Electricity Test

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Electricity Test - 40

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Question 1
1 / -0

What is the approximate resistance setting of a rheostat in which $$650 \ mA$$ of current flows with a $$150 \ V$$ source?

A
$$230 \ \Omega$$

B
$$56 \ \Omega$$

C
$$15 \ \Omega$$

D
$$13 \ \Omega$$

Solution

Given:
Current $$I=650 \ mA=650 \times 10^{-3} \ A$$
Voltage $$V=150 \ V$$

By Ohm's Law,
$$R=\dfrac{V}{I}$$

$$\implies R=\dfrac{150}{650\times 10^{-3}}$$

$$\implies R\approx 230 \ \Omega$$

Answer: Option (A)
Question 2
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In the circuit shown below in Fig., the value of x if the equivalent resistance between A and B is $$4\, \Omega$$ will be:

A
$$2\, \Omega$$

B
$$1\, \Omega$$

C
$$1.5\, \Omega$$

D
$$3\, \Omega$$

Solution

In this case, the 2 resistors of 4 ohms and 8 ohms are connected in series and the combined resistance is given as 4+8 = 12 ohms.
The 2 resistors of x ohms and 5 ohms are connected in series and the combined resistance is given as x+5 ohms.
The above 2 sets of resistors are connected in parallel. Thus, the total resistance is expressed as $$\frac { 1 }{ 12 } +\frac { 1 }{ x+5 } =\frac { 1 }{ 4 } as\quad the\quad total\quad resistance\quad is\quad given\quad as\quad 4\quad ohms\quad in\quad the\quad question.$$
Solving for x in the above equation we get x=0.99. That is. 1 ohm approximately.
Hence, the value of the unknown resistance in the circuit is 1 ohm.
Question 3
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Same current flows in each resistor when the two resistors are joined with a battery in :

A
Series

B
Parallel

C
Series and parallel.

D
none of above.

Solution

Same current flows in each resistor when they are connected in series.
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. In a series circuit, every device must function for the circuit to be complete. One defective bulb in a series circuit breaks the circuit. A circuit composed solely of components connected in series is known as a series circuit.
The total resistance of resistors in series is equal to the sum of their individual resistances. That is, $${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }$$. The current is given as $$I={ I }_{ 1 }={ I }_{ 2 }$$.
Above is the diagrammatic representation of 2 resistors $${ R }_{ 1 }\ and\ { R }_{ 2 }$$ connected in series.
Question 4
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When two resistors with resistances $$R_1$$ and $$R_2$$ are connected in series, then the total equivalent resistance is

A
$$\dfrac{R_1R_2}{R_1+R_2}$$

B
$$R_1+R_2$$

C
$$1/R_1+1/R_2$$

D
$$R_2-R_1$$

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. In a series circuit, every device must function for the circuit to be complete. One bulb burning out in a series circuit breaks the circuit. A circuit composed solely of components connected in series is known as a series circuit.
The total resistance of resistors in series is equal to the sum of their individual resistances. That is, $${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }$$. The current is given as $$I={ I }_{ 1 }={ I }_{ 2 }$$.
Below is the diagrammatic representation of 2 resistors $${ R }_{ 1 }\quad and\quad { R }_{ 2 }$$ connected in series.
Question 5
1 / -0

Equivalent resistance is more than either of the two resistances when they are in:

A
Series

B
Series and parallel

C
Parallel

D
none of these.

Solution

Equivalent resistance is more than either of the two resistances when they are connected in series.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. In a series circuit, every device must function for the circuit to be complete. One bulb burning out in a series circuit breaks the circuit. A circuit composed solely of components connected in series is known as a series circuit.The total resistance of resistors in series is equal to the sum of their individual resistances.
Hence, the equivalent resistance is more than the individual resistances because a sum is taken of all the individual resistances.
That is, $${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }$$. The current is given as $$I={ I }_{ 1 }={ I }_{ 2 }$$.
Diagrammatic representation of 2 resistors $${ R }_{ 1 } \ and \ { R }_{ 2 }$$ connected in series is given.
Question 6
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In the figure the equivalent resistance of the following combination of resistors $$r_1,\, r_2,\, r_3$$ and $$r_4$$, if $$r_1\, =\, r_2\, =\,r_3\, =\, r_4\, =\, 2.0\, \Omega$$, between the points A and B is

A
$$2.0\, \Omega$$

B
$$4.0\, \Omega$$

C
$$6.0\, \Omega$$

D
$$5.0\, \Omega$$

Solution

In this case, the first two resistors of 2 ohms each are connected in series and the third and fourth resistors of 2 ohms each are connected in parallel. The first set is connected in series with the second set of resistors.
The resistances R3 and R4 that are connected in parallel have a resistance given as $$\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =\frac { 1 }{ 1 } =1\quad ohm$$. That is, these two resistances is reduced to a single resistance of 1 ohm.
The resistances R1 and R2 that are connected in series have a resistance given as 2+2 = 4 ohms.
This combined resistance when connected in series with reduced resistance of 1 ohm gives a total resistance of 4+1 = 5 ohms.
Hence, the total resistance in the circuit is 5 ohms.
Question 7
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Four resistors each of resistance $$2\, \Omega$$ are connected in parallel. What is the effective resistance?

A
$$2\, \Omega$$

B
$$5\, \Omega$$

C
$$0.5\, \Omega$$

D
$$8\, \Omega$$

Solution

$$\textbf{Step 1: Calculation of net resistance}$$
For parallel arrangement,

$$\dfrac { 1 }{ { R }_{ eq } } =\dfrac { 1 }{ { R }_{ 1 } } +\dfrac { 1 }{ { R }_{ 2 } } +\dfrac { 1 }{ { R }_{ 3 } } +\dfrac { 1 }{ { R }_{ 4 } } $$.

Resistance of every resistor $$=2 \Omega$$

$$\Rightarrow \dfrac { 1 }{ { R }_{ eq } } =\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } =2$$

$$\Rightarrow R_{eq}=\dfrac { 1 }{ 2 } =0.5\quad \Omega $$.

Hence, Option C is correct.
Question 8
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The equivalent resistance R of three resistors $$R_1$$, $$R_2$$ and $$R_3$$ joined in parallel is

A
$$\displaystyle R= R_1+R_2+R_3$$

B
$$\displaystyle R= 1/R_1+1/R_2+1/R_3$$

C
$$\displaystyle R= \frac{R_1R_2R_3}{R_1R_2+R_2R_3+R_3R_1}$$

D
$$\displaystyle R=1/(R_1+R_2+R_3)$$

Solution

Equivalent resistance $$R$$ is given by:
$$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$$
$$\Rightarrow\ R=\dfrac{R_1R_2R_3}{R_1R_2+R_2R_3+R_3R_1}$$
Question 9
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Two resistances are connected in two configurations- series or parallel. Which represents the parallel combination?

A
A

B
A and B both

C
B

D
None of these.

Solution

The slope of V vs. I graph represents resistance. Steeper the graph, the higher the resistance.
On the other hand, when two resistances are joined in series and in parallel, the equivalent resistance in parallel combination is less.
Thus A being less steep, indicates lower resistance, i.e., one received from the parallel combination.
Question 10
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You have three resistors of values $$2\, \Omega,\, 3\, \Omega$$, and $$5\, \Omega$$. If they are in parallel ,then the total resistance is :

A
$$0.97\, \Omega$$

B
$$2 \, \Omega$$

C
$$10 \, \Omega$$

D
$$3\, \Omega$$

Solution

Components connected in parallel are connected so the same voltage is applied to each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component. In parallel circuits, each light has its own circuit, so all or one light could be burned out, and the last one will still function. A circuit composed solely of components connected in parallel is known as a parallel circuit.
To find the total resistance of all components, the reciprocals of the resistances of each component is added and the reciprocal of the sum is taken. Total resistance will always be less than the value of the smallest resistance That is, $$\frac { 1 }{ { R }_{ total } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } +\frac { 1 }{ { R }_{ 4 } } \quad that\quad is,\quad R={ { R }_{ total } }^{ -1 }$$.
In this case, we have three resistors of values 2 ohms, 3 ohms and 5 ohms. For the resistance to be less than 1 ohm, the resistors should be connected in parallel.
Therefore, $$\frac { 1 }{ { R }_{ total } } =\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 5 } =\frac { 1 }{ 1.03 } =0.97\quad \Omega $$.
Hence, the resistors are connected in parallel and their equivalent resistance is 0.97 ohms.