Electricity Test

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Electricity Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

5,000 alpha particles per minute are passing through a straight tube of radius r. The resulting electric current is (approximately)

A
0.26 10^-16 A

B
1 10¹² A

C
0.26 10¹² A

D
1 10^-12 A

Solution

, Q = nq; where q (= 2×1.6×10^-19) is the charge on the alpha particle

So,
Question 2
1 / -0

A wire has a resistance of 1 . It is stretched to double its length. The resistance of the wire will be

A
1

B
2

C
4

D
16

Solution

As the volume of the wire remains constant,
Question 3
1 / -0

On stretching a wire, its radius decreases by 1%. The resistance will

A
increase by 1%

B
increase by 2%

C
increase by 4%

D
not change

Solution

On stretching the wire, the volume of the wire remains constant.
So,
Using the concept of fractional change,
If, then percentage change in the value of z is:
Question 4
1 / -0

Four equal resistances are connected as shown in the figure. The resistance measures across AB will be

A
R

B
R/4

C
R/2

D
(3/4) R

Solution

Here, the three resistances are connected in series. Hence, their equivalent resistance is R.
Therefore, the equivalent resistance across AB is:
Question 5
1 / -0

Seven identical lamps of resistance 2200 each are connected to 220 V line as shown in the figure. What will be the reading on the ammeter?

A
A

B
A

C
A

D
A

Solution

All the 3 bulbs on the left side of the ammeter and 4 bulbs on the right side of the bulb are connected in parallel. Hence, the circuit diagram can be redrawn as:

The current in resister on right side of ammeter is equal to the current in ammeter.
Using V = IR,
Question 6
1 / -0

An electric bulb is designed to draw P₀power and V₀ voltage. If the voltage is V, itdraws P power. Which of the following relations holds true?

A
P =

B
P =

C
P =

D
P =

Solution

Power is given by:
When the same bulb is connected across the voltage source V, then the power consumed is given by:
Question 7
1 / -0

You have the following appliances each of 500 W, running on 220 V AC:

(i) An electric iron
(ii) An electric toaster
(iii) An electric room heater

The electrical resistance is

A
maximum for the room heater

B
maximum for the electric lamp

C
maximum for the electric iron

D
the same in all the three cases

Solution

All the appliances are connected in parallel and the power of each appliance is the same. So, the resistance of each will be the same as the power is given by:
Question 8
1 / -0

Two heating coils, one of fine wire and the other of thick wire of the same material and of the same length, are connected in series and in parallel. Which of the following statements is correct?

A
In series, the fine wire will liberate more energy; while in parallel, the thick wire will liberate more energy.

B
In series, the fine wire will liberate less energy; while in parallel, the thick wire will liberate more energy.

C
Both will liberate equal energy.

D
In series, the thick wire will liberate more energy; while in parallel, it will liberate less energy.

Solution

In series, the current in the circuit is the same. The heat generated can be calculated by using H = I²Rt.
The resistance of the wire is given by:

The thick wire will have less resistance. Hence, the thick wire will liberate less heat in series in comparison to the thin wire. In parallel, the voltage across the given resistance is the same. The heat generated can be calculated by using .
The thick wire will have less resistance. Hence, the thick wire will liberate more heat in parallel in comparison to the thin wire because its resistance is less.
Question 9
1 / -0

Two electric bulbs A and B are designed for the same voltage. Their power ratings respectively are P_A and P_B, with P_A > P_B. If they are joined in series across a V-volt supply, then

A
A will draw more power than B

B
B will draw more power than A

C
the ratio of power drawn by them will depend on V

D
A and B will draw the same power

Solution

Power is given by:

Or we can say, the lesser is the power, the greater is the resistance.
When we connect the two bulbs in series, the current through the two bulbs is the same. Hence, the power consumed can be calculated by using P = I²R.
The bulb having less resistance (greater power) will consume less power.
Question 10
1 / -0

An ammeter and a voltmeter are joined in series to a cell. Their readings are A and V, respectively. If a resistance is now joined in parallel with the voltmeter, then

A
A will increase and V will decrease

B
both A and V will increase

C
both A and V will decrease

D
A will decrease and V will increase

Solution

As the resistance is connected in parallel to the voltmeter, less current will flow through the voltmeter. Hence, the reading of the voltmeter will decrease. As the equivalent resistance decreases as well, so the reading of the ammeter will increase.

Question 11

1 / -0

A galvanometer is used to measure small currents. A certain galvanometer has a resistance 500 and gives a full-scale deflection for a current of 200 µA. This meter is connected as shown in the figure to make a multi-range current meter. Connection to the circuit is made at the terminals as shown.

The values of current in the external circuit needed to give full-scale deflections when X is connected to A, B and C in turn are shown in the table:

X connected to	Current in the external circuit (mA)
A	1
B	10
C	100

The value of R₃ is

2.25

0.25

1.25

3.25 Solution

When switch is connected to A, Voltage across 500 Ω = Voltage across (R₁+ R₂+ R₃)

--- 1
When switch is connected to B, Voltage across (500 + R₁) = Voltage across (R₂+ R₃)
(500 + R₁) x 0.2 x 10^-3 = (R₂ + R₃) x (10 - 0.2) x 10^-3
(500 + R₁) = 49(R₂+ R₃) --- 2
Voltage across (500 + R₁+ R₂) = Voltage across R₃
(500 + R₁ + R₂) x 0.2 x 10^-3 = (R₃) x (100 - 0.2) x 10^-3
(500 + R₁ + R₂) = 499 (R₃) --- 3
Subtracting 1 from 3,
500 - R₃ =499R₃- 125
500R₃ = 625 R₃ = 1.25

Question 12
1 / -0

How many turns of a nichrom wire, 1mm in diameter, should be wound around a porcelain cylinder with radius 2.5 cm to obtain a heater with resistance of 20 ohms. (Given (Nichrome) = 1.0 × 10^-6 )

A
650

B
1200

C
100

D
5

Solution

For 1 turn, length required = 2R (where R is the radius of cylinder)

1 turn = 2R m

1 m = turns

For , number of turns = turns
turns
Question 13
1 / -0

The heat energy produced by the coil in the given circuit in five minutes is

A
6 10⁵ J

B
5.4 10⁵J

C
6 10⁴ J

D
5.4 10⁴ J

Solution

Energy consumed by the coil =
Voltage across the coil = Voltage across the 60 Ώ resistance
Resistance of the coil,

Equivalent resistance in parallel,

Voltage across the coil,
Question 14
1 / -0

A uniform conductor of resistance 'R' is cut into 20 equal pieces. Half of them are joined in series and remaining half of them are connected in parallel. If two combinations are joined in series, the effective resistance of all pieces is

A
R

B
R/2

C
101R/200

D
201R/100

Solution

Resistance,
Resistance of one piece of the wire,
When half of them are connected in parallel, their equivalent resistance in parallel is given by:
When half of them are connected in series, their equivalent resistance in series is given by:

When the two combinations are further connected in series, the effective resistance of the network is given by:
Question 15
1 / -0

A circuit is shown in the figure. If the switch 'S' is closed, the current in the ammeter

A
does not change

B
increases

C
decreases

D
may decrease or increase

Solution

In upper limb, the resistances R and 2R are in series. They are equal to 3R. The resistance 3R in upper limb and 3R in lower limb are in parallel.

Hence, the equivalent resistance of the circuit is

After closing the switch, the circuit diagram is as shown below:

Here, R and 2R are in parallel and the equivalent resistance is R' = , which is in series with another parallel combination of R and 2R. Hence, the equivalent resistance of the circuit is , which is less as compared to the initial situation. So, the current will increase in the circuit.
Question 16
1 / -0

If the current flowing in a coil of resistance 90 is to be reduced to 90%, what value of resistance should be connected in series with it?

A
90 Ω

B
9 Ω

C
1000 Ω

D
10 Ω

Solution

Let the voltage supply be V volts.

Let the current flowing initially be i amperes.

Now, by Ohm's law,

Let the final resistance be R ohms.

As the final current is 90% of the initial value,

Or R = 100 ohms

Let the resistance added in series be r ohms.

Now, r + 90 = 100

Or r = 10

Thus, 10 ohms resistance ought to be connected in series to the given resistor.
Question 17
1 / -0

The magnitude of l in amperes is

A
0.1

B
0.3

C
0.6

D
None of these

Solution

All the resistances are in parallel order, so voltage across them will be equal.

60 I = (15 + 5) I₁

60 I = 20 I₁
I₁ = 3 I ... (i)
Again (15 + 5) I₁ = 10 (1 - I - I₁)
2 I₁ = 1 - I - I₁
2 (3I) = 1 - I - 3I
6I + 4I = 1
10I = 1
I = = 0.1 A
Question 18
1 / -0

A current of 6 A enters one corner P of an equilateral triangle PQR having three wires of resistances 2 Ω each and leave by the corner R. Then, l₁ and l₂ respectively are:

A
2 A, 4 A

B
4 A, 2 A

C
1 A, 2 A

D
2 A, 3 A

Solution

From Kirchhoff's first law at junction P,
I₁ + I₂ = 6 ... (i)
From Kirchhoff's second law to the closed
circuit PQRP,
-2I₁ - 2I₁ + 2I₂ = 0
-4 I₁ + 2 I₂ = 0
2 I₁ - I₂ = 0 .... (ii)
Adding Eqs. (i) and (ii), we get
3I₁ = 6
I₁ = 2A
From Eq. (i),
I₂ = 6 - 2 = 4A
Question 19
1 / -0

If you are provided three resistances of 2Ω, 3Ω and 6Ω, how will you connect them so as to obtain the equivalent resistance of 4?

A

B

C

D
None of these

Solution

In the given question in option (3), 3Ω and 6Ω resistances are in parallel, hence equivalent resistance is

⇒ R' =2Ω

The two 2Ω resistances are now connected in series, hence equivalent resistance is

R' = 2Ω + 2Ω = 4Ω

Note: The value of the equivalent resistance of the resistances connected in parallel is less than the value of the smallest resistance among those resistances.
Question 20
1 / -0

By using only two resistance coils singly, in series or in parallel, one should be able to obtain resistances of 3, 4, 12 and 16 ohms. The separate resistances of the coil are

A
3 and 4

B
4 and 12

C
12 and 16

D
16 and 3

Solution

If we take R₁ = 4Ω R₂ = 12Ω,
then in series resistance
R = R₁ + R₂
= 4 + 12
= 16Ω
In parallel, resistance R = = 3Ω
So, R₁ = 4Ω and R = 12Ω