Electricity Test

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Electricity Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

Equivalent resistance between $$A$$ and $$B$$ is

A
$$3 \Omega$$

B
$$6 \Omega$$

C
$$1.5 \Omega$$

D
$$4.5 \Omega$$

Solution

The correct option is C.
Question 2
1 / -0

The resistance across A and B is

A
$$\dfrac {R} {2}$$

B
$$R$$

C
$$\dfrac {6R} {5}$$

D
$$4R$$

Solution

The resistances in upper branch are a series combination of $$R$$ and $$2R$$
So their series resistance is $$R_1=R+2R=3R$$

The resistances in lower branch are a series combination of $$R$$ and $$R$$
So their series resistance is $$R_2=R+R=2R$$

Now $$3R$$ and $$2R$$ are in parallel.

$$\dfrac{1}{R_{eq}}=\dfrac{1}{2R}+\dfrac{1}{3R}=\dfrac{5}{6R}$$
$$\therefore R_{eq}=\dfrac{6R}{5}$$.
Question 3
1 / -0

When two resistors are connected in parallel, then the equivalent resistance is

A
$${R_1} + {R_2}$$

B
$$\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$$

C
$$\dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}$$

D
$$\dfrac{{{R_1}}}{{{R_2}}}$$
Question 4
1 / -0

Potential difference across $$AB$$ shown in the figure is:

A
10 V

B
12 V

C
8 V

D
0 V
Solution
- In the given figure, $$A$$ and $$B$$ are on the same potential as there is a direct connection between $$A$$ and $$B$$ without any other element (via $$D$$)
So $$V_A=V_B$$
Potential difference $$=V_B-V_A=0$$
Alternatively,
we can see that
Voltage between $$A$$ and $$C$$ i.e. $$V_{AC}=V_A-V_C= 10 \ V$$ . . . (i)
And,
Voltage between $$C$$ and $$B$$ i.e. $$V_{CB}=V_C-V_B=-10 \ V$$ . . . (ii)

Adding $$(i)$$ and $$(ii)$$,
$$(V_A-V_C)+(V_C-V_B)=10+(-10)$$
$$\Rightarrow V_A-V_B=0$$

So Voltage between $$A$$ and $$B$$ i.e. $$V_{AB}=V_A-V_B= 0 \ V$$
Thus, option D is correct.
Question 5
1 / -0

A wire of resistance R is cut into n equal parts these parts are then connected in parallel with each other. The equivalent resistance of the combination is :

A
$$nR$$

B
$$\cfrac {R}{n}$$

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

D
$$\cfrac {R}{n^2}$$

Solution

when wire of resistance R is cut into n equal parts then Resistance of each part will be $$\dfrac Rn$$;
such $$n$$ parts are joined in parallel
$$\dfrac{1}{R_{eq}}=$$$$\dfrac{1}{(R/n)}+$$$$\dfrac{1}{(R/n)}+$$$$\dfrac{1}{(R/n)}+$$$$\dfrac{1}{(R/n)}+$$...

so $$R_{eq}=\dfrac {R}{n^2}$$.
Question 6
1 / -0

A conductor of resistance $$3 \Omega$$ is stretched uniformly till its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in $$\Omega$$

A
$$\dfrac{9}{2}$$

B
$$\dfrac{8}{3}$$

C
$$2$$

D
$$1$$

Solution

If we double the length of the conductor$$,$$ then its cross$$-$$sectional area must become half of its original value (since$$,$$ $$volume = area \times length$$ is constant).
Therefore$$,$$ we can say that after stretching the new resistance is $$R' = \rho \dfrac{{2l}}{{A/2}} = 4\rho \dfrac{1}{A} = 4{R_0} = 4 \times 3\Omega = 12\Omega $$
Now$$,$$ the wire is bent such that each side has a resistance $$4\Omega ,$$ (since the resistance is divided uniformly).

Now from the diagram, we can see that between two vertices$$.$$ there is an arrangement of two $$4\Omega $$ resistors in series connected in parallel with another $$4\Omega $$ resistor$$.$$
The net resistance becomes$$:-$$ $$\dfrac{{8 \times 4}}{{8 + 4}} = \dfrac{8}{3}\Omega $$
Hence,
option $$(B)$$ is correct answer.
Question 7
1 / -0

When resistors are connected in parallel, the current divides in the ...................

A
Direct ratio of resistances

B
Inverse ratio of resistances

C
Inverse ratio of the potential differences

D
None of these

Solution
Question 8
1 / -0

The equivalent resistance between A and B in the following figure is

A
$$120\Omega $$

B
$$40\Omega $$

C
$$30\Omega $$

D
$$22.5\Omega $$

Solution
Question 9
1 / -0

Equivalent resistance between A and B is.

A
3 $$\Omega $$

B
6 $$\Omega $$

C
1.5 $$\Omega $$

D
4.5 $$\Omega $$

Solution
Question 10
1 / -0

In the circuit shown in the figure below, the magnitude and direction of the current will be

A
$$2.3 \ Amp$$ from A to B via E

B
$$2.3 \ Amp$$ from B to A via E

C
$$1.0 \ Amp$$ from B to A via E

D
$$1.0 \ Amp$$ from A to B via E

Solution

The voltage of the battery $$V_1=10 \ V$$ and that of the battery $$V_2=4 \ V$$ get added as they are connected in series with same polarity.
Net emf $$V_{net}=10\,V+4\,V=14\,V$$
The resistances $$1 \Omega, 3 \Omega$$ and $$2 \Omega$$ are connected in series.
Total resistance $$R_{series}=1+3+2=6\Omega$$

By Ohm's law,
$$V_{net}=IR_{series}$$
$$\Rightarrow I=\dfrac{V_{net}}{R_{series}}$$
Hence, Current $$I=\dfrac {14}{6}=2.3\,A$$

For the direction of current, we can see net emf is in the same sense of the individual batteries because $$10\,V $$ battery and $$4\,V $$ battery are connected back-to-back. So $$2.3\,A$$ current will flow in the direction $$A$$ to $$B$$ via $$E$$.