MCQ Questions for Class 10 Physics Electricity Quiz 13

${ R }_{ 1 },{ R }_{ 2 },{ R }_{ 3 }$ are all connected in parallel, the effective resistance will be:

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$37\Omega$
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$7\Omega$
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$50.7\Omega$
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$8.57\Omega$

Calculate the potential difference between, points A and B current flowing in

$10\Omega$ resistor in the part of the network below:-

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$20 V,2 A$
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$50 V, 1 A$
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$40 V, 1 A$
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$30 V,1 A$

A hollow cylinder

$(\rho =2.2\times 10^{-8}\Omega -m)$ of length 3m has inner and outer diameter as 2mm and 4mm respectively. The resistance of the cylinder is

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$0.35\times 10^{-3}\Omega$
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$3\times 10^{-3}\Omega$
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$7\times 10^{-3}\Omega$
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none of these

Explanation

outer radii

$=2=R$

inner radii

$1=r$

$l=3m$

$p=2.2\times 10^{-8}\Omega-m$

$resistance=resistivity\left(\dfrac{lenght}{area}\right)$

Area of hollow cylinder

$=\pi(R^2-r^2)$

$=\pi (2^{ 2 }-1^{ 2 })=3.14\times 3\times 10^{ -6 }=9.42\times { 10 }^{ -6 }$

$R=2.2\times 10^{ -8 }\left[ \dfrac { 3 }{ 9.42\times { 10 }^{ -6 } } \right]$

$R=0.7\times 10^{-2}\Omega$

$R=7\times 10^{-3}\Omega$

A metal wire of resistance R is cut into three equal pieces which are then connected side by side to form a new wire, the length of which is equal to one third of the original length. The resistance of this new wire is:-

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$R$
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$3R$
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$\cfrac{R}{9}$
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$\cfrac{R}{3}$

5 identical thin conducting plates are kept parallel to each other such that separation between each plate is d. Area of each plate is A. Plates are arranged as shown in figure. All wires are conducting.
Switch S is closed, find the work done by battery.

$\left( A\succ \succ \succ { d }^{ 2 } \right)$

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$\dfrac { { \in }_{ 0 }{ AV }^{ 2 } }{ 5d }$
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$\dfrac { 2{ \in }_{ 0 }{ AV }^{ 2 } }{ 5d }$
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$\dfrac { 3{ \in }_{ 0 }{ AV }^{ 2 } }{ 2d }$
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$\dfrac { 2{ \in }_{ 0 }{ AV }^{ 2 } }{ d }$

Find the potential differences across the

$24\Omega$

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$48 volts$
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$2 volts$
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$4 volts$
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$1 volts$

Explanation

Let the equivalent resistance consisting of six resistances in parallel be

$R_{eq}.$

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}+\dfrac{1}{R_5}+\dfrac{1}{R_6}$

$\Rightarrow \dfrac{1}{R_{eq}}=\dfrac{1}{48}+\dfrac{1}{48}+\dfrac{1}{24}+\dfrac{1}{6}+\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{1+1+2+8+12+24}{48}$

$\Rightarrow \dfrac{1}{R_{eq}}=\dfrac{48}{48}$

$\Rightarrow R_{eq}=1 \Omega$

Let the voltage across

$R_{eq}$ be

$V_{eq}.$

Applying Ohm's law for

$R_{eq}$

$V_{eq}=IR_{eq}$

$V_{eq}=(2 \ A) \times (1 \ \Omega)=2 \ V$

The voltage

$V_{eq}=2 \ V$ also appears across

$R_1$ ,

$R_2, R_3 . . . R_6.$

$\therefore$ Voltage across

$24 \Omega$ is

$V_{24 \ \Omega}=V_{eq}=2 \ V$

Answer is option (B).

In the adjoining figure the equivalent resistance between

$A$ and

$B$ is

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$5\ \Omega$
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$8\ \Omega$
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$2.5\ \Omega$
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$6.8\ \Omega$

Explanation

We are given circuit, we have to calculate equivalent between A and B.

Clearly in 2nd fig.

2, 2 and 2 are in series

So,

$R_1 = 2 + 2 + 2$

$= 6\Omega$

Now,

$6\Omega$ and

$6\Omega$ are in parallel

$\therefore R_2 = \dfrac{6\times 6}{12} = 3\Omega$

Now, Equivalent Resistance (

$R_{AB}$ ) = 1 + 3 + 1

$= 5\Omega$

Hence option A is correct.

2106543_1110284_ans_15d61fd6650d4c3e98cc66f9d93f2685.png

A wire of resistance R divided in 10 equal parts. These parts are connected in parallel. Then resistance equivalent is

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$0.01 R$
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$0.1 R$
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$10 R$
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$100 R.$

Explanation

Resistance of a wire is proportional to its length. The wire is cut into 10 equal parts. So, resistance of each part = $\dfrac{R}{10}$

Then all 10 wires are connected in parallel. Then the equivalent resistance is

$\dfrac{1}{{{R}_{eq}}}=\dfrac{1}{\dfrac{R}{10}}+\dfrac{1}{\dfrac{R}{10}}+.........(10\,\,times)$

$\dfrac{1}{{{R}_{eq}}}=\dfrac{100}{R}$

${{R}_{eq}}=\dfrac{R}{100}$

${{R}_{eq}}=0.01R$

Three equal resistors are connected as shown in figure. the maximum power consumed by each resistor is 18 W. Then the maximum power consumed by the combination is :

0%
$18 W$
0%
$27 W$
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$36 W$
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$54 W$

Explanation

equivalent resistance from

$A\rightarrow B$

$\boxed{(R\parallel R)+R}$

$\Rightarrow \dfrac{1}{\dfrac{1}{R}+\dfrac{1}{R}}+R=\dfrac{1}{\dfrac{2}{R}}+R=\dfrac{R}{2}+R$

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

$\therefore$ equivalent resistance

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

Power across the single resistor

$=18\ W$ ( given)

$P=i^2R=18\ W.......(1)$

Power across the equivalent circuit is

$P_1=i^2\times \dfrac{3R}{2}=\dfrac{3i^2R}{2}\rightarrow .....(2)$

dividing

$(1)$ by

$(2)$

$\dfrac{P}{P_1}=\dfrac{18}{P_1}=\dfrac{i^2R}{\dfrac{3i^2R}{2}}=\dfrac{2}{3}$

$\Rightarrow P_1=\dfrac{3}{2}\times 18=27\ W$

$\therefore$ maximum power consumed by circuit

$=27\ W$

1457480_1101106_ans_e0622a5ac5384cd6b033a38c35ea73a6.png

Effective resistance between

$A$ and

$B$ is

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$6\Omega$
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$3\Omega$
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$\dfrac {2}{3}\Omega$
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$\dfrac {3}{2}\Omega$

In the circuit shown, when R has removed an additional resistance of

$72\ \Omega$ must be introduced in series with the battery in order to keep the current through

$30\ \Omega$ resistance unaltered. Hence R is

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$15\ \Omega$
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$18\ \Omega$
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$25.7\ \Omega$
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$21\ \Omega$

In the following circuit, what is equivalent resistance between A and B points?

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$3\Omega$
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$6\Omega$
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$1.5\Omega$
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$2\Omega$

The value of the resistance

$r$ in the electric circuit given below is (in

$\Omega$ )

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$2$
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$5$
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$1$
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$3$

In the given circuit (as shown in the figure), choose the wrong statement.

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The equivalent resistance between $C$ and $G$ is $3\ K\Omega$
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The current provided by the source is $4\ mA$
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The current provided by the source is $8\ mA$
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Voltage across points $G$ and $E$ is $4V$

Meeta connected 4 resistors in parallel. Each has a resistance of

$2 \Omega$ . Find the effective resistance .

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$8\Omega$
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$4\Omega$
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$0.5\Omega$
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$0.4\Omega$

A bulb rated at (100W-200V) is used on a 100 V line. The current in the bulb is

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1/4 amp
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4 amp
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1/2 amp
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2 amp

Explanation

Given,

$P = 100\ W$ ; power of the bulb

$V = 200\ V$ ; voltage rating

So value of resistance-

$P = \dfrac{V^2}{R}$

$\Rightarrow R = \dfrac{V^2}{P}$

$\Rightarrow R = \dfrac{200^2}{100}$

$\Rightarrow R = 400\ \Omega$

Now applied voltage changes to

$100\ V$ , so current passing through the resistance-

$I = \dfrac{V}{R}$

$\Rightarrow I = \dfrac{100}{400}$

$\Rightarrow I = \dfrac{1}{4} A$

The rate of change of magnetic field inside a solenoid is constant(

$\alpha$ )
. Then the induced electric field inside the solenoid at a distance 'a' from its axis is

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$\dfrac { a\alpha }{ 2 }$
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$a\alpha$
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Along the axis of solenoid
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Forming circular loop

A metal wire of resistance

$3\Omega$ is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle

$60^o$ at the centre, the equivalent resistance between these two points will be?

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$\dfrac{12}{5}\Omega$
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$\dfrac{5}{3}\Omega$
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$\dfrac{5}{2}\Omega$
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$\dfrac{7}{2}\Omega$

Explanation

$R=\dfrac{\rho l^2}{AlD}d=\dfrac{\rho dl^2}{m}$

$R\propto l^2$

$R=12\Omega$ (new resistance of wire)

$R_1=2\Omega$

$R_2=10\Omega$

$R_{eq}=\dfrac{10\times 2}{10+2}=\dfrac{5}{3}\Omega$ .
1247242_1613925_ans_62bfe94f99b64d138ab32fc36afe7554.png

The reading of the ammeter is:

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$\dfrac { 1 }{ 8 }\ A$
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$\dfrac { 3 }{ 4 }\ A$
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$\dfrac { 1 }{ 2 } \ A$
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$2\ A$

A copper wire of resistance

$R$ is cut into ten parts of equal length. Two-piece are joined in series and then five such combinations are joined in parallel. The new combination will have resistance.

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$R$
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$\frac { R } { 4 }$
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$\frac { R } { 5 }$
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$\frac { R } { 25 }$

The wires of same dimension but different resistivities

$\rho_1$ and

$\rho_2$ are connected in parallel. The equivalent resistivity of the combination is

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$\dfrac {\rho_1 + \rho_2}{2}$
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$2 (\rho_1 + \rho_2)$
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$\dfrac {\rho_1 \rho_2}{\rho_1 + \rho_2}$
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$\sqrt {\rho_1\rho_2}$

In a series circuit, the current is different in each resistor.

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True
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False

Most of the times we connect remote speakers to play music in another room along with the built-in speakers. These speakers are connected in parallel with the music system.
At the instant represented in the picture, the voltage across the speakers is

$6.00 V.$ The resistance of the main speaker is

$8 \space \Omega$ and the remote speaker has resistance

$4 \space \Omega$
The equivalent resistance of the speakers is

0%
$12 \space \Omega$
0%
$\dfrac{7}{3 }\space \Omega$
0%
$\dfrac{8}{3}\space \Omega$
0%
$0.375 \space \Omega$

The effective resistance between

$A$ and

$B$ is

0%
$7R/12$
0%
$5R/7$
0%
$5R/12$
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$none\ of\ these$

Explanation

Let us remove the resistance between

$A$ and

$B$ ,

then

$E$ and

$F$ will be at the same potential, we can draw the remaining as:

$M$ is the midpoint of

$DC, M$ will be at the same pot as

$E$ and

$F$ .

Now finf

$R_{AB}=5/7R$ .

The resistance which we had removed will be in parallel to it so

$R_{eq}=\dfrac {R_{AB}R}{R_{AB}+R}=\dfrac {(5/7)RR}{(5/7)R+R}=\dfrac {5}{12}R$

1627395_1753659_ans_955c27e9c64348c9bb7e7160443576e2.png

The equivalent resistance of the circuit across points A and B is equal to

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$22.5 \space \Omega$
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$25 \space \Omega$
0%
$37.5 \space \Omega$
0%
$75 \space \Omega$

Six identical resistors are connected as shown in figure .The equivalent resistance will be

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Maximum between P and R
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Maximum between Q and R
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Maximum between P and Q
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All are equal

For the circuit shown in fig. select the correct statements from the following options.

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x and y are equipotential points
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Effective resistance between A and B is $2 \Omega$
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Effective resistance between A and B is $1 \Omega$
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none of these

The three resistance of equal value are arranged in the different combinations shown below. Arrange them in increasing order of power dissipation

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III < II < IV < I
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II < III < IV < I
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I < IV < III < II
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I < III < II < IV

Instead of metal wires, a jute string can be used to make a circuit.

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True
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False

When resistors with different resistances are connected in parallel, which of the following must be the same for each resistor? Choose the correct answer.

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potential difference
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current
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power delivered
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charge entering each resistor in a given time interval
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none of the above

The current-versus-voltage behavior of a certain electrical device is shown in the figure. When the potential difference across the device is

$2\ V$ , what is its resistance?

0%
$1\ \Omega$
0%
$\dfrac{3}{4}\ \Omega$
0%
$\dfrac{3}{5}\ \Omega$
0%
Undefined
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None of these

Explanation

$R=V/I=2V/2A=1\Omega$

Why rubber or plastic used to cover electric wires?

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To appear good
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They are insulators and do not allow electric current to pass when we touch those wires
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To give strength to the circuit
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All of the above

Several resistors are connected in parallel. Which of the following statements 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

Suppose

$3\Omega$

$4\Omega$

$6\Omega$ are connected in parallel.

$\frac{1}{R_{eq}}= \frac{1}{3}$ +

$\frac{1}{4}$ +

$\frac{1}{6}$

$R_{eq}$ =

$1.33\Omega$ which is less than all three resistances .

And the resistances are properties of conductor they don't depend on current or voltages so option C is wrong.

So correct answer is option

$B$

In an electric circuit, the direction of the current is taken from

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Negative to Positive
0%
Positive to Positive
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Positive to Negative
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None of the above

CBSE Questions for Class 10 Physics Electricity Quiz 13 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

CBSE Questions for Class 10 Physics Electricity Quiz 13 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

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