Also, Since Power $$P = Vi$$
Hence$$\dfrac{P_{2}}{P_{1}}=\dfrac{\left ( 220\times 0.8 \right )i_2}{220i_1},$$
Since $$V_2<V_1$$, voltage has been decreased and is directly proportional to the current, from Ohm's law.
Hence, $$i_{2}< i_{1}$$
$$\Rightarrow \dfrac{i_2}{i_1} <1$$
So $$\dfrac{P_{2}}{P_{1}}< 0.8\Rightarrow P_{2}< (100\times 0.8) \ W$$
Hence the actual power would be between $$100 \times (0.8)^{2} \ W$$ and $$100\times (0.8) \ W$$
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
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$$6\Omega$$
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$$0.6\pi \Omega$$
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$$3\Omega$$
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$$4\Omega$$
An electric lamp is marked $$60 \ W$$, $$230 \ V$$. The cost of $$1 \ kilowatt-hour$$ of power is $$Rs \ 1.25$$. The cost of using this lamp for $$8 \ hours$$ is
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$$Rs. \ 1. 20$$
0%
$$ Rs. \ 4. 00$$
0%
$$Rs. \ 0. 25$$
0%
$$Rs. \ 0. 60$$
Explanation
Total energy consumed is given as $$E=P\times t$$
where $$P$$ is power and $$t$$ is time for which it is used.
Given, $$P=60\,W$$ and $$t=8\,h$$
$$E = 60 \times 8 \ Wh=\dfrac{60\times 8}{1000} = 0.48 \ kWh $$
Given that, cost of $$1 \ kWh$$ is $$Rs. \ 1.25$$
Hence, total cost is $$ \text{Cost}=0.48\times 1.25 = Rs. \ 0.6 $$
The current flowing through a lamp marked as $$50 \ W$$ and $$250 \ V$$ is
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$$5 \ amp$$
0%
$$2.5 \ amp$$
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$$2 \ amp$$
0%
$$0. 2 \ amp$$
Explanation
The power dissipated is given by
$$P=Vi$$,
where $$V$$=voltage across lamp, and $$i$$=current flowing through lamp
Hence,
$$i = \dfrac{P}{V} = \dfrac{50}{250} = 0.2 \ amp$$
If a $$5 \,V$$ battery is provoding $$2 \,A$$ current in a conductor then what is the resistance of conductor ?
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$$3$$ Ohm
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$$2.5$$ Ohm
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$$10$$ Ohm
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$$2$$ Ohm
Explanation
Given that:
Battery supplies a voltage of $$V=5 \ V$$
Current provided by the battery $$I=2 \ A$$
Resistance= ?
By Ohm's law,
$$V=IR$$
$$\Rightarrow R=\dfrac{V}{I}=\dfrac{5}{2}=2.5 \Omega$$
A heater coil is cut into two equal parts and only one part is now use in the heater. The heat generated will now be
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0%
One fourth
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Halved
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Doubled
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Four times
Explanation
The heat generation is given by
$$Heat=Power\times time$$
$$Power=\dfrac{V^2}{R}$$
$$H=\dfrac{V^2 \times t}{R}$$
After cutting equally, length becomes half.
The resistance is directly proportional to $$l$$
It means, resistance becomes half.
$$H=\dfrac{V^2 \times t}{R/2}$$
$$H=2\ \dfrac{V^2 \times t}{R}$$
Therefore the h
eat generated will be doubled.
The conductors of resistance $$1 \Omega, 2 \Omega$$ and $$3\Omega$$ are in series in a circuit. What is the equivalent resistance in the circuit ?
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Less than $$1\Omega$$
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Less than $$3\Omega$$
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More than $$1 \Omega$$
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More than $$3 \Omega$$
Explanation
Given:
$$R_1=1 \Omega$$
$$R_2=2 \Omega$$
$$R_3=3 \Omega$$
All are in series
So $$R_{eq}=R_1+R_2+R_3=1+2+3=6 \Omega$$
Kilowatt hour is unit of
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Energy
0%
Power
0%
Impulse
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Force
Explanation
The kilowatt-hour (symbolized kWh) is a unit of energy equivalent to one kilowatt (1 kW) of power expended for one hour. The kilowatt-hour is commercially used as a billing unit for energy delivered to consumers by electric utilities.
What happens to the current in each resistor when resistors of different values in parallel combination are connected to a source of electricity?
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Values of current and potential difference are different
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Values of current and potential difference are same
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Values of current are different but value of potential difference is same
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Values of current are the same but values of potential difference are different
Explanation
In a parallel combination, all the resistors are connected across the same voltage source. Thus, all of them will have the same potential difference.
But since the values of resistors are different, they will draw different current from the source.
Two wires A and B of same metal are connected in parallel. Wire A has length $$l$$ and radius $$r$$ and wire B has length $$2l$$ and radius $$2r$$. then the ratio of total resistance of parallel combination and the resistance of wire A is
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$$1 : 2$$
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$$1 : 3$$
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$$1 : 4$$
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$$1 : 5$$
If the five equal pieces of a resistance wire having $$5 $$ resistance each is connected in parallel, then their equivalent resistance will be
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0%
$$1/5 $$
0%
$$1 $$
0%
$$5 $$
0%
$$25 $$
Explanation
Number of resistance given $$n=5$$
Resistance of each resistor $$R=5\Omega$$
All resistors are connected in parallel
$$\dfrac 1{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_5}+\dfrac{1}{R_5}=\dfrac{5}{R}$$
$$\dfrac 1{R_{eq}}=\dfrac{5}{5}=1\ \Omega$$
Five resistances of R $$\Omega$$ were taken. First three resistances are connected in parallel combination and rest two are connected in series combination, then the equivalent resistance is :
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0%
$$\frac{3}{7} R \ \Omega$$
0%
$$\frac{7}{3} R \ \Omega$$
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$$\frac{7}{8} R \ \Omega$$
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$$\frac{8}{7} R \ \Omega$$
Explanation
$$R_{equi} = \frac{R}{3} + R + R = \frac{7R}{3} \ \Omega$$
The resistance of one conducting wire is $$10 $$. How much electric current will flow by connecting it with a battery of $$1.5 V$$ ?
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$$0.15 mA$$
0%
$$1.5 mA$$
0%
$$15 mA$$
0%
$$150 mA$$
Explanation
From Ohm's law,
$$V = IR$$
$$ \Rightarrow I = \dfrac{V}{R}= \dfrac{1.5}{10}= 0.15A = 150 mA$$
Materials which allow current to pass through them easily are called?
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Insulator
0%
Conductor
0%
Switch
0%
Battery
Explanation
Materials which allow current to pass through them easily are called as conductor. The most common conductors are metals: copper, silver, gold, platinum and aluminum etc.
So, the correct option is $$(B)$$
The amperage of the fuse wire used in a circuit that works on $$230 \ V$$ is $$2.2 \ A$$. If so. the power of the device is
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0%
less than $$300 \ W$$
0%
$$300 \ W$$ to $$500 \ W$$
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between $$500 \ W$$ and $$510 \ W$$
0%
more than $$510 \ W$$
Explanation
The amperage of the fuse wire used in a circuit that works on $$V = 230 \ V$$ is $$i = 2.2 \ A$$.
Power, $$P = Vi = 230 \times 2.2 = 506 \ W$$
Hence, the power of the device is between $$500 \ W$$ and $$510 \ W$$.
The correct option is C.
Several resistors are connected in series. Which of the following statements is correct? Choose all that are correct.
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The equivalent resistance is greater than any of the resistances in the group.
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The equivalent resistance is less than any of the resistances in the group.
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The equivalent resistance depends on the voltage applied across the group.
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The equivalent resistance is equal to the sum of the resistances in the group.
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None of those statements is correct.
Explanation
According to the relationship for resistors in series,
$$R_{oq}=R_1+R_2+...$$
the sum $$R_{oq}$$ is always larger than any of the resistances $$R_1, R_2,$$ etc.
Answer the question:
Two $$5.0\Omega$$ resistors are connected as shown in the diagram.
What is the total resistance of the combination ?
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less than $$5.0\Omega$$
0%
$$5.0\Omega$$
0%
more than $$5.0\Omega$$ but less than $$10.0\Omega$$
0%
$$10.0\Omega$$
Explanation
As we can see that both resistance are connected in parallel.
For parallel combination:
$$R_{eq}=\dfrac{R_1R_2}{R_1+R_2}$$
$$R_{eq}=\dfrac{5\times 5}{5+5}=2.5\Omega$$
Hence option A is correct, equivalent resistance is less than 5 ohm
If $$500\Omega$$ of resistance is made by adding five $$100\Omega$$ resistance of tolerance $$4\%$$, then the tolerance of the combination is .
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$$5\%$$
0%
$$4\%$$
0%
$$20\%$$
0%
$$10\%$$
Explanation
Individual resisistors are having resistance:
$$R=100\pm 4$$
On combining 5 similar resistors we get:
$$5R=500\pm 20$$
So, tolerance of the combination is also $$4\%$$
In which direction conventional current flow in an electric circuit:
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Negative to positive
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Positive to negative
0%
Both A & B
0%
None of these
The equivalent resistance between points $$A$$ and $$B$$ is
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0%
$$\dfrac {91}{2}\Omega$$
0%
$$\dfrac {65}{2}\Omega$$
0%
$$\dfrac {45}{2}\Omega$$
0%
$$\dfrac {5}{2}\Omega$$
The equivalent resistance of eight equal resistances in series is $$48$$ ohms. What would be the equivalent resistance if they are connected in parallel?
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$$\displaystyle \frac {3}{4} \Omega$$
0%
$$\displaystyle \frac {4}{3} \Omega$$
0%
$$4 \Omega$$
0%
$$3 \Omega$$
Explanation
Let the resistance of each resistor is $$R$$.
In series we know, equivalent resistance is $$R_s=nR=8R=48\Omega$$
$$\implies R=6\Omega$$
In paralle we know, equivalent resistance is $$R_p=\dfrac{R}{n}$$
$$\implies R_p=\dfrac{6}{8}=\dfrac{3}{4}\Omega$$
Answer-(A)
On a bulb it is written $$220\ V$$ and $$60\ W$$. Find out the resistance of the bulb and the value of the current flowing through it:
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$$806.67\ \Omega \ \text{and} \ 0.27\ A$$
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$$500\ \Omega \ \text{and} \ 2\ A$$
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$$200\ \Omega \ \text{and} \ 4\ A$$
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$$100\ \Omega \ \text{and} \ 1\ A$$
Explanation
$$\textbf{Step1:Resistance of bulb}$$
We know,
Power, $$P=\dfrac {V^2}{R}$$
$$\implies R=\dfrac {V^2}{P}$$
$$R=\dfrac {(220)^2}{60}$$
$$R=806.67\ \Omega$$
$$\textbf{Step2:Apply Ohm's law}$$
According to Ohm's law,
$$V=IR$$
$$\implies I=\dfrac {V}{R}$$
$$I=\dfrac {220}{806.67}$$
$$I=0.27\ A$$
1 kWh is equal to
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$$3.6 \times 10^6 MJ$$
0%
$$3.6 \times 10^5 MJ$$
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$$3.6 \times 10^2 MJ$$
0%
$$3.6 MJ$$
Explanation
1 kilowatt hour is the energy produced by 1 kilowatt power source in 1 hour.
$$1kWh=1kW\times 1hour=1000\times 3600 W.s$$
$$\implies 1kWh=3.6\times 10^6J$$
$$\implies 1kWh=3.6MJ$$
Answer-(D)
An electric lamp whose resistance of 10 ohm, and a conductor of 2 ohm resistance are connected in series with a 6 V battery. The total current through the circuit and the potential difference across the electric lamp are :
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3.6 A, 6 V
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0.5 A, 5 V
0%
2.0 A, 0.2 V
0%
0.3 A, 3 V
Explanation
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If the filament of bulb D breaks then does bulb C give more light, less light, the same light or no light?
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0%
more light
0%
less light
0%
same light
0%
no light
Explanation
It is assumed that every bulb is identical and has same resistance($$R$$).
So C and D are in parallel.And A , B and combination of (C and D) all these three are in series.
Initially resistance here was $$R/2$$ (C and D both have $$R$$ in parallel).So resistances in series are $$R$$ ,$$R$$ and $$R/2$$. Since resistance across third element is less than first two elements so less portion of cell voltage will be shared here.
Now bulb D breaks.So combination of C and D will reduced to only C.
Now only C is there so net resistance here is $$R$$ .So now A ($$R$$ resistance) , B ($$R$$ resistance) and C ($$R$$ resistance) are in series and have equal resistances so voltage across C will be $$1/3$$ of cell voltage.
In other words after breaking of D voltage across C has increased from before hence it will glow more brightly than before.
Do the arrows in the diagram indicate the direction of electron flow or conventional flow of electricity?
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Conventional flow of current
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Flow of Electron
0%
Flow of positive charge
0%
None of the above
Explanation
We know that conventional current flows from positive terminal of battery or cell to negative terminal of cell.Since in the figure the current is shown from positive terminal of cell (represented by bigger line) hence it is conventional current shown in the figure.
Direction of electron flow is opposite to conventional current direction.
Two wire of same metal have same length but their cross-sections are in the ratio 3:They are joined in series. The resistance of thick wire is $$10\Omega$$. The total resistance of combination will be
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0%
$$10\Omega$$
0%
$$20\Omega$$
0%
$$40\Omega$$
0%
$$100\Omega$$
For a circuit shown in the figure, the total current in the circuit is:
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0%
$$0.1\ A$$
0%
$$0.2\ A$$
0%
$$18\ A$$
0%
$$1\ A$$
Explanation
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, $$I = \dfrac{V}{R} = \dfrac{6}{1}\times 3$$
$$I = 18\ A$$
Identify the cases where the bulbs are connected totally in series.
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0%
0%
0%
Explanation
A series connection is one in which all elements are in the same branch or the same wire goes through all the elements.
So clearly only in option A and D all bulbs are in the same branch hence they are totally in series.So option A and D are correct.
Arrange the order of power dissipated in the given circuits, if the same current enters at point $$A$$ in all the circuits and resistance of each resistor is $$R$$.
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$$(ii) > (iii) > (iv) > (i)$$
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$$(iii) > (ii) > (iv) > (i)$$
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$$(iv) > (iii) > (ii) > (i)$$
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$$(i) > (ii) > (iii) > (iv)$$
When current is passed through a conductor, the heat produced in it depends upon
I. The material of the conductor
II. The current flowing through the conductor.
III. The time for which the current flows.
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I and II only
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I and III only
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II and III only
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All I, II and III
Two heater coils separately take $$10$$ minute and $$5$$ minute to boil certain amount of water. If both the coils are connected in series, the time taken will be
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$$15$$ min
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$$7.5$$ min
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$$3.33$$ min
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$$2.5$$ min
Explanation
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Practice Class 10 Physics Quiz Questions and Answers
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