Current Electricity Test

Result

Current Electricity Test - 85

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The temperature coefficient of resistance of a wire is $$0.00125$$ per degree celcius. At $$300K$$ its resistance is $$1$$ ohm. The resistance of the wire will be $$2$$ ohms at a temperature:

A
$$1154K$$

B
$$1127K$$

C
$$600K$$

D
$$1400K$$

Solution
Question 2
1 / -0

Most of the times we connect remote speakers to play music in another room along with the built-in speakers. These speakers are connected in parallel with the music system.
At the instant represented in the picture, the voltage across the speakers is $$6.00 V.$$ The resistance of the main speaker is $$ 8 \space \Omega$$ and the remote speaker has resistance $$ 4 \space \Omega$$
The equivalent resistance of the speakers is

A
$$ 12 \space \Omega$$

B
$$ \dfrac{7}{3 }\space \Omega$$

C
$$ \dfrac{8}{3}\space \Omega$$

D
$$ 0.375 \space \Omega$$
Question 3
1 / -0

Directions For Questions

Consider $$12$$ resistors arranged symmetrically in shape of bi-pyramid $$ABCDEF$$. Here $$ABCD$$ is a square. Point $$E$$, paint $$F$$, and center of square are in the same straight line perpendicular to the plane of square. The resistance of each resistor is $$R$$.

...view full instructions

The effective resistance between $$A$$ and $$B$$ is

A
$$7R/12$$

B
$$5R/7$$

C
$$5R/12$$

D
$$none\ of\ these$$

Solution

Let us remove the resistance between $$A$$ and $$B$$,
then $$E$$ and $$F$$ will be at the same potential, we can draw the remaining as:

$$M$$ is the midpoint of $$DC, M$$ will be at the same pot as $$E$$ and $$F$$.

Now finf $$R_{AB}=5/7R$$.
The resistance which we had removed will be in parallel to it so

$$R_{eq}=\dfrac {R_{AB}R}{R_{AB}+R}=\dfrac {(5/7)RR}{(5/7)R+R}=\dfrac {5}{12}R$$
Question 4
1 / -0

In a practical Wheatstone bridge circuit (Fig.), when one more resistance of $$ 100 \Omega $$ is connected in parallel with unknown resistance $$ x $$, then the ratio $$ l_{1} / l_{2} $$ becomes $$ 2 . l $$, is the balance length. $$ A B $$ is a uniform wire. Then the value of $$ x $$ must be

A
$$ 50 \Omega $$

B
$$ 100 \Omega $$

C
$$ 200 \Omega $$

D
$$ 400 \Omega $$

Solution
Question 5
1 / -0

Directions For Questions

Potentiometer is an ideal voltmeter as voltmeter draws some current through the circuit while potentiometer needs no current to work . Potentiometer works on the principle of e.m.f . comparison. In working condition, a constant current flows throughout the wire of potentimeter using standard cell of e,m,f.e The wire of potentiometer is made of uniform material and cross- sectional area and it has uniform resistance per unit length. The potential gradient depends upon the current in the wire.
A potentiometer with a cell of e.m.f. 2 V and internal resistance 0.4$$\Omega $$ is used across the wire AB. A standard cadium cell of e.m.f. 102 V gives a balance point at 66 cm length of wire.The standard cell is then replaced by a cell of unknown e. m. f. e.(internal resistance r) and the balance point found similarly turns out to be 88 cm length of the wire. The length of potentiometer wire AB is 1 m

...view full instructions

The value of e is

A
1.36 V

B
2.63 V

C
1.83 V

D
none

Solution
Question 6
1 / -0

The equivalent resistance of the circuit across points A and B is equal to

A
$$22.5 \space \Omega$$

B
$$25 \space \Omega$$

C
$$37.5 \space \Omega$$

D
$$75 \space \Omega$$

Solution
Question 7
1 / -0

Six identical resistors are connected as shown in figure .The equivalent resistance will be

A
Maximum between P and R

B
Maximum between Q and R

C
Maximum between P and Q

D
All are equal
Question 8
1 / -0

The length of wire of a potentiometer is 100 cm, and the e.m.f. of its standrad cell is E volt. It is employed to measure the e.m.f . of a battery whose internal resistance is 0.5 $$\Omega $$ $$. If the balance point is obtained at l=30 cm from the positive end, e.m.f of the battery is
Where i is the current in the potentiometer wire.

A
$$ \dfrac{30 E}{100} $$

B
$$ \dfrac{30 E}{100.5} $$

C
$$ \dfrac{30 E}{100-0.5} $$

D
$$ \dfrac{30 E-0.5l}{100} $$

Solution
Question 9
1 / -0

When resistors with different resistances are connected in parallel, which of the following must be the same for each resistor? Choose the correct answer.

A
potential difference

B
current

C
power delivered

D
charge entering each resistor in a given time interval

E
none of the above

Solution

When the resistors are connected in parallel the potential difference across all the resistors is the same.
Question 10
1 / -0

A metal wire has a resistance of $$10.0\Omega$$ at a temperature of $$20.0^{o}C$$. If the same wire has a resistance of $$10.6\Omega$$ at $$90.0^{o}C$$, what is the resistance of this wire when its temperature is $$-20.0^{o}C$$?

A
$$0.700\Omega$$

B
$$9.66\Omega$$

C
$$10.3\Omega$$

D
$$13.8\Omega$$

E
$$6.59\Omega$$

Solution

Using $$R_{0}=10.0\Omega$$ at $$T=20.0^{o}C$$, we have $$R=R_{0}(1+\alpha \Delta T)$$ or
$$\alpha=\dfrac{R/R_{0}-1}{\Delta T}=\dfrac{10.6/10.0-1}{(90.0^{o}C-20.0^{o}C)}=8.57\times 10^{-4}$$ $$^{o}C^{-1}$$
At $$T=-20.0^{C}$$, we have
$$R=R_{0}(1+\alpha \Delta T)$$
$$= (10.0\Omega)[1+8.57\times 10^{-4} {^{o}C^{-1}}(-20.0^{o}C-20.0^{o}C)]=9.66\Omega$$