Current Electricity Test

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Current Electricity Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

What happens when negative terminal of cell is connected to other negative terminal of cell in a particular circuit?

A
Current will not flow in circuit

B
Current will flow in circuit

C
temperature of cell will increase

D
cell gets damaged

Solution

$1.$ Current means flow of electrons and electrons are negative charged and so are attracted to the positive end of the battery and repelled by the negative end.
$2.$ When battery is connected in a circuit that lets the electron flow through it, they flow from negative to positive.
$3.$ When negative terminal of cell is connected to other negative terminal of the cell in a particular circuit then, current will not flow in circuit as electrons cannot flow from negative to negative terminal.
Question 2
1 / -0

Which of the following has negative temperature coefficient of resistance?

A
Copper

B
Aluminium

C
Iron

D
Germanium

Solution
Question 3
1 / -0

In the series combination of two or more than two resistances.

A
The current through each resistance is same

B
The voltage through each resistance is same

C
Neither current nor voltage through each resistance is same

D
Both current and voltage through each resistance are same

Solution

For resistances in series, the current through each resistors remains the same.
Question 4
1 / -0

If the length of potentiometer wire is increased, then the length of the previously obtained balance point will

A
Increase

B
Decrease

C
Remains unchanged

D
Becomes two times

Solution

When the length of potentiometer wire is increased, the potential gradient decreases and the length of previous balance point is increased.
Question 5
1 / -0

Combine three resistors $5\Omega, 4.5\Omega$ and $3\Omega$ in such a way that the total resistance of this combination is maximum:

A
$12.5\Omega$

B
$13.5\Omega$

C
$14.5\Omega$

D
$16.5\Omega$

Solution

Given resistance,
$R_1=5\Omega\\R_2=4.5\Omega\\R_1=3\Omega$

In series combination effective resistance will be maximum.
ie, $R_{eq}=R_1+R_2+R_3$
$R_{eq}=5 \Omega + 4.5 \Omega + 3 \Omega = 12.5 \Omega$

Option A
Question 6
1 / -0

A (75 - j40 ) $\Omega$ load is the connected to a coaxial line of $z _ { 0 } = 75 \Omega$ The load matching on the line can be accomplished by connecting

A
A short-circuit stub at the load

B
An inductance at the load

C
A short circuited stub at a specific distance from the load

D
A capacitance at a specific distance from the load

Solution

A (75 - j40 ) Ω load is the connected to a coaxial line of z0=75Ω The load matching on the line can be accomplished by connecting a short circuit stub at the load.
so the option is A.
Question 7
1 / -0

The value of $I$ in the figure shown below is

A
$19 A$

B
$21 A$

C
$8 A$

D
$4 A$

Solution

Krichhoff's 1st law states that the total current flowing towards a junction is equal to the total current flowing away from that junction.
At junction a : $I_1 = 20 - 5 = 15$ A
At junction b : $I_3=I_1 + 4$
$\therefore$ $I_3 =15 + 4 = 19$ A
At junction c : $I_2 = 5-3 =2$ A
At junction d : $I_4 = I_3 + I_2$
$\therefore$ $I_4 = 19+2 =21$ A
Question 8
1 / -0

The Kirchhoff's first law ( $\sum$ i=0) and second law ( $\sum$ iR= $\sum$ E) , where the symbols have their usual meanings, are respectively based on :

A
Conservation of charge, conservation of momentum

B
Conservation of energy, conservation of charge

C
Conservation of momentum, conservation of charge

D
Conservation of charge, conservation of energy

Solution

Kirchhoffs first law says that no charge can be accumulated at a junction which implies conservation of charge Kirchhoffs second law states that the energy liberated in the circuit comes from the battery which implies the conservation of energy
Question 9
1 / -0

The resistance of a wire is 5 ohm at $50^{\circ}$ C and 6 ohm at $100^{\circ}$ C. The resistance of the wire at $0^{\circ}$ C will be:

A
2 ohm

B
1 ohm

C
4 ohm

D
3 ohm

Solution

$R={ R }_{ 0 }(1+\alpha \Delta T)$

$R$ at ${ 100 }^{ 0} C=6\Omega$

$R$ at ${ 50}^{ 0 } c =5\Omega$

$\Rightarrow \dfrac{5}{(1+\alpha (50))}=\dfrac{6}{(1+\alpha (100))}$

$\Rightarrow 6(1+\alpha (50))=5(1+\alpha (100))$

$\Rightarrow \alpha =\dfrac { 1 }{ 200}$

$R={ R }_{0 } (1+\alpha \Delta T)$

$R$ at ${ 50 }^{ 0 } C =5\Omega$

${ R }_{0 }$ at ${ 0}^{ 0 } C ={ R }_{ 0}$

$\Rightarrow R={R }_{0 }(1+\dfrac { 1}{ 200} \times 50)$

$\Rightarrow 5={R }_{0 }(1+\dfrac { 1}{ 200} \times 50)$

$\Rightarrow R_0 = 4\Omega$
Question 10
1 / -0

$n$ identical cells, each of internal resistance $(r)$ are first connected in parallel and then connected in series across a resistance $( R)$ . If the current through $R$ is same in both the cases then:

A
$R = r/2$

B
$r = R/2$

C
$R = r$

D
$r = 0$

Solution

If the emf of each cell be E, then for parallel connection, equivalent emf is

$E_{eq}= \dfrac{\dfrac{E_1}{r_1}+\dfrac{E_2}{r_2}+....}{\dfrac{1}{r_1}+\dfrac{1}{r_2}+....}= \dfrac{\dfrac{E}{r}+\dfrac{E}{r}+....}{\dfrac{1}{r}+\dfrac{1}{r}+.....}=E$

And equivalent internal resistance of the cells, $r_{eq}=\dfrac{r}{n}$

So net current in parallel connection is

$I_p=\dfrac{E_{eq}}{r_{eq}+R}=\dfrac{E}{\dfrac{r}{n}+R}$

And for series connection of cells, eqv emf , $E_{eq}= E_1+E_2+...=nE$

and equivalent internal resistance for series, $r_{eq}= r_1+r_2+....=nr$

So current , $I_s=\dfrac{E_{eq}}{r_{eq}+R}=\dfrac{nE}{nr+R}$

Now according to question, both the currents are equal.

So, $I_p=I_s$ reduces to $\dfrac{E}{\dfrac{r}{n}+R}=\dfrac{nE}{nr+R}$ which gives $R=r$

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Current Electricity Test - 34

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Current Electricity Test - 34

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Question 1

Current will not flow in circuit

Current will flow in circuit

temperature of cell will increase

cell gets damaged

Solution

Question 2

Copper

Aluminium

Iron

Germanium

Solution

Question 3

The current through each resistance is same

The voltage through each resistance is same

Neither current nor voltage through each resistance is same

Both current and voltage through each resistance are same

Solution

Question 4

Increase

Decrease

Remains unchanged

Becomes two times

Solution

Question 5

12.5Ω12.5\Omega12.5Ω

13.5Ω13.5\Omega13.5Ω

14.5Ω14.5\Omega14.5Ω

16.5Ω16.5\Omega16.5Ω

Solution

Question 6

A short-circuit stub at the load

An inductance at the load

A short circuited stub at a specific distance from the load

A capacitance at a specific distance from the load

Solution

Question 7

19A19 A19A

21A21 A21A

8A8 A8A

4A4 A4A

Solution

Question 8

Conservation of charge, conservation of momentum

Conservation of energy, conservation of charge

Conservation of momentum, conservation of charge

Conservation of charge, conservation of energy

Solution

Question 9

2 ohm

1 ohm

4 ohm

3 ohm

Solution

Question 10

R=r/2R = r/2R=r/2

r=R/2r = R/2r=R/2

R=rR = rR=r

r=0r = 0r=0

Solution

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$12.5\Omega$

$13.5\Omega$

$14.5\Omega$

$16.5\Omega$

$19 A$

$21 A$

$8 A$

$4 A$

$R = r/2$

$r = R/2$

$R = r$

$r = 0$