Current Electricity Test

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Current Electricity Test - 58

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Question 1
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A current is flowing in a circuit consisting of a resistor and a battery.
What happens to the power dissipated in the resistor when the resistance is quadrupled and the voltage remains constant?

A
It is doubled

B
It is quadrupled

C
It is halved

D
It is quartered

E
It remains the same

Solution

Let the voltage of the battery be $$V$$ and initial resistance be $$R$$.
Thus, the initial power dissipated is $$P= \dfrac{V^2}{R}$$
Now the resistance is quadrupled keeping the voltage constant i.e $$R' = 4R$$
$$\therefore$$ New power dissipattion is $$P' = \dfrac{V^2}{R'} = \dfrac{V^2}{4R} = \dfrac{P}{4}$$
Hence the power dissipated is quartered.
Question 2
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Calculate the unknown resistance R of the circuit as shown in figure, all resistance are connected in the series. The current is flowing through circuit is 2A and battery is of 20 voltage.

A
$$1 \Omega$$

B
$$2 \Omega$$

C
$$4 \Omega$$

D
$$6 \Omega$$

E
$$12 \Omega$$

Solution

Given : $$V =20$$ volts $$I = 2$$ A
Total resistance of the circuit $$R_{eq} =R + 4+4 = R+8$$
Using $$V = IR_{eq}$$
$$\therefore$$ $$20 =2 (R+8)$$ $$\implies R = 2\Omega$$
Question 3
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Three resistors ($$2, 5$$ and $$7$$ ohm) are wires as shown in the diagram. The equivalent resistance of this combination (in ohm) is:

A
$$\dfrac{59}{70}$$

B
$$\dfrac{70}{59}$$

C
$$\dfrac{1}{14}$$

D
$$14$$

E
Cannot be determined without additional information.

Solution

Given : $$R_1 = 2\Omega$$ $$R_2= 5\Omega$$ $$R_3 = 7\Omega$$
Equivalent resistance of parallel combination
$$\dfrac{1}{R_{eq}} =\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$$
$$\therefore$$ $$\dfrac{1}{R_{eq}} =\dfrac{1}{2}+\dfrac{1}{5}+\dfrac{1}{7}$$
$$\Rightarrow \dfrac{1}{R_{eq}}=\dfrac{35+14+10}{70}$$
$$\Rightarrow R_{eq} =\dfrac{70}{59}$$ $$\Omega$$
Question 4
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This question relates to the DC circuit shown below.
The values for current, voltage, and resistance in the circuit are graphed from point A through point G in the graphs directly below.
Which of the graphs shows the resistance from point A to point G?

A
A

B
B

C
C

D
D

E
E

Solution

Here the resistance from A to G through the circuit is increasing. Since the all resistances have same value so the resistance will be increased same amount at each point of the circuit from A to G. Here graph B shows the resistance increase by the same value from point to point. Thus, option B will be correct.
Question 5
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All resistors in the circuit have the same resistance, $$R$$, and the battery is having a constant voltage of $$V$$.
Find out power dissipated by resistor $$f$$. Given that the power dissipated by resistor $$e$$ is $$P$$:

A
$$\dfrac{P}{6}$$

B
$$\dfrac{P}{3}$$

C
$$\dfrac{P}{2}$$

D
$$P$$

E
$$2P$$

Solution

The power dissipated by a resistor is $$i^2R$$
where $$i$$ is the current through the resistor of resistance $$R$$.
Since both $$i$$ and $$R$$ for the resistors $$e$$ and $$f$$ are equal (they are of equal resistance and connected in series), the power dissipated by both of them is the same and equal to $$P.$$
Question 6
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Three resistors $$R_1, R_2$$, and $$R_3$$ are connected as shown below.
The resistance of $$R_3$$ is $$1 \Omega$$ and it is known that the value of $$R_3$$ is smaller than the values of $$R_1$$ or $$R_2$$. The resistances of $$R_1$$ and $$R_2$$ are unknown.
Select the best statement describing the equivalent resistance of this combination:

A
The equivalent resistance of this combination must be lower than $$1\Omega$$.

B
The equivalent resistance of this combination may be lower than $$1\Omega$$.

C
The equivalent resistance of this combination must be greater than $$1\Omega$$.

D
The equivalent resistance of this combination may be greater than $$1\Omega$$.

E
The equivalent resistance of this combination cannot be estimated without additional information.

Solution

The resistances in given diagram are in parallel combination , therefore equivalent resistance of the given combination ,
$$1/R=1/R_{1}+1/R_{2}+1/R_{3}$$ ,
it is clear from the equation that value of $$R$$ is less than $$R_{1}$$ or$$R_{2}$$ or $$R_{3} =1$$ , therefore equivalent resistance of this combination is less than 1 ohm .
Question 7
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If four resistances, each of value $$1\ ohm$$, are connected in series, what will be the resultant resistance?

A
3 ohm

B
4 ohm

C
5 ohm

D
2 ohm

Solution

In series combination , equivalent resistance is given by-

$$R=R_1+R_2+R_3+R_4+R_5+.....$$

Here there are four resistors.

$$R=1+1+1+1$$

$$\implies R=4\Omega$$

Answer-(B)
Question 8
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Two resistance are connected in series as shown in diagram. What is the current through the $$5\ ohm$$ resistance?

A
$$2 A$$

B
$$3 A$$

C
$$4 A$$

D
$$5 A$$

Solution

Voltage across $$5\Omega$$ resistor is $$V=10V$$

Current, $$I=\dfrac{V}{R}=\dfrac{10}{5}$$

$$\implies I=2A$$

Answer-(A)
Question 9
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Two resistance are connected in series as shown in diagram. What is the value of $$R$$?

A
$$4 \Omega$$

B
$$6 \Omega$$

C
$$5 \Omega$$

D
$$3 \Omega$$

Solution

Voltage across $$5\Omega$$ resistor is $$V=10V$$

Current, $$I=\dfrac{V}{R}=\dfrac{10}{5}$$

$$\implies I=2A$$

Since resistance $$5\Omega$$ and $$R$$ are in series and hence current through both resistance is same.
Hence current through $$R$$ is also $$2A$$.

Now for $$R$$, potential difference is $$V=6V$$ and $$I=2A$$.

Hence , $$R=\dfrac{V}{I}=\dfrac{6}{2}$$

$$\implies R=3\Omega$$

Answer-(D)
Question 10
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The resistance of the bulb filament is $$100\Omega$$ at a temperature of $$100^{\circ}C$$. If its temperature co-efficient of resistance be $$0.005$$ per $$^{\circ}C$$, its resistance will become $$200\Omega$$ at a temperature

A
$$300^{\circ}C$$

B
$$400^{\circ}C$$

C
$$500^{\circ}C$$

D
$$200^{\circ}C$$

Solution

Given : $$R_{100} = 100\Omega$$ $$T_1 = 100^oC$$ $$R' = 200\Omega$$ $$\alpha = 0.005$$ per $$^oC$$
Using $$R' = R_{100}(1+\alpha (T_2-T_1))$$
$$\therefore$$ $$200 = 100(1+0.005 (T_2-100))$$
OR $$T_2 - 100 = 200$$ $$\implies T_2 = 300^oC$$