Current Electricity Test

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Current Electricity Test - 56

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Question 1
1 / -0

Imagine that you have a 100$$\Omega$$ resistor. You want to add a resistor in series with this 100$$\Omega$$ resistor in order to limit the current to 0.5A when $$110 $$ Volts is placed across two resistors in series. How much resistance should you use?

A
$$100\Omega$$

B
$$120 \Omega$$

C
$$250 \Omega$$

D
$$300 \Omega$$

Solution

Given : $$V = 110$$ volts $$I = 0.5$$ A $$R_1 = 100\Omega$$
Let a resistance $$R$$ is connected in series with $$R_1$$.
Using Ohm's law, $$V = I(R_1+R)$$
$$\therefore$$ $$110 = 0.5(100+R)$$ $$\implies R = 120\Omega$$
Question 2
1 / -0

The attached circuit diagram shows connection of 3 resistors. All are connected in parallel . Thus what is the equivalent resistance.

A
1 ohm

B
4 ohm

C
3ohm

D
2 ohm

Solution

Equivalent resistance of resistors connected in parallel $$\dfrac{1}{R_{eq}} =\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$$
$$\therefore$$ $$\dfrac{1}{R_{eq}} =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}$$ $$\implies R_{eq} = 1$$ ohm
Question 3
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What is the highest equivalent resistance that can be secured by combinations of four coils of resistance $$4\ \Omega$$, $$8\ \Omega$$, $$12\ \Omega$$, $$24\ \Omega$$?

A
$$24\ \Omega$$

B
$$48\ \Omega$$

C
$$12\ \Omega$$

D
$$4\ \Omega$$

Solution

Connecting resistances in series with each other increases the equivalent resistance of the combination.
Let $$4\ \Omega$$, $$8\ \Omega$$, $$12\ \Omega$$, and $$24\ \Omega$$ resistors are connected in series.
$$\therefore$$ Equivalent resistance of the circuit
$$R_{eq} = 4+8+12+24$$
$$\Rightarrow R_{eq} = 48$$ ohm
Question 4
1 / -0

The equivalent resistance of parallel combination is:

A
Lower than the value of lowest resistance in the combination

B
Higher than the value of lowest resistance in the combination

C
Both A and B

D
None of these

Solution

Consider two resistors having resistances $$R_1$$ and $$R_2$$.
The equivalent resistance of $$R_1$$ and $$R_2$$ in parallel is,
$$\dfrac{1}{R_{eq}}= \dfrac{1}{R_1}+\dfrac{1}{R_2}$$

$$\implies R_{eq}=\dfrac{R_1R_2}{R_1+R_2}$$
For any values of $$R_1$$ and $$R_2$$, $$R_{eq}$$ will be less than $$R_1$$ and $$R_2$$.
For example,
Let $$R_1=2\ \Omega$$
$$R_2=6\ \Omega$$
Then,
Equivalent resistance of $$R_1$$ and $$R_2$$ in parallel,
$$ R_{eq}=\dfrac{R_1R_2}{R_1+R_2}$$

$$ R_{eq}=\dfrac{2\times 6}{2+8}$$

$$ R_{eq}=\dfrac{12}{8}$$

$$ R_{eq}=1.5\ \Omega$$
$$\implies R_{eq} \lt R_1\ \text{and}\ R_{eq} \lt R_2$$
Question 5
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How much current is drawn by the motor of 1 H.P. from 220 volt supply.

A
3.4A

B
2A

C
6A

D
22A

Solution

We know that $$1$$ H.P $$ = 746$$ watt
Current $$I = \dfrac{P}{V} = \dfrac{746}{220} = 3.4$$ A
Question 6
1 / -0

If n resistance (each R) are connected in parallel then the resultant will be

A
$$nR$$

B
$$R/n$$

C
$$\displaystyle { n }^{ 2 }R$$

D
$$\text{None}$$

Solution

Equivalent resistance of parallel combination

$$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+.........$$

Given, Resistanvce of all resistors $$=R$$
$$\therefore$$ $$\dfrac{1}{R_{eq}} = \dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}+.........n$$ terms

OR $$\dfrac{1}{R_{eq}} = \dfrac{n}{R}$$

$$\implies$$ $$R_{eq} = \dfrac{R}{n}$$
Question 7
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Two cells A and B of emf 2V and 1.5 V respectively, are connected as shown in figure through an external resistance $$10\Omega$$. the internal resistance of each cell is $$5\Omega$$. The potential difference $$E_A$$ and $$E_B$$ across the terminals of the cells A and B respectively are:

A
$$E_A = 2.0 V. E_B = 1.5 V$$

B
$$E_A = 2.125 V, E_B = 1.375 V$$

C
$$E_A = 1.875 V, E_B = 1.625 V$$

D
$$E_A = 1.875 V, E_B = 1.375 V$$

Solution

$$R_{eq} = 10 + 5 + 5 = 20 \Omega$$
$$V_{eq} = 2 - 1.5 = 0.5V$$
$$I = \dfrac{V_{eq}}{R_{eq}} \\ \\ I = \dfrac{0.5}{20} \\ I = 0.025 A$$
$$E_A = 2 - R_AI \\ E_A = 2 - 5 * 0.025 = 2 - 0.125 = 0.875 V $$
$$E_B = 1.5 + R_BI \\ E_B = 1.5 + 5 * 0.025 = 1.5 + 0.125 = 1.625V $$
Hence option 'C' is correct.
Question 8
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A bulb of $$22\ \Omega $$ is producing light when connected to a $$220\ V$$ supply. What is the electric power of the bulb?

A
$$2200\ W$$

B
$$1000\ W$$

C
$$22\ W$$

D
$$220\ W$$

Solution

Given,
Resistance of the bulb, $$R = 22\Omega$$
Voltage, $$V = 220$$ volts
We know,
Power of the bulb, $$P = \dfrac{V^2}{R}$$

$$P = \dfrac{220^2}{22}$$

$$P =2200\ W$$
Question 9
1 / -0

Potentiometer measures the potential difference more accurately than a voltmeter because :

A
it has a wire of high resistance

B
it has a wire of low resistance

C
it does not draw current from external circuit

D
it draws a heavy current from external circuit

Solution

When we measure the emf of a cell by the potentiometer then no current flows in the circuit in zero-deflection condition ie, cell is in open circuit. Thus, in this condition the actual value of a cell is found. In this way, potentiometer is equivalent to an ideal voltmeter of infinite resistance.
Note. The emf in the potentiometer is measured by null method in which zero deflection position is found on the wire.
Question 10
1 / -0

A $$2V$$ battery, a$$990 \Omega$$ resistor and a potentiometer of $$2m$$ length, all are connected in series of the resistance of potentiometer wire is $$10\Omega$$, then the potential gradient of the potentiometer wire is

A
$$0.05Vm^{-1}$$

B
$$0.5Vm^{-1}$$

C
$$0.01Vm^{-1}$$

D
$$0.001Vm^{-1}$$

Solution

Potential gradient $$x=\displaystyle\frac{e}{(R+R_{h}+r)}\cdot \frac{R}{L}$$
$$\Rightarrow \displaystyle x=\frac{2}{(990+10)}\times \frac{10}{2}=0.01Vm^{-1}$$

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Current Electricity Test - 56

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Current Electricity Test - 56

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Question 1

$$100\Omega$$

$$120 \Omega$$

$$250 \Omega$$

$$300 \Omega$$

Solution

Question 2

1 ohm

4 ohm

3ohm

2 ohm

Solution

Question 3

$$24\ \Omega$$

$$48\ \Omega$$

$$12\ \Omega$$

$$4\ \Omega$$

Solution

Question 4

Lower than the value of lowest resistance in the combination

Higher than the value of lowest resistance in the combination

Both A and B

None of these

Solution

Question 5

3.4A

2A

6A

22A

Solution

Question 6

$$nR$$

$$R/n$$

$$\displaystyle { n }^{ 2 }R$$

$$\text{None}$$

Solution

Question 7

$$E_A = 2.0 V. E_B = 1.5 V$$

$$E_A = 2.125 V, E_B = 1.375 V$$

$$E_A = 1.875 V, E_B = 1.625 V$$

$$E_A = 1.875 V, E_B = 1.375 V$$

Solution

Question 8

$$2200\ W$$

$$1000\ W$$

$$22\ W$$

$$220\ W$$

Solution

Question 9

it has a wire of high resistance

it has a wire of low resistance

it does not draw current from external circuit

it draws a heavy current from external circuit

Solution

Question 10

$$0.05Vm^{-1}$$

$$0.5Vm^{-1}$$

$$0.01Vm^{-1}$$

$$0.001Vm^{-1}$$

Solution

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