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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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 $3000\Omega$
$V=I\times 3000=\dfrac{30}{4000}\times 3000=22.5 V$