MCQ Questions for Class 10 Physics Electricity Quiz 9

The resistors of resistances

$2\Omega, 4\Omega, 5\Omega$ are connected in parallel. The total resistance of the combination will be:

0%
$\dfrac {29}{10}\Omega$
0%
$\dfrac {19}{20}\Omega$
0%
$\dfrac {10}{20}\Omega$
0%
$\dfrac {20}{19}\Omega$

Explanation

Given:

$R_1=2\ \Omega$ ,

$R_2=4\ \Omega$ ,

$R_3=5\ \Omega$

In parallel combination-

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

$\Rightarrow\dfrac{1}{R}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{5}$

$\Rightarrow R=\dfrac{20}{19}\Omega$

When resistances are connected in series then the same current flows through each resistance.

0%
True
0%
False

An electric kettle has two heating coils. When one of the coils is connected to an AC source, the water in the kettle boils in

$10\ min$ . When the other coil is used the water boils in

$40\ min$ . If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be

0%
$25\ min$
0%
$15\ min$
0%
$8\ min$
0%
$4\ min$

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

$R$ ?

0%
$1 A$
0%
$2 A$
0%
$3 A$
0%
$4 A$

Explanation

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

$2A$ .

Two resistances are connected in parallel and a current is sent through the combination. The current divides itself:

0%
In the inverse ratio of resistance
0%
In the direct ratio of resistance
0%
Equally in both the resistance
0%
In none of the above manner

Explanation

Current flows in inverse ratio of resistance

$\dfrac{i_1}{i_2}=\dfrac{R_2}{R_1}$

What do you add to distilled water for making it to conduct electricity.

0%
sulphuric acid
0%
water
0%
milk
0%
base

In a circuit two or more cells of the same e.m.f. are connected in parallel in order to

0%
Increase the P.D across a resistance in the circuit
0%
Decrease the P.D across a resistance in the circuit
0%
Facilitate drawing more current from the battery system
0%
Change the e.m.f. across the system of batteries

Explanation

$E_{eff}=\dfrac{E_1}{r_1}+\dfrac{E_2}{r_2}$ where r is internal resistancE

It will increase the current ,

$I=\dfrac{E_{eff}}{R}$

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:

0%
$4.5\ V$
0%
$9\ V$
0%
$3\ V$
0%
$2\ V$

Explanation

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.

$V_2=IR_2$

$V_2=1.5\times 3$

$V_2=4.5\ V$

2109570_523815_ans_f37e19d78a5349a8b06803f5989c2e0f.png

Human body is a

0%
good conductor of electricity
0%
poor conductor of electricity
0%
semiconductor
0%
good insulator

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of

$4/3$ and

$2/3$ , then the ratio of the currents passing through the wire will be

0%
$3$
0%
$\dfrac{1}{3}$
0%
$\dfrac{8}{9}$
0%
$2$

Explanation

$R=\rho\dfrac{l}{A}$

$\implies R=\rho\dfrac{l}{\pi r^2}$

$\implies R\propto \dfrac{l}{r^2}$

Hence,

$\dfrac{R'}{R}=\dfrac{l'}{r'^2}\dfrac{r^2}{l}$

$\dfrac{R'}{R}=\dfrac{l'}{l}\dfrac{r^2}{(r')^2}$

$\implies \dfrac{R'}{R}=\dfrac{4/3}{(2/3)^2}$

$\implies \dfrac{R'}{R}=3$

Since,

$I\propto \dfrac{1}{R}$

Hence,

$\dfrac{I'}{I}=\dfrac{1}{3}$

Answer-(B)

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

0%
$5$ %
0%
$10$ %
0%
$12$ %
0%
$15$ %

Explanation

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

$= \dfrac{1k\Omega}{10k\Omega} \times 100 =10$ %

The resistance of the series combination of two resistances is S. When they are joined in parallel the total resistance is P. If S = nP, then the minimum possible value of n is

0%
4
0%
3
0%
2
0%
1

Explanation

Let two resistor be

$R_{1} \ and \ R_{2}$

In series combination the equivalent resistance is given by

$S = R_{1} + R_{2}$

In parallel combination the equivalent resistance is given by

$P = \dfrac{R_{1} \times R_{2}}{R_{1} + R_{2}}$

By relation S = nP

We can write it

$R_{1} + R_{2} = n\dfrac{R_{1} \times R_{2}}{R_{1} + R_{2}}$

Rearranging the term

$(R_{1} + R_{2})^2 = n R_{1} \times R_{2}$

$R_1^2 + R_2^2+ 2R_1R_2=nR_1R_2$

$R_1^2 + R_2^2 + (2-n)R_1R_2 = 0$

Dividing the whole equation with

$R_2^2$ , we will get

$(\dfrac{R_1}{R_2})^2 + 1 + (2-n) \dfrac{R_1}{R_2} = 0$

Put

$\dfrac{R_1}{R_2} = k$

$k^2 + (2-n)k +1 = 0$

For real value of k,

$D \geq 0$

$(2-n)^2 - 4 \geq 0$

For minimum value of n,

$(2-n)^2 - 4 = 0$

$4 + n^2 - 4n-4=0$

$n^2 - 4n = 0$

$n(n-4)=0$

$\implies n=0 \ or \ 4$

Since

$n$ can't be 0,

$n = 4$

Four bulbs marked

$40W, \ 250V$ are connected in series with

$250\ V$ mains, the total power consumed is

0%
$10W$
0%
$40W$
0%
$320W$
0%
$160W$

Explanation

Power of each bulb,

$P = 40 W$

Total power consumed when four bulbs are connected in series

$\dfrac{1}{P_s} = \dfrac{1}{P_1}+$

$\dfrac{1}{P_2} + \dfrac{1}{P_3}+\dfrac{1}{P_4}$

$\therefore$

$\dfrac{1}{P_s} = \dfrac{1}{P}+$

$\dfrac{1}{P} + \dfrac{1}{P}+\dfrac{1}{P}$

We get

$P_s =\dfrac{P}{4}$

$\Rightarrow P_s = \dfrac{40}{4}$

$\Rightarrow P_s =10 W$

Two resistors

$400\ \Omega$ and

$800\ \Omega$ are connected in series with a 6 V battery. the potential difference measured by voltmeter across

$400\ \Omega$ resistor is:

0%
$2\ V$
0%
$1.95\ V$
0%
$3.8\ V$
0%
$4\ V$

Explanation

Total resistance of the circuit (series combination)

$R_s = R_1+R_2$

$\therefore$

$R_s = 400+800 =1200\Omega$

Current,

$I=\dfrac{E}{R_s}=\dfrac{6}{1200}$

$I=0.005\ A$

Potential difference across

$R_1$ ,

$V' = IR_1$

$\therefore$

$V' = 0.005 \times 400$

$V' =2\ V$

What is the minimum number of bulbs, each marked

$60W, \ 40V,$ that can work safely when connected in series with a

$240V$ mains supply?

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

Resistance of each bulb

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

$\Rightarrow R = \dfrac{(40)^2}{60}$

$\Rightarrow R = 26.67\Omega$

Maximum current that can flow through the bulb without damaging it

$I_{max}= \dfrac{P}{V}$

$\Rightarrow I_{max} = \dfrac{60}{40}$

$\Rightarrow I_{max} = 1.5\ A$

Thus minimum resistance required in the circuit

$R_{min} = \dfrac{V_s}{I_{max}}$

$\Rightarrow R_{min} = \dfrac{240}{1.5}$

$\Rightarrow R_{min} = 160 \Omega$

Let

$n$ number of bulbs are connected in series in the circuit.

Equivalent resistance of

$n$ bulbs

$R_{eq} = nR = 26.67 n$

But,

$26.67 n \leq 160$

$\Rightarrow n \leq 6$

Thus minimum number of bulbs required is

$6$ .

Two wire of same metal have same length but their cross-sections are in the ratio

$3:1$ . They are joined in series. The resistance of thick wire is

$10$ . The total resistance of combination will be

0%
$10 \Omega$
0%
$20 \Omega$
0%
$40 \Omega$
0%
$100 \Omega$

Explanation

Resistance of the wire

$R = \dfrac{\rho L}{A}$

where

$\rho$ is the resistivity and

$L$ and

$A$ is the length and cross-section area of the wire, respectively.

$\implies$

$R\propto \dfrac{1}{A}$ (for same

$L$ and

$\rho$ )

Ratio of cross-sectional area of thick wire to thin wire

$\dfrac{A_1}{A_2} = \dfrac{3}{1}$

Thus ratio of resistance of thin wire to thick wire

$\dfrac{R_2}{R_1} = \dfrac{A_1}{A_2} = \dfrac{3}{1}$

$\dfrac{R_2}{10}= \dfrac{3}{1}$

$\implies$

$R_2 = 30\Omega$

Total resistance of the (series) combination

$R_{eq} = R_1 + R_2$

$\implies$

$R_{eq} = 10 + 30 = 40 \Omega$

$100W, 200V$ bulb is connected to a

$160V$ supply. The actual power consumption would be

0%
$185W$
0%
$100W$
0%
$54W$
0%
$64W$

Explanation

Resistance of the bulb

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

where

$P = 100 W$ and

$V = 200$ V.

$\therefore$

$R = \dfrac{(200)^2}{100} = 400\Omega$

Actual power consumption

$P' = \dfrac{V_1^2}{R}$

where

$V_1 = 160$ V

$\therefore$

$P' = \dfrac{(160)^2}{400} = 64 W$

$1$ kilowatt hr

$=$ __________ joules.

0%
$10.6 \times {10}^{6}$
0%
$3.6 \times {10}^{6}$
0%
$30.6 \times {10}^{6}$
0%
$3.6 \times {10}^{5}$

Explanation

$kWh$ is the commercial unit of energy.

$1\ kWh =1\times 1000\times 3600 = 3.6\times 10^6$ joules

You are given several identical resistance each of value

$R=10$ and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistance of

$5$ which can carry a current of

$4$ ampere. The minimum number of resistance's of the type

$R$ that will be required for this job is

0%
$4$
0%
$10$
0%
$8$
0%
$20$

Explanation

Resistance of each resistor

$R = 10$

Equivalent resistance of each segment

$R' = R+R = 2R$

$\therefore$

$R' = 2\times 10 = 20$

Such

$4$ segments are connected in parallel to each other.

Thus equivalent resistance of the circuit

$R_p = \dfrac{R'}{4}$

$\implies$

$R_p = \dfrac{20}{4} = 5$

Maximum current flowing through each resistor is one ampere i.e.

$i =1 A$

Thus total current flowing through the circuit

$I = i+i+i+i$

$\therefore$

$I = 4i = 4 A$

Thus minimum

$8$ resistors are required to satisfy the above conditions.

$30\ V,\ 90\ W$ lamp is to be operated on a

$120\ V\ DC$ line. For proper glow, a resistor of ..... should be connected in series with the lamp.

0%
$10$
0%
$20$
0%
$30$
0%
$40$

Explanation

Resistance of the lamp

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

$\therefore$

$R = \dfrac{30^2}{90}$

$\implies$

$R = 10 \Omega$

Current flowing through the lamp

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

$\therefore$

$I = \dfrac{90}{30} =3 A$

DC line voltage

$V_d = 120$ V

Resistance of the line must be

$R' = \dfrac{V_d}{I}$

We get

$R' = \dfrac{120}{3} = 40 \Omega$

Let

$R_{1}$ be the external resistance to be connected in series with the lamp for proper glow.

$\therefore$

$R + R_1 = R'$

$10 + R_1 = 40$

$\implies$

$R_1 = 30 \Omega$

$1 Wh$ (Watt hour) is equal to :

0%
$36\times { 10 }^{ 5 }J$
0%
$36\times { 10 }^{ 4 }J$
0%
$3600 J$
0%
$3600 { Js }^{ -1 }$

Explanation

Watt is a unit of power such that

$1W =1\dfrac{J}{s}$

Also we know

$1h = 3600$

$s$

$\therefore$

$1Wh = 1\dfrac{J}{s} \times 3600$

$s = 3600$

$J$

Aluminium is preferred in overhead power cables because:

0%
It is a good conductor
0%
It acts as an insulator, in case of lightening
0%
It prevents accidents
0%
It is lightweight

A 30V-90W lamp is operated on a 120 V DC line. A resistor is connected in series with the lamp in order to glow it properly. The value of resistance is

0%
$10 \Omega$
0%
$30 \Omega$
0%
$20 \Omega$
0%
$40 \Omega$

Explanation

Resistance of lamp

$R_0 = \dfrac{V^2}{P} = \dfrac{(30)^2}{90} = 10 \Omega$
Current in the lamp

$I = \dfrac{V}{R_0} = \dfrac{30}{10} = 3A$
As the lamp is operated on 120V DC, then resistance becomes

$R' = \dfrac{V'}{i} = \dfrac{120}{3} = 40 \Omega$
For proper glow, a resistance R is joined in series with the bul

$R' = R + R_0$

$\Rightarrow R^{\alpha} = R' - R_0 = 40 - 10 = 30 \Omega$

In the following circuit, the 1

$\Omega$ resistor dissipates power P. If the resistor is replaced by 9

$\Omega$ , the power dissipated in it is

0%
$P$
0%
$3P$
0%
$9P$
0%
$P/3$

Explanation

Equivalent resistance in the circuit

$= (3+1)=4\Omega$

By Ohm's law,

$10 \ V=(4 \ \Omega)(i \ Ampere)$

$i=2.5 \ A$

Power through a resistance is given by

$P=I^2R$

Power through

$1 \Omega,$

$P_{1 \Omega}=i^2\times 1=6.25 \ W$

Given,

$P_{1\Omega}=P$

When

$1\Omega$ is replaced by

$9\Omega$ , equivalent resistance

$=(3+9)=12\Omega$

By Ohm's law,

$10 \ V=(12 \ \Omega)(i' \ Ampere)$

Current

$i'=\dfrac{10}{12} \ A$

Power dissipated through replaced resistor,

$P_{9\Omega}=(\dfrac{10}{12})^2\times 9=6.25=P_{1\Omega}=P$

Option A is correct.

The equivalent resistance between A and B in the following figure is

0%
$120\Omega$
0%
$40\Omega$
0%
$30\Omega$
0%
$22.5\Omega$

Two electric bulbs have ratings respectively of

$25 \ W, 220 \ V$ and

$100 \ W, 220 \ V$ . If the bulbs are connected in series with a supply of

$440\ V$ , which bulb will fuse?

0%
$25\ W$ bulb
0%
$100\ W$ bulb
0%
Both of these
0%
None of these

Explanation

Power is given by

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

$P$ is Power,

$V$ is voltage across bulb and

$R$ is resistance of bulb.

Hence,

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

Resistance of

$25\ W\ bulb = \dfrac {220\times 220}{25} = 1936\ \Omega$

Using Ohm's law,

$V = iR$
Value of current that can pass through it

$i= \dfrac{V}{R} = \dfrac {220}{1936} = 0.11\ A$

Resistance of

$100\ W\ bulb = \dfrac {220\times 220}{100} = 484\ \Omega$
Value of current that can pass through it

$i = \dfrac{V}{R} = \dfrac {220}{484} = 0.45\ A$

When connected in series to

$440 \ V$ supply, the equivalent resistance is given by,

$R_{equi} = R_1+R_2 = 1936+484=2420 \ \Omega$

Calculating the value of the current

$i = \dfrac {440}{2420} = 0.18\ A$

Thus

$0.18 \ A >0.11 \ A$

The value of current flowing in the final circuit is greater than that allowed by the

$25W \ bulb$ , so it will fuse.

The total electrical resistance between the points

$A$ and

$B$ of the circuit shown is:

0%
$9.23\Omega$
0%
$15\Omega$
0%
$30\Omega$
0%
$100\Omega$

Explanation

$\cfrac { 1 }{ Req. } =\cfrac { 1 }{ 30\Omega } +\cfrac { 1 }{ 20\Omega } +\cfrac { 1 }{ 40\Omega }$

$Req.=\cfrac { 120 }{ 13 } \Omega =9.23\Omega$

Which of the following are the properties of fuse wire?

0%
Made of alloy of tin
0%
Has a low melting point
0%
Connected in series with main supply
0%
All of the above

When the electric current through the fuse exceeds a certain limit, the fuse wire melts and breaks.

0%
True
0%
False

The resistivity of the material of potentiometer wire is

$5\times 10^{-5}\Omega m$ and its area of cross-section is

$5\times 10^{-6} m^{2}$ . If

$0.2\ A$ current is flowing through the wire, then the potential drop per metre length of the wire is

0%
$0.1\ Vm^{-1}$
0%
$0.5\ Vm^{-1}$
0%
$0.25\ Vm^{-1}$
0%
$0.2\ Vm^{-1}$
0%
$0.001\ Vm^{-1}$

Explanation

Given,

$i = 0.2\ A$
Resistivity,

$\rho = 5\times 10^{-6}\Omega - m$
and

$A = 5\times 10^{-6} m^{2}, l = 1\ m$
We know that,

$V = iR$

$\Rightarrow V = \dfrac {i\rho L}{A}\left [\because R = \dfrac {\rho l}{A}\right ]$

$\Rightarrow V = \dfrac {0.2\times 5\times 10^{-6}\times 1}{5\times 10^{-6}}$

$\Rightarrow V = 0.2\ Vm^{-1}$ .

The ratio of resistance of two copper wire of the same length and of same cross-sectional area, when connected in series to that when connected in parallel, is

0%
$1 : 1$
0%
$1 : 2$
0%
$2 : 1$
0%
$4 : 1$
0%
$1 : 4$

Explanation

Let

$R$ be the resistance of copper wire. In the first condition,

$R_{series} = R + R = 2R .... (i)$
In the second condition,

$\dfrac {1}{R_{parallel}} = \dfrac {1}{R} + \dfrac {1}{R} = \dfrac {1}{R_{parallel}} = \dfrac {1 + 1}{R}$

$R_{parallel} =\dfrac {R}{2} ..... (ii)$

On dividing Eq. (i) by Eq. (ii), we get

$\dfrac {R_{series}}{R_{parallel}} = \dfrac {2R}{\dfrac {R}{2}} = \dfrac {4}{1}$ .

Two bulbs of

$250 \ V$ and

$100 \ W$ are first connected in series and then in parallel with a supply of

$250 \ V$ . Total power in each of the case will be respectively

0%
$100 \ W, 50 \ W$
0%
$50\ W, 100 \ W$
0%
$200 \ W, 150 \ W$
0%
$50 \ W, 200 \ W$

A wire of resistance 'R' is cut into 'n' equal parts. These parts are then connected in parallel with each other. The equivalent resistance of the combination is?

0%
nR
0%
$R/n$
0%
$n/R^2$
0%
$R/n^2$

A student's

$9.0V, 7.5W$ portable radio was left on from

$9:00\ P.M.$ until

$3:00 A.M.$ How much charge passed through the wires?

0%
$18000C$
0%
$24000C$
0%
$6000C$
0%
$12000C$

Explanation

As we know,
Power,

$P = V\cdot I$

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

$\Rightarrow I = \dfrac {Q}{t} = \dfrac {P}{V}$

$\Rightarrow Q = \dfrac {P}{V} \times t = \dfrac {7.5}{9}\times 3600\times 6$

$= 18000\ C$

$[\because t = 9\ PM$ to

$3\ AM = 6\ hr = 6\times 3600s]$

The diagram shows identical lamps X and connected in series with a battery. The lamps light with normal brightness.If a third lamp Z is connected in parallel with lamp X, then what will happens to the brightness of the lamp Y?

0%
Brighter than normal
0%
Normal
0%
Dimmer than normal
0%
Very dim (cannot be seen)

Observe the given figure and answer the following questions. The bulb will glow when a ______ is placed in between the probes.

0%
Crayon
0%
Comb
0%
Key
0%
Piece of stone

Fuse is the most important safety device, used for protecting the circuits due to

0%
Short circuiting and overloading
0%
Overloading and earthing
0%
Earthing only
0%
Short circuiting, overloading and earthing.

$4$ bulbs rated

$100$ W each, operate for

$6$ hours per day. What is the cost of the energy consumed in

$30$ days at the rate of Rs.

$5$ /kWh?

0%
Rs. $360$
0%
Rs. $90$
0%
Rs. $120$
0%
Rs. $400$

Explanation

$\displaystyle E= Power(in kW)\times time=\frac{4\times 100\times 6}{1000}$ kWh

$=2.4$ kWh
Consumed in

$30$ days

$=30\times 2.4=72$
Total cost

$=72\times 5=360$ Rs.

Consider four circuits shown in the figure below. In which circuit power dissipated is highest?

Explanation

We know that

$\rho =\dfrac { { \varepsilon }^{ 2 } }{ R } \\ \rho \propto \dfrac { 1 }{ R } (\varepsilon = constant)$ (1)

so we calculate

${ R }_{ eq }$ in all options as shown in fig

we know that

$\rho \propto \dfrac { 1 }{ R }$

In (a) option resistance is minimum, so power dissipated will be highest

Equivalent resistance between A and B is.

0%
3 $\Omega$
0%
6 $\Omega$
0%
1.5 $\Omega$
0%
4.5 $\Omega$

Ravi connected three bulbs with the cells and a switch as shown. When switch is moved to ON position.

0%
The bulb X will glow first.
0%
The bulb Y will glow first.
0%
The bulbs Z and X will glow first.
0%
All the bulbs will glow simultaneously.

Two wires of same material have lengths

$L$ and

$2L$ cross-sectional areas

$4A$ and

$A$ respectively. The ratio of their resistances would be

0%
$1:1$
0%
$1:8$
0%
$8:1$
0%
$1:2$

Explanation

$R=\cfrac { \rho l }{ A }$

$R\propto \cfrac { l }{ A }$

$\cfrac { { R }_{ 1 } }{ { R }_{ 2 } } =\cfrac { { l }_{ 1 } }{ { A }_{ 1 } } \times \cfrac { { A }_{ 2 } }{ { l }_{ 2 } }$

$\cfrac { { R }_{ 1 } }{ { R }_{ 2 } } =\cfrac { L }{ 4A } \times \cfrac { A }{ 2L } =1:8$

$\begin{bmatrix} { l }_{ 1 }=L \\ { l }_{ 2 }=2L \\ { A }_{ 1 }=4A \\ { A }_{ 2 }=A \end{bmatrix}$

$n$ resistors each of resistance

$R$ first combine to give maximum effective resistance and then combine to given minimum effective resistance. The ratio of the maximum to minimum resistance is:

0%
$n$
0%
$n^2$
0%
$n^2-1$
0%
$n^3$

Explanation

Maximum effective resistance will be when all the resistors are in series combination. So,

$R_{eff_{max}}=n\ R$

Minimum effective resistance will be when all the resistors are in parallel combination. So,

$R_{eff_{min}}=R/n$

$\therefore \ \dfrac {R_{eff_{max}}}{R_{eff_{min}}}=\dfrac {n\ R}{R/ n}=n^2$

Which of the following I-V graph represents ohmic conductors?

Explanation

Ohm's law

$I=\dfrac{V}{R}$ is an equation of straight line. Hence I-V characteristics for ohmic conductors is also straight line and its slope gives,

$m=\dfrac{1}{resistance}$ of the conductor.

Option A

A wire of resistance

$12\: ohm/meter\:$ is bent to form a complete circle of radius

$10cm$ . The resistance between its two diametrically opposite points, A and B as shown in the figure is?

0%
$3\Omega$
0%
$6\pi \Omega$
0%
$6\Omega$
0%
$0.6\pi \Omega$

Explanation

Total length of wire,

$l= 2\pi\times 0.1=0.2\pi\ m$

Resistance per unit length,

$r=12\Omega/m$

Let us assume two semicircular part of wire connected in parallel between

$A$ and

$B$

Resistance of each part

$,R= r \times l = 12\times \cfrac{0.2\pi}{2}= 1.2 \pi\ \Omega$

Two equal resistors (R) are connected in parallel between A and B so equivalent resistance is equal to

$R/2$ .

$\therefore$ Resistance between

$A$ and

$B = \dfrac R2 = \dfrac{1.2 \pi}2= 0.6\pi \ \Omega$

Hence, option

$(D)$ is correct.

An infinite ladder network of resistances is constructed with

$1\Omega$ and

$2\Omega$ resistance as shown in figure. The

$6$ V battery between A and B has negligible internal resistance. The equivalent resistance between A and B is?

0%
$1\Omega$
0%
$2\Omega$
0%
$3\Omega$
0%
$4\Omega$

Explanation

$\textbf {Step 1: Find Smallest Repeating unit [Refer Fig. 1]}$

$\textbf{Step 2: Replace with equivalent resistance R. [Refer Fig. 2]}$

After the first smallest repeating unit, the remaining circuit is also same as the initial circuit.

Therefore, if we assume the equivalent resistance between A & B as

$R$ , then the resistance of the remaining circuit will also be

$R$ as shown in the figure.

$\textbf {Step 3: Find equivalent resistance between A and B and equate with R. [Refer Fig. 3]}$

From figure, the equivalent resistance between A & B will be equal to the equivalent resistance of the remaining circuit. So,

$R_{AB} = 1 + \dfrac {2R}{2+R}$

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

$\Rightarrow\ \ 3R + 2 = R^2 + 2R$

$\Rightarrow\ \ R^2 - R - 2 = 0$

$\therefore\ \ \ \ \boxed{R = 2 \Omega}$ (Here, Negative value can is rejected)

Hence Equivalent resistance between points A and B is 2

$\Omega$ .

2102459_941868_ans_8d9f7ce88e6548a1b906216eaea5c3b4.png

Two coils have a combined resistance of

$12 \Omega$ , when combined series and

$5/3 \Omega$ , when connected in parallel. Their respective resistances are:

0%
$6$ and $6$ ohms
0%
$10$ and $2$ ohms
0%
$5$ and $7$ ohms
0%
$4$ and $8$ ohms

Explanation

Let us assume the two resistances to be

$R_1$ and

$R_2$ .

Let resistances in series and parallel be

$R_s$ and

$R_p$ respectively.

Given:

$R_{s}=12 \Omega=R_1+R_2$

So,

$R_1+R_2=12$ . . . . (i)

$R_2=12-R_1$ . . . . (ii)

And,

$R_{p}=\dfrac{5}{3}\Omega \Rightarrow \dfrac{1}{R_{p}}=\dfrac{3}{5}$

So,

$\dfrac{3}{5}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

$\Rightarrow \dfrac{3}{5}=\dfrac{R_1+R_2}{R_1R_2}$ . . . (iii)

Using

$eq. (i)$ and

$eq. (ii)$ in

$eq. (iii)$

$\dfrac{3}{5}=\dfrac{12}{R_1(12-R_1)}$

$\Rightarrow 12R_1-R_1^2=20$

$\Rightarrow R_1^2-12R_1+20=0$

This is a quadratic equation in

$R_1$ . Let's solve it by factorisation.

$R_1^2-10R_1-2R_1+20=0$

$R_1(R_1-10)-2(R_1-10)=0$

$(R_1-10)(R_1-2)=0$

$\Rightarrow R_1-10=0$ or

$R_1-2=0$

$\Rightarrow R_1=10$ or

$R_1=2$

Putting

$R_1=10$ in

$eq.(i)$ ,

$R_2=2$

Putting

$R_1=2$ in

$eq.(i)$ ,

$R_2=10$

So option B is correct.

You have been provided with four

$400$

$\ \Omega$ resistors each. The number of ways in which these can be combined to have different equivalent resistances is

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Seven different combinations and seven different equivalent resistances
0%
Eight different combinations and seven different equivalent resistances
0%
Nine different combinations and eight different equivalent resistances
0%
Ten different combinations and nine different equivalent resistances.

A copper cylindrical tube has inner radius a and outer radius b. The resistivity is

$\rho$ . The resistance of the cylinder between the two rends is?

0%
$\dfrac{\rho l}{b^2-a^2}$
0%
$\dfrac{\rho l}{2\pi (b-a)}$
0%
$\dfrac{\rho l}{\pi (b^2-a^2)}$
0%
$\dfrac{\pi(b^2-a^2)}{\rho l}$

Explanation

$R=\cfrac{\rho l}{A}\\=\cfrac{\rho l}{\pi(b^2-a^2)}\\A=\pi b^2-\pi a^2\\=\pi (b^2-a^2)$

Find the equivalent resistance between points A and B in the given circuit

0%
$6\Omega$
0%
$12\Omega$
0%
$4\Omega$
0%
$18\Omega$

Explanation

using wheatstone bridge,

$\dfrac{6}{3}=\dfrac{12}{6}\Rightarrow$ It is a balanced wheatstone bridge, hence no current in

$12\Omega$ i.e

$PQ$ branch or equivalent resistance b/w

$A\times B$ .

$6$ &

$12\Omega$ are in series

$=6\Omega +12\Omega$

$=18\Omega$

similarly,

$3\Omega$ &

$6\Omega$ , equi

$=3+6=9\Omega$

$18\Omega$ &

$9\Omega$ are in parallel

$\dfrac{1}{R_e}=\dfrac{1}{18}+\dfrac{1}{9}=\dfrac{3}{18}$

$\boxed{R_e=6\Omega}$

1432517_1016671_ans_467d99a857e24a2282687dff5b48605c.png

CBSE Questions for Class 10 Physics Electricity Quiz 9 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

CBSE Questions for Class 10 Physics Electricity Quiz 9 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

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