Electricity Test

Result

Electricity Test - 61

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

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

A
$$\varepsilon, \dfrac{r}{n}$$

B
$$\varepsilon, nr$$

C
$$n\varepsilon, \dfrac{r}{n}$$

D
$$n\varepsilon, nr$$

Solution

$$\cfrac{1}{Net_r}=\cfrac{1}{r}+\cfrac{1}{r}....n times$$
$$\cfrac{1}{r_{net}}=\cfrac{n}{r}\\r_{net}=\cfrac{v}{n}$$
and $$emf=\epsilon$$
Answer $$\epsilon,\cfrac{v}{n}$$
Question 2
1 / -0

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.

A
$$n\varepsilon, (nr-3r)$$

B
$$(n\varepsilon -2\varepsilon), nV$$

C
$$(n\varepsilon -4\varepsilon), nV$$

D
$$(n\varepsilon - 6\varepsilon), nV$$

Solution

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.
Question 3
1 / -0

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

A
$$4P$$

B
$$7P$$

C
$$13P$$

D
$$17P$$

Solution

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=$$I^24R_1$$
$$=4I^2R_1$$
$$=4P$$
Question 4
1 / -0

The conventional electric current flows from

A
higher potential to lower potential

B
lower potential to higher potential

C
Both (A) and (B)

D
None of these

Solution

Conventional current or simply current behaves as if positive charge carriers cause current flow. Conventional current flows from the positive terminal to the negative terminals that means it flows from higher potential to lower potential.
Question 5
1 / -0

If two bulbs of $$25W$$ and $$100W$$ rated at $$200$$ volts are connected in series across a $$440$$ volts supply, then

A
$$100$$ watt bulb will fuse

B
$$25$$ watt bulb will fuse

C
none of the bulb will fuse

D
both the bulbs will fuse

Solution

$${ R }_{ 1 }=\cfrac { { v }^{ 2 } }{ P } =\cfrac { { 200 }^{ 2 } }{ 25 } =1600\Omega $$
$${ R }_{ 2 }=\cfrac { { 200 }^{ 2 } }{ 100 } =400\Omega $$
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
Question 6
1 / -0

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?

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

$$n$$ is minimum when two resistances are the same.
So,
$$S=2R$$
and,
$$P=\dfrac{R}{2}$$

$$\Rightarrow S = nP$$

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

$$\Rightarrow n = 4$$
Question 7
1 / -0

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

A
$$6\Omega$$

B
$$12\Omega$$

C
$$4\Omega$$

D
$$18\Omega$$

Solution

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
$$\dfrac{1}{R_e}=\dfrac{1}{18}+\dfrac{1}{9}=\dfrac{3}{18}$$
$$\boxed{R_e=6\Omega}$$
Question 8
1 / -0

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

A
$$10R$$

B
$$(6/11)R$$

C
$$(12/15)R$$

D
$$(50/11)R$$

Solution

$$2R, R$$ 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
$$\dfrac{6R}{11}+4R=\dfrac{6R+44R}{11}=\left(\dfrac{50}{11}\right)R$$
$$\boxed{Req_0=\left(\dfrac{50}{11}\right)R}$$
Question 9
1 / -0

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:

A
$$nR$$

B
$$\dfrac{R}{n}$$

C
$$\dfrac{R}{n^2}$$

D
$$\dfrac{n^2}{R}$$

Solution
Question 10
1 / -0

Three resistance $$R, 2R$$ and $$3R$$ are connected in parrallel to a battery ,then

A
The potential drop across $$3R$$ is maximum

B
The current through each resistance is same

C
The heat developed in $$3R$$ is maximum

D
The heat developed in $$R$$ is maximum

Solution

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 $$R$$.