Electricity Test

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Electricity Test - 69

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Weekly Quiz Competition

Question 1
1 / -0

The power consumed by $$4$$V battery in the circuit as shown is?

A
$$8$$W

B
$$7$$W

C
$$6$$W

D
$$5$$W

Solution

$$E_{\text{net}} = 10 - 4 = 6\ V $$
$$R = 3\ \Omega $$
$$i = \dfrac{V}{R} = \dfrac{6}{3} = 2A $$

$$\therefore $$ Power consumed by $$4V$$ battery
$$= VI = 4 \times 2 $$
$$ = 8 VA = 8W$$
Question 2
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An electrical conductor has a resistance of $$5.6k\Omega$$. A potential difference (p.d.) of $$9.0V$$ is applied
across its ends.
How many electrons pass a point in the conductor in one minute?

A
$$6.0\times 10^{20}$$

B
$$1.0\times 10^{19}$$

C
$$6.0\times 10^{17}$$

D
$$1.0\times 10^{16}$$

Solution

We know that
$$V = IR$$
$$\implies I = \dfrac{V}{R}$$
$$\implies I = \dfrac{9}{5.6 \times 10^3}$$A
This means that no.of electron passes through a point in one second = $$\dfrac{9}{5.6 \times 10^3 \times 1.6 \times 10^{-19}}$$
$$\implies $$ no. of electron passes through a point in one minute = $$\dfrac{9}{5.6 \times 10^3 \times 1.6 \times 10^{-19}} \times 60$$

=$$6.0 \times 10^{17 } $$ electrons
Question 3
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The current in a kettle is $$10\ A$$ and the kettle is protected by a $$13\ A$$ fuse. The owner of the kettle replaces the $$13\ A$$ fuse with a $$3\ A$$ fuse. What happens when the kettle is switched on?

A
The fuse melts and the kettle might be damaged

B
The fuse melts and the kettle is undamaged

C
The fuse does not melt and the kettle works correctly

D
The fuse does not melt but the kettle fails to work

Solution

The current rating of the new fuse is $$3\ A$$. Therefore, if the current passing through it exceeds $$3\ A$$, the fuse wire will melt and the electric device will be safe.
So, when a $$10\ A$$ is passing, the fuse wire will melt and the kettle will remain undamaged.
Question 4
1 / -0

If two identical heaters each rated as ($$1000$$W-$$220$$V) are connected in parallel to $$220$$V, then the total power consumed is?

A
$$200$$W

B
$$2500$$W

C
$$250$$W

D
$$2000$$W

Solution
Question 5
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Directions For Questions

A network of resistance is constructed with $$R_{1}$$ and $$R_{2}$$ as shown in Fig $$5.243$$. The potential at the points $$1,2,3 ... N$$ are $$V_{1}, V_{2}, V_{3},.... V_{n}$$, respectively, each having a potential $$k$$ times smaller than the previous one.

...view full instructions

The ratio $$R_{2}/R_{3}$$ is

A
$$\dfrac{(k-1)^{2}}{k}$$

B
$$k^{2}-\dfrac{1}{k}$$

C
$$\dfrac{k}{k-1}$$

D
$$k-\dfrac{1}{k^{2}}$$

Solution

Current in $$R_1$$ and $$R_2$$ will be the same:

$$\dfrac {V_{n-1}-V_n}{R_1}=\dfrac {V_n}{R_3}$$

$$\dfrac {V_{n-1}-\dfrac {V_{n-1}}{k}}{R_1}=\dfrac {V_{n-1}}{k\ R_3}$$

$$\Rightarrow \ R_1 =R_3 (k-1)$$

Put the value of $$R_1$$ in the above expression, we get:
$$\dfrac {R_2}{R_3}=\dfrac {k}{k-1}$$
Question 6
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The equivalent resistance between $$A$$ and $$B$$ ( of the circuit as shown ) is

A
$$4.5\Omega$$

B
$$12\Omega$$

C
$$5.4\Omega$$

D
$$20\Omega$$

Solution

Clearly from the figure,
Resistors $$7\Omega$$ and $$3\Omega$$ are in parallel; $$6\Omega$$
and $$4\Omega $$ are in parallel and both in series.

So $$R_{eq}=\dfrac{7\times 3}{7+3}+\dfrac{4\times 6}{4+6}=4.5\Omega$$
Question 7
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The equivalent resistance between $$A$$ and $$B$$ in the network in figure is

A
$$\dfrac{4}{3}\Omega$$

B
$$\dfrac{3}{2}\Omega$$

C
$$3\Omega$$

D
$$2\Omega$$

Solution

Above circuit can be reduced as shown ,
Now,
The equivalent of the network is given in figure.
The equivalent of the above network is a parallel combination of $$3\Omega, 4\Omega $$ and $$6\Omega$$, i.e,
$$\dfrac{1}{R}=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{6}\Rightarrow R=\dfrac{4}{3}\Omega$$
Question 8
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Directions For Questions

A network of resistance is constructed with $$R_{1}$$ and $$R_{2}$$ as shown in Fig $$5.243$$. The potential at the points $$1,2,3 ... N$$ are $$V_{1}, V_{2}, V_{3},.... V_{n}$$, respectively, each having a potential $$k$$ times smaller than the previous one.

...view full instructions

The ratio $$R_{1}/R_{2}$$ is

A
$$k^{2}-\dfrac{1}{k}$$

B
$$\dfrac{k}{k-1}$$

C
$$k-\dfrac{1}{k^{2}}$$

D
$$\dfrac{(k-1)^{2}}{k}$$

Solution

Given ,

$$V_1 =\dfrac {V_0}{k}, V_2 =\dfrac {V_1}{k}, V_3 =\dfrac {V_2}{k},$$

$$\ I=I_1 +I_2$$

$$\dfrac {V_0-V_0 /k}{R_1}=\dfrac {V_0/ k-V_0 /k^2}{R_1}+\dfrac {V_0 /k}{R_2}$$

$$\dfrac {R_1}{R_2}=\dfrac {(k-1)^2}{k}$$
Question 9
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A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if

A
both the length and the radius of the wire are halved

B
both the length and the radius of the wire are doubled

C
the radius of the wire is doubled

D
the length of the wire is doubled

Solution

The heat produced in the metallic wire is given using
$$H=\dfrac{V^{2}t}{R}$$
As $$R=\dfrac{\rho L}{A}$$
or
$$H=\dfrac{V^{2}t\pi r^{2}}{\rho L}$$
Therefore, heat produced will be doubled if both length and the radius of the wire are doubled.
Question 10
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In Fig. when an ideal voltmeter is connected across 4000$$\Omega $$ resistance, it reads 30 V. If the voltmeter is connected across 3000$$\Omega $$ resistance, it will read

A
$$20 V$$

B
$$22.5 V$$

C
$$35 V$$

D
$$40 V$$

Solution

Let $$I$$ be the current in the circuit then, $$I\times R=V$$
$$\Rightarrow I\times 4000=30 V$$
$$\Rightarrow I=\dfrac{30}{4000}$$
Reading of voltmeter across $$3000\Omega$$
$$V=I\times 3000=\dfrac{30}{4000}\times 3000=22.5 V$$