Current Electricity Test

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Current Electricity Test - 64

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Question 1
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An electric heater is connected to the voltage supply. After a few seconds, current get its steady value then its initial current will be?

A
Equal to its steady current

B
Slightly higher than its steady current

C
Slightly less than its steady current

D
Zero

Solution

Resistance increases with temperature and hence steady current decreas.So, Initial current will be slightly higher than its steady current.
Question 2
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A wire of resistance $$12\: ohm/meter\: $$is bent to form a complete circle of radius $$10cm$$. The resistance between its two diametrically opposite points, A and B as shown in the figure is?

A
$$3\Omega$$

B
$$6\pi \Omega$$

C
$$6\Omega$$

D
$$0.6\pi \Omega$$

Solution

Total length of wire, $$l= 2\pi\times 0.1=0.2\pi\ m$$
Resistance per unit length, $$r=12\Omega/m$$
Let us assume two semicircular part of wire connected in parallel between $$A$$ and $$B$$

Resistance of each part $$,R= r \times l = 12\times \cfrac{0.2\pi}{2}= 1.2 \pi\ \Omega$$

Two equal resistors (R) are connected in parallel between A and B so equivalent resistance is equal to $$R/2$$.

$$\therefore$$ Resistance between $$A$$ and $$B = \dfrac R2 = \dfrac{1.2 \pi}2= 0.6\pi \ \Omega$$

Hence, option $$(D)$$ is correct.
Question 3
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An infinite ladder network of resistances is constructed with $$1\Omega$$ and $$2\Omega$$ resistance as shown in figure. The $$6$$V battery between A and B has negligible internal resistance. The equivalent resistance between A and B is?

A
$$1\Omega$$

B
$$2\Omega$$

C
$$3\Omega$$

D
$$4\Omega$$

Solution

$$ \textbf {Step 1: Find Smallest Repeating unit [Refer Fig. 1]} $$

$$ \textbf{Step 2: Replace with equivalent resistance R. [Refer Fig. 2]} $$
After the first smallest repeating unit, the remaining circuit is also same as the initial circuit.

Therefore, if we assume the equivalent resistance between A & B as $$R$$, then the resistance of the remaining circuit will also be $$R$$ as shown in the figure.

$$ \textbf {Step 3: Find equivalent resistance between A and B and equate with R. [Refer Fig. 3]} $$
From figure, the equivalent resistance between A & B will be equal to the equivalent resistance of the remaining circuit. So,

$$ R_{AB} = 1 + \dfrac {2R}{2+R} $$

$$\Rightarrow\ \ R = 1 + \dfrac{2R}{2+R} $$

$$\Rightarrow\ \ 3R + 2 = R^2 + 2R $$

$$\Rightarrow\ \ R^2 - R - 2 = 0 $$

$$\therefore\ \ \ \ \boxed{R = 2 \Omega}$$ (Here, Negative value can is rejected)

Hence Equivalent resistance between points A and B is 2$$\Omega$$.
Question 4
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In the series combination of n cells each cell having emf $$\varepsilon$$ and internal resistance r. If three cells are wrongly connected, then total emf and internal resistance of this combination will be.

A
$$n\varepsilon, (nr-3r)$$

B
$$(n\varepsilon -2\varepsilon), nV$$

C
$$(n\varepsilon -4\varepsilon), nV$$

D
$$(n\varepsilon - 6\varepsilon), nV$$

Solution

Since in a combination of $$n$$ cells, there are $$3$$ cells connected with opposite polarity. It means that there are $$(n-3)$$ cells connected with correct polarity and remaining $$3$$ with opposite polarity.

The net emf of the combination is:
$$E_{net}=(n-3)E-3E$$

$$E_{net}=(n-6)E$$

On the other hand, the total internal resistance of all the batteries remains the same for the whole circuit, even for the batteries connected in opposite polarity.
So, the net internal resistance for the circuit will be $$nr$$.

So, option $$(D)$$ is correct.
Question 5
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In a Wheatstone bridge if the battery and galvanometer are interchanged then the deflection in galvanometer will:

A
Change in previous direction

B
Not change

C
Change in opposite direction

D
None of these

Solution

The changed configuration is still a wheat stone bridge hence there will be no change in deflection.
Answer B
Question 6
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Point out the right statements about the validity of Kirchhoff's junction rule.

A
It is based on conservation of charge

B
Outgoing currents add up and are equal to incoming currents at a junction

C
Bending or reorienting the wire does not change the validity of Kirchhoff's junction rule

D
All of above

Solution

By Kirchhoff's junction rule. Incoming current $$=$$ Outgoing current
The net charge is conserved and it is based on conservation of charge.
Also bending or reorienting the wire does not invalidate the conservation of charge principle. So, option D is correct.
Question 7
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A circuit has a section ABC as shown in figure. If the potentials at points A, B and C are $$V_1, V_2$$ and $$V_3$$ respectively. The potential at point O is?

A
$$V_1+V_2+V_3$$

B
$$\left[\dfrac{V_1}{R_1}+\dfrac{V_2}{R_2}+\dfrac{V_3}{R_3}\right]\left[\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\right]^{-1}$$

C
Zero

D
$$\left[\dfrac{V_1}{R_1}+\dfrac{V_2}{R_2}+\dfrac{V_3}{R_3}\right](R_1+R_2+R_3)$$

Solution

According to kirchhoff's first rule to $$O$$
$$-l_1 -l_2 - l_3 = 0$$ or
$$l_1 + l_2+l_3 = 0$$ ...(i)
If $$V_0$$ is the electric potential at $$O$$
$$V_1, V_2$$ and $$V_3$$ be the potential at $$A, B$$ and $$C$$ respectively then $$V_0 - V_1 = l_1R_1$$.
$$V_0 - V_2 = l_2R_2$$ and $$V_0 - V_3 = l_3R_3$$
This, $$l_1 = \left(\dfrac{V_0-V_1}{R_1}\right), l_2=\left(\dfrac{V_0 - V_2}{R_2}\right)$$, and $$l_3 = \left(\dfrac{V_0 - V_3}{R_3}\right)$$

putting volume of $$l_1, l_2$$ and $$l_3$$ in (I) we get
$$\dfrac{V_0 - V_1}{R_1} + \dfrac{V_0-V_2}{R_2} + \dfrac{V_0 - V_3}{R_3} = 0$$

or $$V_0 \left[\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\right] - \left[ \dfrac{\frac{V_1}{R_1} + \frac{V_2}{R_2} + V_3}{R_3}\right] = 0$$

or $$V_0 = \left[ \dfrac{\frac{V_1}{R_1} + \frac{V_2}{R_2} + V_3}{R_3}\right]\left[\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\right] ^{-1}$$
Question 8
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In a potentiometer the balancing with a cell is at length of $$220$$cm. On shunting the cell with a resistance of $$3\Omega$$ balance length becomes $$130$$cm. What is the internal resistance of this cell?

A
$$4.5\Omega$$

B
$$7.8\Omega$$

C
$$6.3\Omega$$

D
$$2.08\Omega$$

Solution

Given that,
Initial balancing length of potentiometer with a cell, $$l_1= 220\ cm$$
Now the cell is shunted with resistance, $$ R=3\Omega$$
Final balancing length, $$l_2= 130\ cm$$

Internal resistance of the cell, $$r= \bigg( \dfrac{l_1-l_2}{l_2} \bigg) R$$

$$\implies r= \bigg( \dfrac{90}{130}\bigg) \times 3= \dfrac{270}{130}=2.08 \Omega$$

Hence, option $$(D)$$ is correct.
Question 9
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Which of the following statements is false?

A
Kirchhoff's second law represents energy conservation.

B
Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude.

C
In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed.

D
A rheostat can be used as a potential divider.

Solution

The balanced condition is given by $$\dfrac{P}{Q}=\dfrac{R}{S}$$; When battery and Galvanometer are exchanged, it become $$\dfrac{P}{R}=\dfrac{Q}{S}$$; which is same as previous.
Question 10
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Find the equivalent resistance between points A and B in the given circuit

A
$$6\Omega$$

B
$$12\Omega$$

C
$$4\Omega$$

D
$$18\Omega$$

Solution

using wheatstone bridge,
$$\dfrac{6}{3}=\dfrac{12}{6}\Rightarrow$$ It is a balanced wheatstone bridge, hence no current in $$12\Omega$$ i.e $$PQ$$ branch or equivalent resistance b/w $$A\times B$$.
$$6$$ & $$12\Omega$$ are in series
$$=6\Omega +12\Omega$$
$$=18\Omega$$
similarly, $$3\Omega$$ & $$6\Omega$$, equi $$=3+6=9\Omega$$
$$18\Omega$$ & $$9\Omega$$ are in parallel
$$\dfrac{1}{R_e}=\dfrac{1}{18}+\dfrac{1}{9}=\dfrac{3}{18}$$
$$\boxed{R_e=6\Omega}$$