Current Electricity Test

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Current Electricity Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

Three resistors of $4.0\ \Omega$ , $6.0\ \Omega$ and $10.0\ \Omega$ are connected in series. What is their equivalent resistance?

A
$20\ \Omega$

B
$7.3\ \Omega$

C
$6.0\ \Omega$

D
$4.0\ \Omega$

Solution

The equivalent resistance of N resistors connected in series is given by,
$R^{}_{Equivalent} = R^{}_{1} + R^{}_{2} + R^{}_{3} + . . . +R^{}_{N}$
Hence,
$R^{}_{Equivalent} = 4.0 + 6.0 + 10.0 = 20.0 \Omega$
Question 2
1 / -0

Two students set up their circuits for finding the equivalent resistance of two resisto connected in series in two different ways as shown.
The circuit (s) likely to be labelled as incorrect :

A
are neither of the two circuits

B
is only circuit I

C
is only circuit II

D
are both one circuits

Solution

In correct circuit, Ammeter should linked in series whereas voltmeter should linked in parallele.
Hence circuit II is wrong
Option C
Question 3
1 / -0

Two wires of same metal have the same length but their cross-sections area in the ratio 3 : 1. They are joined in series. The resistance of the thicker wire is $10\Omega$ .The total resistance of the combination will be:

A
$40 \Omega$

B
$40/3 \Omega$

C
$5/2 \Omega$

D
$100 \Omega$

Solution

(Let R' denote thicker and R denote thinner resistor)
We know,
$R = \dfrac{\rho \times L}{A} \implies \dfrac{R'}{R} = \dfrac{A}{A'}$
$\implies \dfrac{10}{R} = \dfrac{1}{3} \implies R = 30 \Omega$

Since they are connected in series,the equivalent resistance is given by,
$R^{}_{eq} = R + R' \implies R^{}_{eq} = 30 + 10 = 40 \Omega$
Question 4
1 / -0

The resistors of resistances $2 \Omega, 4 \Omega, 5 \Omega$ are connected in parallel. The total resistance of the combination will be :

A
$\displaystyle \frac{29}{10} \Omega$

B
$\displaystyle \frac{19}{20} \Omega$

C
$\displaystyle \frac{10}{20} \Omega$

D
$\displaystyle \frac{20}{19} \Omega$

Solution

According to parallel combination.
$\displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}+\frac{1}{R_3}$
$=\displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{5}$
$=0.5 + 0.25 + 0.2$
$=0.95 \Omega$
$\therefore R_{eq} \displaystyle =\frac{1}{0.95} =\frac{20}{19} \Omega$
Question 5
1 / -0

What is the maximum resistance one can make with ten $1 \Omega$ resistors?

A
$1 \Omega$

B
$2 \Omega$

C
$5 \Omega$

D
$10 \Omega$

Solution

Given ten $1 \Omega$ resistors.
If all these resistors are connected in series a maximum value of resistance is obtained then
$R_{eq} = (1+1+1+1+1+1+1+1+1+1) \Omega = 10 \Omega$
Question 6
1 / -0

The value of current $i$ in the given circuit is:

A
Zero

B
5 Amp.

C
7 Amp.

D
11 Amp.

Solution

$\textbf{Hint:}$
Kirchhoff's Current Laws states that net current coming towards a junction is zero.

$\textbf{Step 1: Note the values of current coming to and leaving the junction.}$

Current coming to junction is $= 5+7+2 = 14A$
Current leaving the junction is $= (3+i)A$

$\textbf{Step 2: Use Kirchhoff's Current Law and find the value of $$i$$.}$
Kirchhoff's First Law is also known as Kirchhoff's Current Law. It says that current coming towards junction is equal to the current leaving the junction.
By using KCL we have
$\text{Current coming to junction} = \text {Current leaving the junction}$
$\Rightarrow 14 = 3+i$
$\Rightarrow i=11A$

Thus, the value of $i$ is $11A$ .
Option D is correct.
Question 7
1 / -0

A battery of 20 cells (each having e.m.f. 1.8 volt and internal resistance 0.1 ohm) is charged by 220 volts and the charging current is 15A. The resistance to be put in the circuit is

A
10.27 ohms

B
12.27 ohms

C
8.62 ohms

D
16.24 ohms

Solution

The total emf of the 20 cells connected in series are $1.8\times 20=36 V$ .
Hence, the total voltage in the circuit is $220V - 36V = 184 V.$
From the Ohm's law, $V=IR$ . Substituting the values of Voltage and current in the equation

$R=\dfrac { V }{ I } ,\quad we\quad get\quad R=\dfrac { 184 }{ 15 } =12.27\quad ohms$ .

The internal resistance of the 20 cells is given as $20\times 0.1ohms=2 ohms$ .
So, the total resistance R in the circuit is $=12.27-2 ohms = 10.27 ohms.$
Hence, the resistance in the circuit is 10.27 ohms.
Question 8
1 / -0

If a resistance $5\Omega$ is connected in the left gap of a meter bridge and $15\Omega$ in the other gap then position of balancing point is

A
10 cm

B
20 cm

C
25 cm

D
75 cm

Solution

Consider the figure , it is given that $R=5 \ ohm$ and $S= 15 \ ohm$
Now, at the balance point the circuit will be like a Wheatstone bridge .
That is,

$R/S = l_{1}/ (100 - l_{1})$

$5/15 = l_{1}/ (100 - l_{1})$

$4*l_{1} = 100$
$l_{1} = 25cm$
Question 9
1 / -0

Five resistors each of 1 ohm are connected in parallel. The resultant resistance of the combination is :

A
$5 \Omega$

B
$1 \Omega$

C
$0.2 \Omega$

D
$2 \Omega$

Solution

$\displaystyle R_{eq} = \frac{R}{5}$
Given $R= 1 \Omega$
$R_{eq} =\frac{1}{5} = 0.2 \Omega$
Question 10
1 / -0

Kirchhoff's second law is based on the law of conservation of

A
charge

B
energy

C
momentum

D
sum of mass and energy

Solution

Hint:
Kirchhoff's second law states that the algebraic sum of the potential difference across all the elements in a closed loop must be zero.

Explanation :
Kirchhoff's Second Law is also known as Kirchhoff's Voltage Law. It says that the algebraic sum of the potential differences across all the elements in a closed loop is zero. In other words, it says that voltage supplied is always equal to the voltage consumed in a closed-loop, that is voltage/energy is conserved.

Thus, it is based on the Law of Conservation of Energy

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Current Electricity Test - 42

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Current Electricity Test - 42

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Question 1

20 Ω20\ \Omega20 Ω

7.3 Ω7.3\ \Omega7.3 Ω

6.0 Ω6.0\ \Omega6.0 Ω

4.0 Ω4.0\ \Omega4.0 Ω

Solution

Question 2

are neither of the two circuits

is only circuit I

is only circuit II

are both one circuits

Solution

Question 3

40Ω40 \Omega40Ω

40/3Ω40/3 \Omega40/3Ω

5/2Ω5/2 \Omega5/2Ω

100Ω100 \Omega100Ω

Solution

Question 4

2910Ω\displaystyle \frac{29}{10} \Omega1029​Ω

1920Ω\displaystyle \frac{19}{20} \Omega2019​Ω

1020Ω\displaystyle \frac{10}{20} \Omega2010​Ω

2019Ω\displaystyle \frac{20}{19} \Omega1920​Ω

Solution

Question 5

1Ω1 \Omega1Ω

2Ω2 \Omega2Ω

5Ω5 \Omega5Ω

10Ω10 \Omega10Ω

Solution

Question 6

Zero

5 Amp.

7 Amp.

11 Amp.

Solution

Question 7

10.27 ohms

12.27 ohms

8.62 ohms

16.24 ohms

Solution

Question 8

10 cm

20 cm

25 cm

75 cm

Solution

Question 9

5Ω 5 \Omega5Ω

1Ω 1 \Omega1Ω

0.2Ω 0.2 \Omega0.2Ω

2Ω 2 \Omega2Ω

Solution

Question 10

charge

energy

momentum

sum of mass and energy

Solution

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$20\ \Omega$

$7.3\ \Omega$

$6.0\ \Omega$

$4.0\ \Omega$

$40 \Omega$

$40/3 \Omega$

$5/2 \Omega$

$100 \Omega$

$\displaystyle \frac{29}{10} \Omega$

$\displaystyle \frac{19}{20} \Omega$

$\displaystyle \frac{10}{20} \Omega$

$\displaystyle \frac{20}{19} \Omega$

$1 \Omega$

$2 \Omega$

$5 \Omega$

$10 \Omega$

$5 \Omega$

$1 \Omega$

$0.2 \Omega$

$2 \Omega$