MCQ Questions for Class 10 Physics Electricity Quiz 10

A solenoid is at potential of difference

$60\ V$ and current flows through it is

$15$ ampere, then the resistance of coil will be

0%
$4\ \Omega$
0%
$8\ \Omega$
0%
$0.25\ \Omega$
0%
$2\ \Omega$

Explanation

Given,

Potential difference

$V=60\ V$

Current

$i=15\ A$

Resistance

$R=?$

Resistance

$=\dfrac {Potential\ difference}{Current}=\dfrac{60V}{15A}=4\Omega$

Match the Column I with Column II.

	Column I		Column II
(A)	Smaller the resistance greater the current in a circuit	(p)	If the same voltage is applied across a resistance
(B)	Greater or smaller the resistance the current is same	(q)	If the same current is passed
(C)	Greater the resistance, smaller the power	(r)	When resistances are connected in series
(D)	Greater the resistance, greater the power	(s)	When resistances are connected in parallel

0%
A-r, B-p, C-q, D-s
0%
A-p, B-r, C-q, D-s
0%
A-r, B-p, C-s, D-q
0%
A-s, B-r, C-p, D-q

Explanation

In a parallel combination of resistances, the voltage drop across all the resistors is the same and therefore, the current across the resistances depends on the value of resistance. If the resistance is small, the current flowing through it will be more.

In a series combination of the resistances, the current through all the resistors is constant whereas the voltage differs for the resistors.

Similarly, if the same voltage is applied across all the resistors, the power is inversely proportional to the resistance. So, greater the resistance, smaller will be the power and if the current is the same in the circuit, the power will be greater for the higher resistance.

So,

$A\rightarrow s$

$B\rightarrow r$

$C\rightarrow p$

$D\rightarrow q$

A metal plate can be heated by

0%
passing either a direct or alternating current through the plate.
0%
placing in a time varying magnetic field.
0%
placing in a space varying magnetic field, but does not vary with time.
0%
both (a) and (b) are correct

In parallel combination of n cells, we obtain:

0%
More voltage
0%
More current
0%
Less voltage
0%
Less current

Explanation

In parallel combination of n cells the net resistance of cells becomes

$\cfrac{1}{n}$ times. Resistance reduces and current increases.
1380612_941936_ans_b101955f7e9147c1846144d5c58a8d36.png

If two bulbs of

$25W$ and

$100W$ rated at

$200$ volts are connected in series across a

$440$ volts supply, then

0%
$100$ watt bulb will fuse
0%
$25$ watt bulb will fuse
0%
none of the bulb will fuse
0%
both the bulbs will fuse

Explanation

${ R }_{ 1 }=\cfrac { { v }^{ 2 } }{ P } =\cfrac { { 200 }^{ 2 } }{ 25 } =1600\Omega$

${ R }_{ 2 }=\cfrac { { 200 }^{ 2 } }{ 100 } =400\Omega$
When both bulbs are connected in series

${ V }_{ 1 }=440\times \cfrac { { R }_{ 1 } }{ { R }_{ 1 }+{ R }_{ 2 } } =440\times \cfrac { 1600 }{ 1600+400 } =352V;{ V }_{ 2 }=440\times \cfrac { 400 }{ 2500 } =88V$
Voltage across

$25W$ bulb is greater than specified voltage, hence it will fuse
1051071_1016514_ans_5e06dbe8e35949ccb12ffc83f2c4e064.PNG

The resistance of the series combination of two resistance is

$S$ . When they are joined in parallel through total resistance is

$P$ . If

$S=nP$ , then the minimum possible value of

$n$ is?

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

Explanation

$n$ is minimum when two resistances are the same.
So,

$S=2R$

and,

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

$\Rightarrow S = nP$

$\Rightarrow 2R = n \dfrac{R}{2}$

$\Rightarrow n = 4$

If n cells each of emf

$\varepsilon$ and internal resistance r are connected in parallel, then the total emf and internal resistances will be:

0%
$\varepsilon, \dfrac{r}{n}$
0%
$\varepsilon, nr$
0%
$n\varepsilon, \dfrac{r}{n}$
0%
$n\varepsilon, nr$

Explanation

$\cfrac{1}{Net_r}=\cfrac{1}{r}+\cfrac{1}{r}....n times$

$\cfrac{1}{r_{net}}=\cfrac{n}{r}\\r_{net}=\cfrac{v}{n}$

and

$emf=\epsilon$

Answer

$\epsilon,\cfrac{v}{n}$

1175792_941938_ans_6ff5763ed8a14de3b3773ae743778f15.png

In the series combination of n cells each cell having emf

$\varepsilon$ and internal resistance r. If three cells are wrongly connected, then total emf and internal resistance of this combination will be.

0%
$n\varepsilon, (nr-3r)$
0%
$(n\varepsilon -2\varepsilon), nV$
0%
$(n\varepsilon -4\varepsilon), nV$
0%
$(n\varepsilon - 6\varepsilon), nV$

Explanation

Since in a combination of

$n$ cells, there are

$3$ cells connected with opposite polarity. It means that there are

$(n-3)$ cells connected with correct polarity and remaining

$3$ with opposite polarity.

The net emf of the combination is:

$E_{net}=(n-3)E-3E$

$E_{net}=(n-6)E$

On the other hand, the total internal resistance of all the batteries remains the same for the whole circuit, even for the batteries connected in opposite polarity.

So, the net internal resistance for the circuit will be

$nr$ .

So, option

$(D)$ is correct.

Refer to teh circuit shown. What will be the total power dissipation in the circuit if

$P$ is the power dissipated in

${R}_{1}$ ? It is given that

${R}_{2}=4{R}_{1}$ and

${R}_{3}=12{R}_{1}$

0%
$4P$
0%
$7P$
0%
$13P$
0%
$17P$

Explanation

Total equivalent resistance=

$\frac { { R }_{ 2 }{ R }_{ 3 } }{ { R }_{ 2 }{ +R }_{ 3 } } +{ R }_{ 1 }\\ =\frac { 48{ { R }_{ 1 } }^{ 2 } }{ 16{ R }_{ 1 } } +{ R }_{ 1 }\\ =3{ R }_{ 1 }+{ R }_{ 1 }=4{ R }_{ 1 }$

Power dissipation at

$R_1=I^2R_1=P$

$\therefore$ Power dissipation in the total circuit=

$I^24R_1$

$=4I^2R_1$

$=4P$

In the circuit shown in the figure

${ E }_{ 1 }=7V,{ E }_{ 2 }=7V,{ R }_{ 1 }={ R }_{ 2 }=1\Omega ,{ R }_{ 3 }=3\Omega$ . The current through the resistance

${R}_{3}$ and potential difference between the (+)ve terminal of one battery and the (-ve) terminal of second battery are

0%
$i=0$
0%
$i=2A$
0%
$V=14V$
0%
$V=7V$

Explanation

Let point A is at zero potential as shown in figure (1).

So, on moving towards

$E_{2}$ potential of

$E_{2}$ is 7 V.

So, the required difference will be 7 V.

Now for current,

$E_{eq} = \cfrac{7\times 1 + 7 \times 1}{2}$

$E_{eq} =$ 7 V

$R_{eq} = \cfrac{1}{2}$

Now, circuit will reduce to (shown in figure (2))

So, i=2 Amp.

1105537_1016810_ans_38907d0257d349e59fa09b651c608586.png

The current in the given circuit will be:

0%
$\cfrac{1}{45}$ ampere
0%
$\cfrac{1}{15}$ ampere
0%
$\cfrac{1}{10}$ ampere
0%
$\cfrac{1}{5}$ ampere

Explanation

Two

$30 \Omega$ resistor are in series and one

$30 \Omega$ is parallel to two

$30 \Omega$ resistors.

$R_s=30+30=60 \Omega$

$\begin{array}{l} \dfrac { 1 }{ { { R_{ p } } } } =\dfrac { 1 }{ { 30 } } +\dfrac { 1 }{ { 60 } } \\ { R_{ p } }=20\Omega \end{array}$

From Ohm"s Law,

$V=IR$

$I = \dfrac{2}{{20}} = \dfrac{1}{{10}}A$

In the network shown in the figure, the equivalent resistance between A and C is

0%
$10R$
0%
$(6/11)R$
0%
$(12/15)R$
0%
$(50/11)R$

Explanation

$2R, R$ and

$3R$ are in parallel, as they all are between same potential

$\dfrac{1}{R}+\dfrac{1}{2R}+\dfrac{1}{3R}-\dfrac{1}{R_p}\Rightarrow \dfrac{1}{R_p}=\dfrac{6+3+2}{6R}=\dfrac{11}{6R}$

$\dfrac{6R}{11}$ and

$4R$ is in parallel, so

$\dfrac{6R}{11}+4R=\dfrac{6R+44R}{11}=\left(\dfrac{50}{11}\right)R$

$\boxed{Req_0=\left(\dfrac{50}{11}\right)R}$

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%
$\dfrac{R}{n}$
0%
$\dfrac{R}{n^2}$
0%
$\dfrac{n^2}{R}$

Three resistance

$R, 2R$ and

$3R$ are connected in parrallel to a battery ,then

0%
The potential drop across $3R$ is maximum
0%
The current through each resistance is same
0%
The heat developed in $3R$ is maximum
0%
The heat developed in $R$ is maximum

Explanation

Given,

Three resistances

$(R,\ 2R, \ 3R)$ are connected in parallel.

For a parallel circuit, voltage is constant.

Heat developed,

$H=\dfrac{V^2}{R}t$

$H \propto \dfrac 1R$

$\therefore$ Heat developed will be maximum in

$R$ .

2113598_1035641_ans_4ee32946fcdd44588d3514d66f9bd25f.png

The net resistance between ${\text{X and Y}}$ is

0%
${\text{4}}\Omega {\text{ }}$
0%
${\text{4}}{\text{.55}}\Omega {\text{ }}$
0%
${\text{2}}\Omega$
0%
${\text{20}}\Omega$

In the following part of the circuit, potential difference between points A and B equals to

0%
$30V$
0%
$45V$
0%
$-15V$
0%
$-45V$

Two wires of same dimensions but resistivities

${\rho _1}$ and

${\rho_2}$ are connected in series. The equivalent resistivity of the combination is:

0%
${\rho _1} + {\rho_2}$
0%
$\frac{1}{2}\left( {\,{\rho _1} + {\rho_2}} \right)$
0%
$\sqrt {{\rho _1}{\rho _2}}$
0%
$2\left( {\,{\rho _1} + {\rho_2}} \right)$

Explanation

$\textbf{Step 1: Calculation of resistance of both wires}$

We know that Resistance,

$R=\rho$

$l/A$

As, the physical dimensions are same then Length

$l$ and Area

$A$ are constant

So,

$R_{1}=\rho_{1}$

$l/A$

$\&$

$R_{2}=\rho_{2}$

$l/A$

$\textbf{Step 2: Equivalent resistance and resistivity}$

For series combination

$R_{eq} = R_{1}+R_{2}=(\rho_{1}+ \rho_{2})l/A$

Compare this with

$R_{eq} = (\rho_{eq} )l/A$

So,

$\rho_{eq} =\rho_1+\rho_{2}$

Hence, Option A is correct

The equivalent resistance between the points A and B is:

0%
$\dfrac{36}{7} \Omega$
0%
$10 \Omega$
0%
$\dfrac{85}{7} \Omega$
0%
none of these

In the circuit shown in figure, the ammeter reads a current of?

0%
1A
0%
2A
0%
0.3A
0%
0.2A

Explanation

$(1) \& (1)$ points same potential $\& (2) \& (2)$ points are same potential. So both the corner Resistance shorted.

Most above with has no resistance. So all are shorted.

$I=\dfrac{2}{1}=2A$

1007120_1046330_ans_bdabecade31a4baca819feea0f1f4b48.PNG

The current through the ammeter shown in the figure is

$\dfrac{10}{9}A$ . If each of the

$4 \Omega$ resistor is replaced by

$2\Omega$ resistor, the current in the circuit will become

0%
(a) $\frac{10}{9} A$
0%
(b) $\frac{5}{4} A$
0%
(c) $\frac{9}{8} A$
0%
(d) $\frac{5}{8} A$

Resistors $R^{}_{1}$ and $R^{}_{2}$ have an equivalent resistance of 6 ohms when connected in the circuit shown below. The resistance of $R_1$ could be:

0%
1
0%
5
0%
8
0%
4

Explanation

In parallel combination the equivalent resistance is always less than either resistance used,

$\Rightarrow R_{1}>6$ and

$R_{2}>6$

which means

$R_{1}=8 {Ohm}$ is feasible answer out of given options.

An electric motor operates on a

$50 \ V$ supply and a current of

$1 \ A$ . If the efficiency of the motion is

$30 \ \%$ , what is the resistance of winding of motion ?

0%
$35 \ \Omega$
0%
$4 \ \Omega$
0%
$2.9 \ \Omega$
0%
$3.1\ \Omega$

$9$ Identical wires each of Resistance

$R$ are connected to form a closed polygon. The equivalent resistance between the ends of any side is

0%
$9R$
0%
$R/9$
0%
$8R/9$
0%
$9R/8$

In the given circuit the current through the cell is:

0%
$\cfrac{4}{3}A$
0%
$\cfrac{5}{3}A$
0%
$\cfrac{1}{3}A$
0%
$\cfrac{2}{3}A$

Explanation

Equivalent resistance in series is given by

$R_{eq} = R_1+R_2$

Hence, equivalent resistance in the two branches is given by

$R_{eq} = 1 + 1 = 2 \ \Omega$

Equivalent resistance in parallel is given by

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

Hence, equivalent resistance in the two branches is given by

$\dfrac{1}{R_{eq}} = \dfrac{1}{2}+\dfrac{1}{2} = 1 \ \Omega$

Equivalent resistance in series in the final circuit is given by

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

Using Ohm's law

$I=\cfrac{V}{R_{eq}}=\cfrac{2}{3}A$

1157105_1079515_ans_770907393cf541e881c2e72301f34abc.png

If the resistance in the circuit is increased four times by keeping the potential difference the same, the current in the circuit is ___________.

0%
Remains same
0%
Four times
0%
One fourth
0%
Half

Explanation

According to ohm's law-

$V = IR$

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

$V$ : Applied potential difference

$I$ : Current in the circuit

$R$ : Resistance of the circuit

Now, new resistance becomes four times of initial

$R' = 4R$

So new current will be-

$I' = \dfrac{ V }{ 4R }$

$V$ , remains same

$\Rightarrow I' = \dfrac{ I }{ 4 }$

So, the current becomes one-fourth of the initial current when resistance becomes four times.

Mohan uses one television of

$100$ W for

$10$ hrs. How much energy is consumed by Mohan?

0%
$E=100 kWh$
0%
$E=0.1 kWh$
0%
$E=1 kWh$
0%
$E=10 kWh$

Explanation

Energy consumption is given by

$E=P\times t$ ,where P= power and t= time

Given power

$P=100\ W=0.1 KW$ and time

$t=10 hour$

$E=1 kWh$

Two bulbs take

$50$ watts each when connected in parallel to

$100\ V$ source. The total power consumed by them when they are connected in series with the same source is

0%
$25\ W$
0%
$50\ W$
0%
$75\ W$
0%
$100\ W$

Explanation

$P = \dfrac {V^{2}}{R} \Rightarrow P_{1} = \dfrac {V^{2}}{R_{1}} \ P_{2} = \dfrac {V^{2}}{R_{2}}$

$50 = \dfrac {100^{2}}{R_{1}}\ 50 = \dfrac {100^{2}}{R_{2}}$

$R_{1} = 200\Omega \ R_{2} = 200\Omega$

$V = IR \Rightarrow I = \dfrac {V}{R}\ I = \dfrac {100}{200 + 200} = \dfrac {100}{400} = 0.25\ A$

$P_{2} = P_{1} = I^{2}R_{1} = I^{2}R_{2} = 0.25^{2}\times 200$

$= \dfrac {200}{10} = \dfrac {100}{8} = \dfrac {50}{4} = \dfrac {25}{2} watt$

$P_{Total} = P_{1} + P_{2} = \dfrac {25}{2} + \dfrac {25}{2} = 25\ watt$

Option A.

2108347_1077869_ans_cac319bf29bb46049b307adad903debf.png

Resistor

$R_1$ and

$R_2$ have an equivalent resistance of

$6 \ \Omega$ when connected in the circuit shown. The resistance of

$R_1$ could be (in

$\Omega$ )

$(R_1<R_2)$

0%
$1$
0%
$5$
0%
$8$
0%
$4$

Explanation

Two resistors when connected in parallel always have equivalent resistance less than or equal to the resistance of the smaller resistor. $\\ \dfrac 1R_{eq} = \dfrac 1R_1 + \dfrac 1R_2$
Since the smaller resistance is $6 \Omega$ ,the resistance of $R_1$ has to be greater than or equal to $6 \Omega$ .

Hence, the resistance of

$R_1$ could be

$8 \ \Omega$

What will be the effective resistance between the points P and Q in the following circuit

0%
$13 \Omega$
0%
$2/3 \Omega$
0%
$2 \Omega$
0%
$7 \Omega$

Explanation

The three resistors between P and Q are connected in series with each other.

Therefore, equivalent resistance

$=2+3+2=7 \Omega$

A block has dimensions

$1\ cm, 2\ cm$ and

$3\ cm$ . Ratio of the maximum resistance to minimum resistance between any pair of opposite faces of this block is:

0%
$9:1$
0%
$1:9$
0%
$18:1$
0%
$1:6$

Explanation

Given that,

The dimension of the block is 1cm, 3cm and 3cm.

The relation of the resistance of a conductor is

$R=\rho \dfrac{l}{A}....(I)$

Where,

$\rho$ = resistivity of the material of the conductor

$A$ = area of the cross section of the conductor

$l$ = length of the conductor

Now, the resistance is maximum

${{R}_{\max }}=\rho \times \dfrac{3}{1\times 2}$

${{R}_{\max }}=\dfrac{3\rho }{2}...(II)$

Now, the resistance is minimum

${{R}_{\min }}=\rho \dfrac{1}{2\times 3}$

${{R}_{\min }}=\dfrac{\rho }{6}...(III)$

Now, from equation (II) and (III)

The ratio of maximum resistance and minimum resistance is

$\dfrac{{{R}_{\max }}}{{{R}_{\min }}}=\dfrac{\frac{3\rho }{2}}{\dfrac{\rho }{6}}$

$\dfrac{{{R}_{\max }}}{{{R}_{\min }}}=\dfrac{3}{2}\times \dfrac{6}{1}$

$\dfrac{{{R}_{\max }}}{{{R}_{\min }}}=\dfrac{9}{1}$

${{R}_{\max }}:{{R}_{\min }}=9:1$

Hence, the ratio is $9:1$

Two wires

$A$ and

$B$ of same material and same mass have radius

$2r$ and

$r$ and the length of the wire B is half the length of the wire

$A$ . If resistance of wire

$A$ is

$34\Omega$ , then resistance of

$B$ will be:

0%
$544\Omega$
0%
$272\Omega$
0%
$68\Omega$
0%
$17\Omega$

Explanation

Resistance in wire, $R=\rho \dfrac{L}{A}=\dfrac{\rho L}{\pi {{r}^{2}}}$

${{R}_{1}}=\dfrac{\rho {{L}_{1}}}{\pi r_{1}^{2}}=\dfrac{\rho {{L}_{1}}}{\pi {{(2r)}^{2}}}=\dfrac{1}{4}\dfrac{\rho {{L}_{1}}}{\pi {{r}^{2}}}=34\,\Omega$

$\Rightarrow \dfrac{\rho {{L}_{1}}}{\pi {{r}^{2}}}=34\times 4=136\ \Omega \ ......\ (1)$

$L_2=L_1/2$ $r_2=r$

${{R}_{2}}=\rho \dfrac{{{L}_{2}}}{{{A}_{2}}}=\dfrac{\rho \dfrac{{{L}_{1}}}{2}}{\pi r_{2}^{2}}=\dfrac{1}{2}\dfrac{\rho {{L}_{1}}}{\pi {{r}^{2}}}=\dfrac{1}{2}\times 136=\ 68\,\Omega$

Resistance of B is $68\,\Omega .$

Find the current supplied by the battery as shown in the figure.

0%
4
0%
5
0%
54
0%
3

Explanation

According to question given,

One end of first group of two resistance 4 and 6 ohms joined with positive terminal of battery and another end at the common potential point similarly one end of other group of 6 and 4 ohms resistance is jointed with negative terminal of battery and another end at a common potential point.

In the above diagram we have reduced circuit,

For parallel combination,

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

For first group

$\dfrac { 1 }{ R_1' } =\quad \dfrac { 1 }{ 4 } +\dfrac { 1 }{ 6 } =\dfrac { 5 }{ 12 }$

${ R }_{ 1}' =\dfrac { 12 }{ 5 }$

Similarly

For second group

${ R }_{ 2}'=\dfrac { 12 }{ 5 }$

Both groups are connected in series

$R_{eq}={ R }_{ 1 }'+{ R }_{ 2 }'=\dfrac { 24 }{ 5 }$

$\text{And we know from ohms law Voltage (V)}=\quad Current(I)\times Resistance(R)\\ I=\dfrac { V }{ R }=\dfrac { 24 }{ 24/5 } =\quad \dfrac { 24\times 5 }{ 24 } =\quad 5\ Amp$

2085497_1088480_ans_84f3e86fe423410f99c2ebfd9bc47e41.png

The mains power supply of a house is through a 20 A fuse. When a 1500 W heater is used in this house, how many 100 W bulbs can be used simultaneously?(The supply is at 220 V and the appliances are rated for 220 V)

0%
27
0%
29
0%
30
0%
33

Explanation

$\large\textbf{Step -1: Calculate current For Electric bulbs}$

Given,

Power of heater $P_h= 1500W,$

Power of bulbs, $P_b = 100W,$

We know,

Power, $P = VI$

The current drawn by the heater,

Or $Electrical\;Current,\;I\; = \dfrac{P_h}{V} = \dfrac{{1500}}{{220}} = 6.81\;A$

The mains power supply of a house is through a 20 A fuse

Maximum Electric current, which can pass through bulbs $= 20- 6.81 =13.19 A$

$\large\textbf{Step -2:Calculate number of bulbs connected in parallel}$

Let number of bulb $= n$

and Current withdraw by a single bulb $, I_b$

$Electrical\;Current,\;{I_b}\; = \dfrac{P_b}{V} = \dfrac{{100}}{{220}} = 0.45\;A$

Maximum number of bulbs, $n= 13.19/0.45 = 29$

Hence maximum, $29$ bulbs can be connected in parallel with heater.

When three resistors of resistances

$3\Omega ,4\Omega$ and

$5\Omega$ are connected in parallel, the currents through them are in the ratio

0%
$5:4:3$
0%
$20:15:12$
0%
$3:4:5$
0%
$9:8:7$

Explanation

Given:

Resistance

$R_{1} = 3\Omega$

Resistance

$R_{2} = 4\Omega$

Resistance

$R_{3} = 5\Omega$

$R_{1} / R_{2} / R_{3}$

Since the resistors are parallel, voltage across them is same

$-V$

Let

$I_{1}, I_{2}$ and

$I_{3}$ be current flowing through

$R_{1}|R_{2}, R_{3}$ respectively.

Then;

$I_{1} = \dfrac {V}{3} A; I_{3} = \dfrac {V}{4}A; I_{3} = \dfrac {V}{5}A$

$I_{1} : I_{2} : I_{3} = \dfrac {V}{3} : \dfrac {V}{4} : \dfrac {V}{5} = \dfrac {1}{3} : \dfrac {1}{4} : \dfrac {1}{5}$

i.e.

$I_{1} : I_{2} : I_{3} = 60 \left (\dfrac {1}{3} : \dfrac {1}{4} : \dfrac {1}{5}\right ) = 20 : 15 : 12$

$\therefore$ Option B is correct.

2110229_1120355_ans_a3a9b45e209a496f83067a7ff42999a0.jpg

Find equivalent resistance between

$A$ and

$B$

0%
$\dfrac{R}{4}$
0%
$\dfrac{2R}{5}$
0%
$\dfrac{3R}{5}$
0%
$\dfrac{5R}{6}$

Explanation

$\textbf{Step 1 - Draw network diagram and simplify it}$

Here

$C, A, D$ are at same potential

and

$G, F, B$ are at same potential

So we can redraw the circuit as

$\textbf{Step 2 - Simplify further}$

Equivalent resistance of parallel combination

$\dfrac{1}{R^{\prime}}=\dfrac{1}{R_1} +\dfrac{1}{R_2}$

$\Rightarrow R^{\prime}=\dfrac{R_1\times R_2}{R_1+R_2}$

$\Rightarrow R^{\prime}=\dfrac{R \times R}{R+R}=\dfrac{R}{2}$

Equivalent resistance of series combination is

$R^{\prime\prime}=R_1 +R_2$

Now

$\dfrac{3R}{2} \text{and} R$ are in series.

$R_{AB} = \dfrac {\dfrac {3R}{2}\times R}{\dfrac {3R}{2} + R} = \dfrac {\dfrac {3R^{2}}{2}}{\dfrac {5R}{2}} = \dfrac {3R}{5}$

Hence option (c) correct.

When a piece of aluminium wire of finite length is drawn through a series of dies to reduce its diameter to half its original value, its resistance will become:-

0%
Two times
0%
Four times
0%
Eight times
0%
Sixteen times

Explanation

$\textbf{Step 1: Calculation of new area}$

As Diameter is reduced to half, so radius is also reduced to half

Let

$D$ and

$A$ are initial diameter and area.

$D'$ and

$A'$ are final diameter and area.

$R'$ is final resistance

Resistivity will be constant, As resistivity is a property of a material

$A = \pi r^2 = \dfrac{\pi D^2}{4}$

$D' = \dfrac{D}{2}$

$\Rightarrow$

$A' = \pi r'^2 = \dfrac{\pi D'^2}{4} = \dfrac{\pi D^2}{16} = \dfrac{A}{4}$

$....1$

$\textbf{Step 2: Calculation of new resistance}$

From

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

$R$ is initial resistance)

$\Rightarrow$

$R = \rho \dfrac{L\times A}{A^2}$

$\Rightarrow$

$R = \rho \dfrac{V}{A^2}$

New resistance

$R' = \rho \dfrac{V}{A'^2}$

From equation (1)-

$R' = 16 \rho \dfrac{V}{A^2}$

$= 16 R$

Option

$D$ is correct

The current in the given circuit will be :

0%
$\dfrac{1}{45}A$
0%
$\dfrac{1}{15}A$
0%
$\dfrac{1}{10}A$
0%
$\dfrac{1}{5}A$

Explanation

Two 30 ohm are in series equivalent resistance is 60 ohm

This 60 ohm and 30 ohm are in parallel its equivalent resistance is ,

${ R }_{ eq }=\frac { 30\times 60 }{ 30+60 } =\frac { 1800 }{ 90 } =20\Omega$

$i=\frac { V }{ { R }_{ eq } } =\frac { 2 }{ 20 } =\frac { 1 }{ 10 } \Omega$

Hence, option (C) is correct option

Two resistors

$R_1 \, (24 \pm 0.5) \Omega$ and

$R_2 \, (8 \pm 0.3) \Omega$ are joined in series. The equivalent resistance is

0%
$32 \pm 0.33 \Omega$
0%
$32 \pm 0.8 \Omega$
0%
$32 \pm 0.2\Omega$
0%
$32 \pm 0.5 \Omega$

Two resistors

$R_1$ and

$R_2$ are connected in parallel. The relative error in their equivalent resistance is [If

$R_1=(60\pm 3)\Omega$ and

$R_2=(30\pm 2)\Omega$

0%
$\dfrac{1}{20}$
0%
$\dfrac{1}{30}$
0%
$\dfrac{11}{180}$
0%
$\dfrac{7}{180}$

Three resistances

$R_1,R_2$ &

$R_3$

$(R_1> R_2> R_3)$ are connected in series. If current

$I_1,I_2$ &

$I_3$ respectively is flowing through them, the correct relation will be

0%
$I_1=I_2=I_3$
0%
$I_1> I_2> I_3$
0%
$I_1< I_2< I_3$
0%
$I_1> I_2< I_3$

Explanation

In series, same amount of current will flow through all the resistances.

Current

$I_1=I_2=I_3=I=\dfrac{V}{R_{eq}}=\dfrac{V}{R_1+R_2+R_3}$

Option A is correct.

1935723_1130018_ans_05714a7879634f9c8f0afca430e4af69.jpeg

The value equlvalent resistance between the points

$A$ and

$B$ in the given circuit will be?

0%
$6\ R$
0%
$\dfrac{4R}{11}$
0%
$\dfrac{11R}{4}$
0%
$\dfrac{R}{6}$

If two resistances

${R_1} = \left( {10.0 \pm 0.1} \right)\Omega \;and\;{R_2} = \left( {20.0 \pm 0.1} \right)\Omega \;are\;connected$ in series, then the maximum percentage error in its equivalent resistance will be

0%
1/5%
0%
2/3%
0%
2%
0%
4%

Explanation

Given,

$R_1=(10\pm0.1)\Omega$

$R_2=(20\pm0.1)\Omega$

In series combination,

$R_{eq}=R_1+R_2$

$R_{eq}=(10\pm0.1)+(20\pm0.1)=10+20+(\pm0.1)+(0.1)$

$R_{eq}=(30\pm0.2)\Omega$

The maximum percentage error

$\dfrac{0.2}{30}\times 100=\dfrac{2}{3}$ %

The correct option is B.

A uniform wire of resistance

$36\ \Omega$ is bent in the form of a circle. The effective resistance across points A and B is___.

0%
$5 \Omega$
0%
$6 \Omega$
0%
$7.2 \Omega$
0%
$30 \Omega$

The lowest resistance which can be obtained by connecting

$10$ resistors each of

$\dfrac{1}{10}\Omega$ is

0%
$1/250\ \Omega$
0%
$1/100\ \Omega$
0%
$1/200\ \Omega$
0%
$1/10\ \Omega$

Explanation

The lowest resistance will be in the case when all the resistors are connected in parallel.

In parallel combination, we calculate the equivalent resistance

$(R_{eq})$ as

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}} +\dfrac{1}{R_{3}}.........$

Where

$R_1, R_2, R_3.....$ are the resistances that we are connecting in the parallel.

In our question, All the 10 resistors have a resistance of

$\dfrac{1}{10}\Omega = 0.1\Omega$

$\dfrac{1}{R_{eq}}=\dfrac{1}{0.1}+\dfrac{1}{0.1}.........10\ times$

$\dfrac{1}{R_{eq}}=10+10......10\ times$

$\dfrac{1}{R_{eq}}=100$

$R_{eq}=\dfrac{1}{100}\Omega$

An electric bulb is rated

$220\ V$ and

$100\ W$ , when it is operated on

$110\ V$ , the power consumed will be

0%
$100\ W$
0%
$75\ W$
0%
$50\ W$
0%
$25\ W$

Explanation

$25W$

Formulae,

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

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

$=\dfrac{220^2}{100}=484\Omega$

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

$P=\dfrac{110^2}{484}=25W$

$P. Q$ and

$R$ are three resistances. When

$P, Q$ and

$R$ are connected in parallel, their total resistance is

$3 \Omega.$ When

$P$ and

$Q$ are connected in parallel their total resistance is

$10/ 3\ \Omega.$ When

$P$ and

$R$ is connected in series, their total resistance is

$15\ \Omega.$ Find the correct value for

$P, Q$ and

$R.$

0%
$P \rightarrow 5 \Omega ; Q \rightarrow 30 \Omega ; R \rightarrow 10 \Omega$
0%
$P \rightarrow 8 \Omega ; Q \rightarrow 10 \Omega ; R \rightarrow 7 \Omega$
0%
$P \rightarrow 5 \Omega ; Q \rightarrow 10 \Omega ; R \rightarrow 10 \Omega$
0%
$P \rightarrow 5 \Omega ; Q \rightarrow 10 \Omega ; R \rightarrow 30 \Omega$

Equivalence Resistance?

0%
2R
0%
5R
0%
3R
0%
7R

When 25 A current with 10 ohm resistance flowing per second then the power lost in the cable during transmission is

0%
$12.5 kW$
0%
$6.25 kW$
0%
$25 kW$
0%
$3.15 kW$

Explanation

Power lost in cable

$= I^2R =10 \times (25)^{2} = 6250 W = 6.25 kW$

In the network atom in figure, the equivalent resistance between the point A and B is :-

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

Explanation

Between points A and B ,

$R_3$ is parallel to

$R_4$ and

$R_5$ whereas

$R_4$ and

$R_5$ are in series.

We know that equivalent resistance in series is given as,

$R_{eq}=R_1+R_2$

Equivalent resistance in parallel is given as,

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

Since

$R_4$ and

$R_5$ are in series,

$R_{45} = 2+4 = 6 \ \Omega$

Since,

$R_{45}$ and

$R_3$ are in parallel,

$\dfrac{1}{R_{eq}}=\dfrac{1}{3}+\dfrac{1}{6}=\dfrac{2+1}{6}=\dfrac{3}{6}$

$R_{eq}=2 \ \Omega$

2112513_1217668_ans_7ed791360dd04938a946442cf2f2674a.png

Two copper wires are of same length one of the twice as thick as the other. Then the resistance of the two wires are in the ratio of :

0%
$1:16$
0%
$1:8$
0%
$1:4$
0%
$1:2$

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

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Practice Class 10 Physics Quiz Questions and Answers

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

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Practice Class 10 Physics Quiz Questions and Answers

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