Electricity Test

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Electricity Test - 62

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Question 1
1 / -0

The net resistance between $${\text{X and Y}}$$ is

A
$${\text{4}}\Omega {\text{ }}$$

B
$${\text{4}}{\text{.55}}\Omega {\text{ }}$$

C
$${\text{2}}\Omega $$

D
$${\text{20}}\Omega $$

Solution
Question 2
1 / -0

In the following part of the circuit, potential difference between points A and B equals to

A
$$30V$$

B
$$45V$$

C
$$-15V$$

D
$$-45V$$

Solution
Question 3
1 / -0

Two wires of same dimensions but resistivities $${\rho _1}$$ and $${\rho_2}$$ are connected in series. The equivalent resistivity of the combination is:

A
$${\rho _1} + {\rho_2}$$

B
$$\frac{1}{2}\left( {\,{\rho _1} + {\rho_2}} \right)$$

C
$$\sqrt {{\rho _1}{\rho _2}} $$

D
$$2\left( {\,{\rho _1} + {\rho_2}} \right)$$

Solution

$$\textbf{Step 1: Calculation of resistance of both wires}$$
We know that Resistance, $$R=\rho$$ $$l/A$$

As, the physical dimensions are same then Length $$l$$ and Area $$A$$ are constant
So, $$R_{1}=\rho_{1}$$ $$l/A$$
$$ \&$$ $$R_{2}=\rho_{2}$$ $$l/A$$

$$\textbf{Step 2: Equivalent resistance and resistivity}$$
For series combination $$R_{eq} = R_{1}+R_{2}=(\rho_{1}+ \rho_{2})l/A$$
Compare this with $$R_{eq} = (\rho_{eq} )l/A$$
So, $$\rho_{eq} =\rho_1+\rho_{2}$$

Hence, Option A is correct
Question 4
1 / -0

In the circuit shown in figure, the ammeter reads a current of?

A
1A

B
2A

C
0.3A

D
0.2A

Solution

$$(1) \& (1)$$ points same potential $$\& (2) \& (2)$$ points are same potential. So both the corner Resistance shorted.

Most above with has no resistance. So all are shorted.

$$I=\dfrac{2}{1}=2A$$
Question 5
1 / -0

The current in the given circuit will be:

A
$$\cfrac{1}{45}$$ ampere

B
$$\cfrac{1}{15}$$ ampere

C
$$\cfrac{1}{10}$$ ampere

D
$$\cfrac{1}{5}$$ ampere

Solution

Two $$30 \Omega$$ resistor are in series and one $$30 \Omega$$ is parallel to two $$30 \Omega$$ resistors.
$$R_s=30+30=60 \Omega $$
$$\begin{array}{l} \dfrac { 1 }{ { { R_{ p } } } } =\dfrac { 1 }{ { 30 } } +\dfrac { 1 }{ { 60 } } \\ { R_{ p } }=20\Omega \end{array}$$
From Ohm"s Law,
$$V=IR$$
$$I = \dfrac{2}{{20}} = \dfrac{1}{{10}}A$$
Question 6
1 / -0

The equivalent resistance between the points A and B is:

A
$$\dfrac{36}{7} \Omega$$

B
$$10 \Omega$$

C
$$\dfrac{85}{7} \Omega$$

D
none of these

Solution
Question 7
1 / -0

Resistors $$R^{}_{1}$$ and $$R^{}_{2}$$ have an equivalent resistance of 6 ohms when connected in the circuit shown below. The resistance of $$R_1$$ could be:

A
1

B
5

C
8

D
4

Solution

In parallel combination the equivalent resistance is always less than either resistance used,
$$\Rightarrow R_{1}>6$$ and $$R_{2}>6$$
which means $$R_{1}=8 {Ohm} $$ is feasible answer out of given options.
Question 8
1 / -0

The current through the ammeter shown in the figure is $$ \dfrac{10}{9}A$$. If each of the $$ 4 \Omega$$ resistor is replaced by $$2\Omega$$ resistor, the current in the circuit will become

A
(a)$$\frac{10}{9} A$$

B
(b)$$\frac{5}{4} A$$

C
(c)$$\frac{9}{8} A$$

D
(d)$$\frac{5}{8} A$$

Solution
Question 9
1 / -0

$$9$$ Identical wires each of Resistance $$R$$ are connected to form a closed polygon. The equivalent resistance between the ends of any side is

A
$$9R$$

B
$$R/9$$

C
$$8R/9$$

D
$$9R/8$$
Question 10
1 / -0

In the given circuit the current through the cell is:

A
$$\cfrac{4}{3}A$$

B
$$\cfrac{5}{3}A$$

C
$$\cfrac{1}{3}A$$

D
$$\cfrac{2}{3}A$$

Solution

Equivalent resistance in series is given by $$R_{eq} = R_1+R_2$$
Hence, equivalent resistance in the two branches is given by
$$R_{eq} = 1 + 1 = 2 \ \Omega$$

Equivalent resistance in parallel is given by $$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_2}$$
Hence, equivalent resistance in the two branches is given by
$$\dfrac{1}{R_{eq}} = \dfrac{1}{2}+\dfrac{1}{2} = 1 \ \Omega$$

Equivalent resistance in series in the final circuit is given by
$$R_{eq}=1+1+1= 3 \ \Omega$$

Using Ohm's law
$$I=\cfrac{V}{R_{eq}}=\cfrac{2}{3}A$$