Current Electricity Test

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Current Electricity Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

The four wires from a larger circuit intersect at junction A a shown. What is the magnitude and direction of the current between points A and B?

A
2 A from A to B

B
2 A from B to A

C
3 A from A to B

D
3 A from B to A

Solution

Kirchhoff's junction rule states that the algebraic sum of all currents into and out of branch point is zero i.e,$$\Sigma I=0$$. By convention, the sign of current entering a junction is positive and current leaving a junction is negative.
Thus, at junction A, $$4+5-6=3A$$. It is positive so the current 3A will flow from A to B.
Question 2
1 / -0

The resistance of a copper wire and an iron wire at 20C are 4.$$\Omega $$ and 3.9$$\Omega $$ respectively. Neglecting any thermal expansion, find the temperature at which resistances of bolh are equal.
$$\alpha _{ cu }=4.0\times { 10 }^{ -3 }{ K }^{ -1 }$$. and $$\alpha _{ fe }=4.0\times { 10 }^{ -3 }{ K }^{ -1 }$$.

A
$${ 65 }^{ o }C$$

B
$${ 75 }^{ o }C$$

C
$${ 270 }^{ o }C$$

D
$${ 95 }^{ o }C$$

Solution
Question 3
1 / -0

Two wires of same metal have the same length but their cross-sections are 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 is:

A
$$\displaystyle \frac {5}{2} \Omega$$

B
$$\displaystyle \frac {40}{3} \Omega$$

C
$$40\Omega$$

D
$$100\Omega$$

Solution

Resistance of a wire $$R=\dfrac{\rho l}{A}$$ where $$\rho=$$ resistivity, $$l=$$ length, $$A=$$ cross section of the wire.
As both have same material and length so $$R \propto \dfrac{1}{A}$$
Thus, $$\dfrac{R_2}{R_1}=\dfrac{A_1}{A_2}=\dfrac{3}{1} \Rightarrow R_2=3R_1$$.
here $$R_1$$ is the resistance of thicker wire so its resistance $$R_1=10 \Omega$$ (given)
so, $$R_2=3(10)=30 \Omega$$
As they are connected in series so the equivalent resistance is
$$R_{eq}=R_1+R_2=10+30=40 \Omega$$
Question 4
1 / -0

The resistance of a wire at room temperature $$30^oC$$ is found to be $$10\Omega$$. Now to increase the resistance by 10%, the temperature of the wire must be: [The temperature coefficient of resistance of the material of the wire is 0.002 per $$^oC$$]

A
$$36^oC$$

B
$$83^oC$$

C
$$63^oC$$

D
$$33^oC$$

Solution

Given: Temperature coefficient of resistance $$=\alpha=0.002 /^oC$$
We know that, $$R_T=R_0(1+\alpha T)$$
Initially, $$10=R_0(1+\alpha .30) ...(1)$$
After $$10\%$$ increase the resistance becomes $$=10+10\times \dfrac{10}{100}=11 \Omega$$ and corresponding temp is $$T$$.
Now, $$11=R_0(1+\alpha T) ...(2)$$

$$\displaystyle (2)/(1) \Rightarrow \frac{11}{10}=\frac{R_0(1+\alpha T)}{R_0(1+30 \alpha)}$$

$$\displaystyle 1+\alpha T=1.1+33\alpha$$

$$\displaystyle T=\frac{1.1+33\alpha -1}{\alpha}=\frac{0.1+33\times 0.002}{0.002}=83 ^oC$$
Question 5
1 / -0

In an experiment to measure the internal resistance of a cell by a potnetiometer, it is found that the balance point is at a length of 2 m, when the cell is shunted by a 5 $$\Omega$$ resistance; and is at a length of 3 m, when the cell is shunted by a 10 $$\Omega$$ resistance. The internal resistance of the cell is, then

A
$$1.5 \Omega$$

B
$$10 \Omega$$

C
$$15 \Omega$$

D
$$1 \Omega$$

Solution

In case of internal resistance measurement by potentiometer,
$$ \displaystyle \frac{V_1}{V_2} = \frac{l_1}{l_2} = \frac{\{ ER_1/(R_1 + r)\}}{\{ ER_2/(R_2 + r)\}} = \frac{R_1 (R_2 + r)}{R_2 (R_1 + r)} $$

Here, $$l_1=2m,l_2=3m,R_1=5\Omega $$ and $$R_2=10\Omega $$

$$ \implies \displaystyle \frac{2}{3} = \frac{5 (10 + r)}{10 (5 + r)} $$

$$\implies 20 + 4r = 30 + 3 r $$

$$\implies r = 10 \Omega $$
Question 6
1 / -0

A bulb is connected to a cell. How is the resistance of circuit affected if another identical bulb is connected in series ?

A
Resistance is doubled

B
Resistance is same.

C
Resistance is tripled

D
Resistance is one half.

Solution

Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. The total resistance of resistors in series is equal to the sum of their individual resistances. That is,
$${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$$.
When a bulb of the same resistance is connected in the series, the total resistance in the circuit is given as sum of both resistances connected. Hence the resistance is doubled.
Question 7
1 / -0

A potentiometer consists of a wire of length 4m and resistance 10$$\Omega$$. It is connected to a cell of e.m.f. 3V. The potential gradient of wire is

A
0.75 V/m

B
2 V/m

C
6 V/m

D
10 V/m

Solution

$$\textbf{Step 1: Potentiometer Figure} $$

Using KVL in above simple circuit of 1 loop, we get current $$ I = \dfrac{V}{R} = \dfrac{3}{10} = 0.3\ A$$

AB is potentiometer wire
From figure we can say that potential difference across AB will be equal to e.m.f of cell.

$$V_{AB} = E = 3\ V$$, Also $$V_{AB} = IR$$

$$\Rightarrow\ \ V_{AB} = 0.3 \times 10 = 3\ V $$

$$\textbf{Step 2: Potential gradient calculation}$$

Potential gradient $$K = \dfrac{V_{AB}}{L}$$

$$= \dfrac{3}{4} = 0.75\ V/m$$

Hence, Option $$A$$ is correct.
Question 8
1 / -0

Value of Current i in the following circuit is :-

A
13 A

B
12 A

C
9 A

D
None of these

Solution

According to Kirchhoff's first law, total entering current into the circuit $$=$$ total leaving current from the circuit. i.e, $$10+1+2=i $$ or $$ i=13 A$$
Question 9
1 / -0

Which device is used to measure the potential difference between two points of a conductor in the laboratory?

A
Voltameter

B
Ammeter

C
Potentiometer

D
Galvanometer

Solution

The null method gives the most accuracy result in potential difference measurement. So the potentiometer is used to measure potential difference between two points of a conductor.
Question 10
1 / -0

The principle involved in potentiometer is

A
variation of current with variati wire on in the diameter of the potentiometer

B
similar to the principle of Wheatstone bridge

C
variation of resistance with temperature

D
both (a) and (b)

Solution

The working principle of potentiometer depends upon the galvanometer null detection process.The functionality of a Wheatstone bridge is similar to the original potentiometer.

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Current Electricity Test - 49

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Current Electricity Test - 49

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SHARING IS CARING

Question 1

2 A from A to B

2 A from B to A

3 A from A to B

3 A from B to A

Solution

Question 2

$${ 65 }^{ o }C$$

$${ 75 }^{ o }C$$

$${ 270 }^{ o }C$$

$${ 95 }^{ o }C$$

Solution

Question 3

$$\displaystyle \frac {5}{2} \Omega$$

$$\displaystyle \frac {40}{3} \Omega$$

$$40\Omega$$

$$100\Omega$$

Solution

Question 4

$$36^oC$$

$$83^oC$$

$$63^oC$$

$$33^oC$$

Solution

Question 5

$$1.5 \Omega$$

$$10 \Omega$$

$$15 \Omega$$

$$1 \Omega$$

Solution

Question 6

Resistance is doubled

Resistance is same.

Resistance is tripled

Resistance is one half.

Solution

Question 7

0.75 V/m

2 V/m

6 V/m

10 V/m

Solution

Question 8

13 A

12 A

9 A

None of these

Solution

Question 9

Voltameter

Ammeter

Potentiometer

Galvanometer

Solution

Question 10

variation of current with variati wire on in the diameter of the potentiometer

similar to the principle of Wheatstone bridge

variation of resistance with temperature

both (a) and (b)

Solution

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