Current Electricity Test

Result

Current Electricity Test - 82

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Weekly Quiz Competition

Question 1
1 / -0

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:-

A
$$R$$

B
$$3R$$

C
$$\cfrac{R}{9}$$

D
$$\cfrac{R}{3}$$

Solution

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Question 2
1 / -0

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

A
The equivalent resistance between $$C$$ and $$G$$ is $$3\ K\Omega$$

B
The current provided by the source is $$4\ mA$$

C
The current provided by the source is $$8\ mA$$

D
Voltage across points $$G$$ and $$E$$ is $$4V$$

Solution
Question 3
1 / -0

The diagram shows three capacitor with their capacitances with breakdown voltages. what should be the maximum value of external emf of source such that no capacitor breaks down :-

A
$$ \frac{ 33 }{ 2 } $$volt

B
$$ \frac{ 11 }{ 3 } $$volt

C
$$ \frac{ 13 }{ 3 } $$volt

D
$$ \frac{ 11 }{ 2 } $$volt

Solution

Maximum charge on $$1\mu F$$, $$Q_1 = CV = 1\times 2= 2\mu C$$
Maximum charge on $$2\mu F$$, $$Q_2 = CV = 2\times 1= 2\mu C$$
Maximum charge on $$3\mu F$$, $$Q_3 = CV = 3\times 3= 9\mu C$$
So, maximum charge on the system is $$Q = 2\mu C$$
Equivalent capacitance $$\dfrac{1}{C_{eq}} = \dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$$
So, $$\dfrac{1}{C_{eq} }=\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}$$
$$\implies \ C_{eq} = \dfrac{6}{11} \ \mu F$$
External emf $$V = \dfrac{Q}{C_{eq}} = \dfrac{2}{6/11} = \dfrac{11}{3} \ $$ volt
Question 4
1 / -0

Identify which of the following diagram represent the internal construction of the coils wound in a resistance box or PO box?

A

B

C

D

Solution
Question 5
1 / -0

In the figure the magnet is moved along the axis of coil from one position to another position in $${10^{ - 3}}$$ sec.Now magnet is at rest for 2 sec, in its new position.The duration of induced emf in the coil is:

A
$${10^{ - 3}}$$ sec

B
2 sec

C
2x$${10^{ - 3}}$$ sec

D
0.5x2x$${10^{ - 3}}$$ sec sec

Solution
Question 6
1 / -0

Calculate the value of current I in given circuit .

A
2 A

B
1 A

C
1 .5 A

D
5 A

Solution

REF.Image
in the given figure inclined wire is shorting the circuit,
and due to this shorting No current will flow through
Resister B,D,E,F (as marked in figure)
so for analysis parpose we can remove this resistor and
modified figure would be :
So Total resistance = $$5+5= 10\Omega $$
so current $$(I)=\frac{V}{R}= \frac{15}{10}= 1.5A$$
Question 7
1 / -0

Dimension of electromotive force (e.m.f) is -

A
$$M{L}^{2}{T}^{-2}$$

B
$$M{L}^{2}{T}^{-2}{I}^{-3}$$

C
$$ML{T}^{-2}$$

D
$$M{L}^{2}{T}^{-3}{I}^{-1}$$

Solution
Question 8
1 / -0

EMF of a battery, when compared to its potential difference, is

A
less

B
more

C
equal

D
none of these
Question 9
1 / -0

Kirchhoff's equation is

A
$$log \frac{K_2}{K_1} = \frac{E_a}{2.303 R}[\frac{1}{T_1} - \frac{1}{T_2}]$$

B
$$log \frac{P_2}{P_1} = \frac{\Delta H_v}{2.303 R}[\frac{T_2 - T_1}{T_1 T_2}]$$

C
$$\Delta C_p = \frac{\Delta H_2 - \Delta H_1}{T_2 - T_1}$$

D
$$log \frac{K_2}{K_1} = \frac{\Delta H_v}{2.303 R}[\frac{1}{T_1} - \frac{1}{T_2}]$$

Solution
Question 10
1 / -0

Diagram shows three capacitors with capacitance and breakdown voltage mentioned. What should be maximum value of the external emf of source such that no capacitor breaks down?

A
1 V

B
2 V

C
1.5 V

D
4 V

Solution

We know that
$$q=cv$$
Total charge of two parallel circuit is ,
$${ q }_{ T }=CV+2CV\\ \quad \quad q_T=3CV$$

Maximum charge stored by 6C capacitor
$$q=6C\times V=6CV$$

Hence q < q$$_T$$ So Maximum charge flow through the battery $$3CV$$
And
$${ C }_{ eq }=\dfrac { 3\times 6 }{ 3+6 } =2C$$

$$Source\quad voltage=\dfrac { 3C }{ 2C } =1.5V$$

CORRECT OPTION IS C