Current Electricity Test

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Current Electricity Test - 60

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Question 1
1 / -0

The $$n$$ rows each containing in cells in series are joined in parallel. Maximum current is taken from this combination across an external resistance of $$3\Omega$$. If the total number of cells used is $$24$$ and internal resistance of each cell is $$0.5\Omega$$, then

A
$$m=8, n=3$$

B
$$m=6, n=4$$

C
$$m=12, n=2$$

D
$$m=2, n=12$$

Solution

Current is maximum when equivalent resistance is equal to external resistance of $$3$$ ohm.
Now total resistance is $$\dfrac{m\times0.5}{n}$$. (Resistances are parallel so they are directly added to each other)

According to question:
$$\dfrac{m\times0.5}{n}=3$$

$$\Rightarrow \dfrac{m}{n}=6$$
As ratio of $$m$$ and $$n$$ is 6,then best option which satisfy this relation is option C
Question 2
1 / -0

Brown,black and gold is the color band on the resistor,find its resistance?

A
$$10\times 10^{-2}\Omega$$

B
$$1\Omega$$

C
$$100 \Omega$$

D
$$10\times 10^{0}\Omega$$

Solution

The first two colours are brown and black.

For first two significant digits we need to consider digits signified by first two colours.
And, brown colour signifies $$1$$, and black colour signifies $$0$$.
And for multiplier the third colour is considered which is here gold meaning $$0.1$$.

Hence, resistor is $$10\times 0.1=1\Omega$$

Answer-(B)
Question 3
1 / -0

Blue,green,yellow and yellow is the color band on the resistor,find its resistance?

A
$$(75\times 10^4 )\Omega \pm 5$$%

B
$$(65\times 10^4 )\Omega \pm 5$$%

C
$$(52\times 10^4 )\Omega \pm 5$$%

D
$$(65\times 10^4 )\Omega \pm 10$$%

Solution

The first two colours are blue and green.

For first two significant digits we need to consider digits signified by first two colours.
And, blue colour signifies $$6$$, and green colour signifies $$5$$.
And for multiplier the third colour is considered which is here yellow meaning $$10^4$$.

And, for tolerance , fourth colour is given yellow meaning $$5\%$$.

Hence resistance is $$ (65\times 10^4)\Omega \pm5\%$$

Answer-(B)
Question 4
1 / -0

Electromotive force represents

A
force

B
energy

C
energy per unit charge

D
current

Solution

Electromotive force is the potential difference (or voltage $$V$$) of the cell.
Using $$V = \dfrac{W}{q}$$
where $$W$$ is the work done in moving a charge $$q$$ from one point to another having potential difference $$V$$.
Thus electromotive force represents energy per unit charge.
Question 5
1 / -0

A straight conductor of uniform cross-section carries a current $$I$$. Let $$S$$ be the specific charge of an electron. The momentum of all the free electron per unit length of the conductor due to their drift velocity only, is

A
$$I.S$$

B
$$\dfrac{I}{S}$$

C
$$\sqrt{\dfrac{I}{S}}$$

D
$$(\dfrac{I}{S})^2$$

Solution

Specific charge is charge per unit mass, $$S=\dfrac{e}{m_e}$$
Momentum of electrons per unit length $$=\dfrac{m_ev}{l}$$
$$=\dfrac{(e/S)v}{l}$$
$$=\dfrac{(e/S)(l/t)}{l}$$
$$=\dfrac{e/S}{t}$$
$$=\dfrac{e/t}{S}$$
$$=\dfrac{I}{S}$$
Question 6
1 / -0

The tolerance for silver color in color coding of a resistor is:

A
$$10$$%

B
$$5$$%

C
$$1$$%

D
$$15$$%

Solution

The tolerance of silver colour is $$10\%$$.
And that of gold colour is $$5\%$$.

Answer-(A)
Question 7
1 / -0

Which color band in the color coding of a resistor has the multiplier $$10^{-1}$$?

A
Silver

B
Gold

C
Yellow

D
Orange

Solution

The $$gold$$ colour has the color coding as the multiplier of $$10^{-1}=0.1$$.

These data need to be remembered.

Answer-(B)
Question 8
1 / -0

What is the minimum number of bulbs, each marked $$60W, \ 40V,$$ that can work safely when connected in series with a $$240V$$ mains supply?

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

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 $$6$$.
Question 9
1 / -0

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

A
$$10 \Omega$$

B
$$20 \Omega$$

C
$$40 \Omega$$

D
$$100 \Omega$$

Solution

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 $$\dfrac{R_2}{R_1} = \dfrac{A_1}{A_2} = \dfrac{3}{1}$$
Or $$\dfrac{R_2}{10}= \dfrac{3}{1}$$
$$\implies$$ $$R_2 = 30\Omega$$
Total resistance of the (series) combination $$R_{eq} = R_1 + R_2$$
$$\implies$$ $$R_{eq} = 10 + 30 = 40 \Omega$$
Question 10
1 / -0

Kirchoffs 1st and 2nd laws are based on conservation of

A
Energy and charge respectively

B
Charge and energy respectively

C
Mass and charge respectively

D
None of these above

Solution

Kirchhoff's $$1$$st law states that the total amount of current coming at a junction must be equal to the total amount of current going away from it i.e. the total charge must be conserved. Thus Kirchhoff's $$1$$st law is based on the conservation of charge.
Kirchhoff's $$2$$nd law states that the sum of the potential drop across all the components in a loop must be zero. Thus Kirchhoff's $$2$$nd law is based on the conservation of energy.

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Current Electricity Test - 60

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Current Electricity Test - 60

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Question 1

$$m=8, n=3$$

$$m=6, n=4$$

$$m=12, n=2$$

$$m=2, n=12$$

Solution

Question 2

$$10\times 10^{-2}\Omega$$

$$1\Omega$$

$$100 \Omega$$

$$10\times 10^{0}\Omega$$

Solution

Question 3

$$(75\times 10^4 )\Omega \pm 5$$%

$$(65\times 10^4 )\Omega \pm 5$$%

$$(52\times 10^4 )\Omega \pm 5$$%

$$(65\times 10^4 )\Omega \pm 10$$%

Solution

Question 4

force

energy

energy per unit charge

current

Solution

Question 5

$$I.S$$

$$\dfrac{I}{S}$$

$$\sqrt{\dfrac{I}{S}}$$

$$(\dfrac{I}{S})^2$$

Solution

Question 6

$$10$$%

$$5$$%

$$1$$%

$$15$$%

Solution

Question 7

Silver

Gold

Yellow

Orange

Solution

Question 8

$$2$$

$$4$$

$$6$$

$$8$$

Solution

Question 9

$$10 \Omega$$

$$20 \Omega$$

$$40 \Omega$$

$$100 \Omega$$

Solution

Question 10

Energy and charge respectively

Charge and energy respectively

Mass and charge respectively

None of these above

Solution

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