Current Electricity Test

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Current Electricity Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

Potentiometer measures potential more accurately because

A
it measures potential in the open circuit

B
it uses sensitive galvanometer for null deflection

C
it uses high resistance potentiometer wire

D
it measures potential in the closed circuit

Solution

Potentiometer measures voltage when galvanometer shows zero current rating, mean it takes zero current (open circuit) while measuring voltage across any component, that is why it is more accurate as all the current passing through that component only.
Question 2
1 / -0

For driving current of $$2A$$ for $$6$$ minute in a circuit. $$1000 J$$ of work is to be done. The e.m.f. of the source in the circuit is

A
$$1.38V$$

B
$$1.68V$$

C
$$2.03V$$

D
$$3.10V$$

Solution

Current passing through the circuit $$I = 2 A$$
Time of operation $$t = 6$$ minute $$ = 360$$ s
Work done $$W = 1000 J$$
Using $$W = VIt$$
$$\therefore$$ $$1000 = V (2)(360)$$
$$\implies$$ $$V = 1.38$$ volts
Question 3
1 / -0

Which of the following devices has a source of emf inside it?

A
Voltmeter

B
Ammeter

C
Ohm-meter

D
Rectifier

Solution

Ohmmeter works either by applying a voltage from an internal battery to the unknown and measuring current (analog ohmmeters) or by applying a current and measuring voltage (digital ohmmeters).
Question 4
1 / -0

The figure shows in apart of an electric circuit, then the current I is

A
$$1 A$$

B
$$3A$$

C
$$2 A$$

D
$$4 A$$

Solution

According to the Kirchhoff's first law, amount of current entering is equal to amount of current leaving.
From the circuit,
$$I+6+1=5+3+2$$
$$I+7=10$$
$$I=3A$$
Question 5
1 / -0

A potentiometer circuit shown in the figure is set up to measure emf of cell E. As the point P moves from X to Y, the galvanometer G shows deflection always in one direction, but the deflection decreases continuously until Y is reached. The balance point between X and Y maybe obtained by

A
Decreasing the resistance R and decreasing V

B
Decreasing the resistance R and increasing V

C
Increasing the resistance R and increasing V

D
Increasing the resistance R and decreasing V

Solution

Decreasing R increase in XY and there by increases the potential drop across XP and the balance point may be obtained. The current may be increased also by increasing V.
Question 6
1 / -0

In specific resistance measurement of a wire using a meter bridge, the key k in the main circuit is kept open when we are not taking readings. The reason is

A
the emf of cell will decrease.

B
the value of resistance will change due to joule heating effect.

C
the galvanometer will stop working.

D
none of these.

Solution

The key k in the main circuit is kept open when we are not taking the readings because the value of resistance will change due to joule heating effect. When an electric current passes through the resistor, heat is generated which rises the temperature of resistor which in result changes the resistance of resistor.
Question 7
1 / -0

Which of the following shows energy changes that takes place when an incandescent bulb glows using a battery?

A
Chemical>Electrical>heat>Light energy

B
Electrical>Chemical>Light>heat energy

C
Chemical>Electrical>Heat energy

D
Electrical>Chemical>Light energy

Solution

Energy stored in the battery in the form of chemical energy. Once the bulb is switched on the chemical energy changes to electrical, followed by heating of the filament then the bulb glows bright and gives light energy.
Question 8
1 / -0

If $$i_{1}=3\sin\omega t$$ and $$i_{2}=4\cos\omega t$$, then $$i_{3}$$ is

A
$$\displaystyle5\sin \left ( \omega t+53^{\circ} \right )$$

B
$$\displaystyle5\sin \left ( \omega t+37^{\circ} \right )$$

C
$$\displaystyle5\sin \left ( \omega t+45^{\circ} \right )$$

D
$$\displaystyle5\cos \left ( \omega t+53^{\circ} \right )$$

Solution

From Kirchoff's current law,
$$\displaystyle i_{3}=i_{1}+i_{2}=3\sin wt + 4\sin\left ( wt+90^{0} \right )$$
$$\displaystyle =\sqrt{3^{2}+4^{2}+2(3)(4)cos90^{0}}\sin (wt+\phi)$$
where $$\displaystyle \tan \phi=\frac{4\sin90^{0}}{3+4\cos 90^{0}}=\frac{4}{3}$$
$$\displaystyle \therefore i_{3}=5\sin \left ( wt+53^{0} \right )$$
Question 9
1 / -0

Steady current flows in a metallic conductor of non-uniform cross-section. Which of the following quantities is a constant along the conductor ?

A
Current

B
Current density

C
Electric field

D
Drift speed

Solution
Question 10
1 / -0

An electron in potentiometer wire experiences a force $$2.4\times 10^{-19}$$N. The length of potentiometer wire is $$6m$$. The e.m.f. of the battery connected across the wire is (electronic charge $$=1.6\times 10^{-19}$$C).

A
$$6$$V

B
$$9$$V

C
$$12$$V

D
$$15$$V

Solution

Electric field in the wire $$E = \dfrac{F}{e}$$
$$\therefore$$ $$E = \dfrac{2.4\times 10^{-19}}{1.6\times 10^{-19}} = 1.5$$ $$N/C$$
Length of the wire $$l = 6$$ m
Thus e.m.f of the battery $$\mathcal{E} = El $$
$$\therefore$$ $$\mathcal{E} = 1.5\times 6 = 9$$ V

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Re-Attempt Weekly Quiz Competition

Current Electricity Test - 61

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Current Electricity Test - 61

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SHARING IS CARING

Question 1

it measures potential in the open circuit

it uses sensitive galvanometer for null deflection

it uses high resistance potentiometer wire

it measures potential in the closed circuit

Solution

Question 2

$$1.38V$$

$$1.68V$$

$$2.03V$$

$$3.10V$$

Solution

Question 3

Voltmeter

Ammeter

Ohm-meter

Rectifier

Solution

Question 4

$$1 A$$

$$3A$$

$$2 A$$

$$4 A$$

Solution

Question 5

Decreasing the resistance R and decreasing V

Decreasing the resistance R and increasing V

Increasing the resistance R and increasing V

Increasing the resistance R and decreasing V

Solution

Question 6

the emf of cell will decrease.

the value of resistance will change due to joule heating effect.

the galvanometer will stop working.

none of these.

Solution

Question 7

Chemical>Electrical>heat>Light energy

Electrical>Chemical>Light>heat energy

Chemical>Electrical>Heat energy

Electrical>Chemical>Light energy

Solution

Question 8

$$\displaystyle5\sin \left ( \omega t+53^{\circ} \right )$$

$$\displaystyle5\sin \left ( \omega t+37^{\circ} \right )$$

$$\displaystyle5\sin \left ( \omega t+45^{\circ} \right )$$

$$\displaystyle5\cos \left ( \omega t+53^{\circ} \right )$$

Solution

Question 9

Current

Current density

Electric field

Drift speed

Solution

Question 10

$$6$$V

$$9$$V

$$12$$V

$$15$$V

Solution

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