Electricity Test

Result

Electricity Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is a conductor ?

A
Sugar

B
Graphite

C
Ceramic

D
Wood

Solution
Question 2
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 3
1 / -0

A cell of emf $$6\ V$$ and resistance $$0.5\ ohm$$ is short circuited. The current in the cell is

A
$$3\ amp$$

B
$$12\ amp$$

C
$$24\ amp$$

D
$$6\ amp$$

Solution

$$i=\dfrac Er=\dfrac{6}{0.5}=12\ amp$$
Question 4
1 / -0

The current in a car headlamp is $$3.0$$A when connected to a $$12$$V battery. What is the resistance of the lamp when it is lit?

A
$$0.25\Omega$$

B
$$4.0\Omega$$

C
$$15\Omega$$

D
$$36\Omega$$

Solution

Given that,
Current in a car headlamp ,$$I=3.0\ A$$
Voltage applied across it , $$V= 12\ V$$

Resistance of the lamp ,$$R=\dfrac VI= \dfrac{12}{3}= 4.0\ \Omega$$
Question 5
1 / -0

If four resistances are connected as shown in the fig. between A and B, the effective resistance is:

A
$$3\Omega $$

B
$$1\Omega $$

C
$$2.4\Omega $$

D
$$2\Omega $$

Solution

Simplified, or equivalent circuit diagram is shown in the figures.

From figure, effective resistance:

$$ R_{eq} = \dfrac{4 \times4}{2 \times 4}$$ $$ = 2 \Omega$$
Question 6
1 / -0

The resistivity of a conductor with a current density of $$2.5 Am$$$$^{-2}$$ when an electric field $$5\times 10^{-8}Vm^{-1}$$ is applied to it, is

A
$$1.0\times 10^{-8}\Omega m $$

B
$$2.0\times 10^{-8}\Omega m $$

C
$$0.5\times 10^{-8}\Omega m $$

D
$$12.5\times 10^{-8}\Omega m $$

Solution

$$\textbf{Step 1:}$$ $$\textbf{Relation between Electric field(E), Resistivity and Current density(J)} $$
Suppose length of conductor $$ = l$$
We know that, Potential difference, $$V=E.l$$
$$\textbf{&}$$ Current Density $$J=I/A$$
So, $$ I = J A$$
and $$ R=\dfrac{\rho l}{A}$$

From $$ V = IR$$
$$\Rightarrow E.l = J A \times \dfrac{\rho l}{A}$$
$$\Rightarrow E = J(\rho)$$
$$\therefore \rho = \dfrac{E}{J}$$

$$\textbf{Step 2: Putting the values in above equation}$$
$$\Rightarrow \rho =\dfrac{E}{J}= \dfrac{5 \times 10^{-8}Vm^{-1}}{2.5Am^2}$$ $$ = 2 \times 10^{-8} \Omega m$$

Hence, Option B is correct
Question 7
1 / -0

A wire of resistance $$20\Omega $$ is bent in the form of a square. The resistance between the ends of diagonal is:

A
$$10\Omega $$

B
$$5\Omega $$

C
$$20\Omega $$

D
$$15\Omega $$

Solution

Since the wire is bend in form of the square and we know that sides of square are equal so each side will have same resistance i.e 5 ohm. Now if we make diagonal two faces in each side of the diagonal will have two sides and resistance will be 10 ohm.
$$R=20 \Omega$$
So, $$\dfrac{R}{4}=5 \Omega$$
$$\Rightarrow$$ $$ R_{1} = 10 \Omega$$; $$R_{2} = 10\Omega$$
So, resistance between the ends of a diagonal = $$ R_{D} = \dfrac{10 \times10}{2 \times10}\Omega = 5\Omega$$
Question 8
1 / -0

The resistance of a conductor is 1.08$$\ \Omega $$. To reduce it to 1$$\ \Omega $$, the resistance that must be connected is:

A
$$0.08$$ $$\Omega $$ in series

B
$$13.5$$ $$\Omega $$ is parallel

C
$$0.08$$ $$\Omega $$ in parallel

D
$$13.5$$ $$\Omega $$ in series.

Solution

To reduce the resistance value we should connect another resistance in parallel.
So, $$ R = \dfrac{R_{1}R_{2}}{R_{1}+R_{2}}$$
or, $$ 1 = \dfrac{1.08 \times R_{2}}{1.08 + R_{2}}$$
or, $$ 1.08 + R_{2} = 1.08 R_{2}$$
or, $$ 1.08 = 0.08 R_{2}$$
$$\therefore$$ $$ R_{2} = \dfrac{108}{8} = 13.5\ \Omega$$
Question 9
1 / -0

Three resistances each of $$3\Omega $$ are connected as shown in figure. The resultant resistance between A and F is:

A
$$9\Omega $$

B
$$2\Omega $$

C
$$4\Omega $$

D
$$1\Omega $$

Solution

To draw the circuit bring point B to D and Point C to point E. From the diagram, inner two resistances are in parallel, and its equivalent is also parallel to the third resistance.
$$\therefore R_{AF} = \dfrac{\dfrac{R}{2} \times R}{\dfrac{3R}{2}}$$
$$ = \dfrac{R}{3}$$
$$\Rightarrow R_{AF} = 1 \Omega$$
Question 10
1 / -0

The least resistance that one can have from six resistors each of 0.1 ohm resistance is:

A
0.167$$\ \Omega $$

B
0.00167 $$\Omega $$

C
1.67 $$\Omega $$

D
0.0167 $$\Omega $$

Solution

Least resistance is achieved when all the resistors are connected in parallel.
$$ \therefore R_{eq} = \dfrac{R}{6} = \dfrac{0.1}{6} = 0.0167\ \Omega$$