Current Electricity Test

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Current Electricity Test - 65

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Question 1
1 / -0

Two cells of emf's approximately $$5$$V and $$10$$V are to be accurately compared using a potentiometer of length $$400$$cm.

A
The battery that runs the potentiometer should have voltage of $$8$$V

B
The battery of potentiometer can have a voltage of $$15$$V and R adjusted so that the potential drop across the wire slightly exceeds $$10$$V

C
The first portion of $$50$$cm of wire itself should have a potential drop of $$10$$V

D
Potentiometer is usually used for comparing resistances and not voltages

Solution
Question 2
1 / -0

In the network shown in the figure, the equivalent resistance between A and C is

A
$$10R$$

B
$$(6/11)R$$

C
$$(12/15)R$$

D
$$(50/11)R$$

Solution

$$2R, R$$ and $$3R$$ are in parallel, as they all are between same potential
$$\dfrac{1}{R}+\dfrac{1}{2R}+\dfrac{1}{3R}-\dfrac{1}{R_p}\Rightarrow \dfrac{1}{R_p}=\dfrac{6+3+2}{6R}=\dfrac{11}{6R}$$
$$\dfrac{6R}{11}$$ and $$4R$$ is in parallel, so
$$\dfrac{6R}{11}+4R=\dfrac{6R+44R}{11}=\left(\dfrac{50}{11}\right)R$$
$$\boxed{Req_0=\left(\dfrac{50}{11}\right)R}$$
Question 3
1 / -0

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

A
$$5\sin{(\omega t+{53}^{o})}$$

B
$$5\sin{(\omega t+{37}^{o})}$$

C
$$5\sin{(\omega t+{45}^{o})}$$

D
$$5\cos{(\omega t+{53}^{o})}$$

Solution

Usin KCL at the node
$$i_1+i_2=i_3$$
$$i_3= 3sin\ \omega t+4 cos\ \omega t $$
$$i_3= 5(3/5 sin\ \omega t+4/5 cos\ \omega t)$$
$$i_3= 5\ sin(\omega t+\ 53^o)$$
Question 4
1 / -0

If the figure shows a part of an electric circuit, then the current $$I$$ is

A
$$1A$$

B
$$3A$$

C
$$2A$$

D
$$4A$$

Solution

Outgoing Current$$=3+2+5=10A$$
Incoming Current $$=6+1+I=(7+I)A$$
By comparing ,we get $$\Rightarrow\,10=7+I$$
$$\Rightarrow I=3$$
Question 5
1 / -0

Consider a cube as shown in the fig-1; with uniformly distributed charge within its volume. The potential at one of its vertex P is $${V_0}$$.A cubical portion of half the size (half edge length) of the original cube is cut and removed as shown in the fig-2. Find the modulus of potential at the point P in the new structure.

A
$${{7} \over 8}{V_0}$$

B
$${{{V_0}} \over 2}$$

C
$${{3{V_0}} \over 4}$$

D
$${{{V_0}} \over 4}$$

Solution

Potential at one vertex $$=V-0$$
When one of the vertex out of $$8$$ is removed
Portion of vertex$$=\cfrac{1}{8}$$
Potential of the vertex$$=\cfrac{1}{8}(V_0)$$
Potential of remaining cube $$=(V_0-\cfrac{1}{8}V_0)=\cfrac{7}{8}V_0$$
Question 6
1 / -0

The number of turns in th coil of an $$ac$$ generator is $$5000$$ and the area of the coil is $$0.25\ {m}^{2}$$. The coil is rotated at the rate of $$100$$ cycles/s in a magnetic field of $$0.2\ T$$. The peak value of emf generated is nearly:

A
$$786\ kV$$

B
$$440\ kV$$

C
$$220\ kV$$

D
$$157.1\ kV$$

Solution

As given,
$$N=5000 turns$$
$$B=0.2 w/m^{2}$$
$$A=0.25m^{2}$$
$$V=1000sec^{-1}$$
as $$E_{0}=NBAW$$
$$=5\times { 10 }^{ 3 }\times 2.5\times 0.25\times 100\times 2\times 3.14$$
$$=3.14\times 0.5\times 10^{6}V$$
$$E_{0}=157kV$$
Question 7
1 / -0

The net resistance between $${\text{X and Y}}$$ is

A
$${\text{4}}\Omega {\text{ }}$$

B
$${\text{4}}{\text{.55}}\Omega {\text{ }}$$

C
$${\text{2}}\Omega $$

D
$${\text{20}}\Omega $$

Solution
Question 8
1 / -0

In the following part of the circuit, potential difference between points A and B equals to

A
$$30V$$

B
$$45V$$

C
$$-15V$$

D
$$-45V$$

Solution
Question 9
1 / -0

If current $$I_1 = 3A$$ sin $$\omega t$$ and $$I_2 = 4A$$ cos $$\omega t$$, then $$I_3$$ is :

A
$$5A sin (\omega t + 53^0)$$

B
$$5A sin (\omega t + 37^0)$$

C
$$5A sin (\omega t + 45^0)$$

D
$$5A sin (\omega t + 30^0)$$

Solution

Given,
$$I_{1}=3Asin\omega t$$ and $$I_{2}=4Acos\omega t$$

$$I_{3}=I_{1}+I_{2}$$

$$\Rightarrow I_{3}=5\times(\dfrac{3}{5}sin\omega t+\dfrac{4}{5}cos\omega t)=5Asin(\omega t+53^0)$$.

Hence the correct option is A
Question 10
1 / -0

Two wires of same dimensions but resistivities $${\rho _1}$$ and $${\rho_2}$$ are connected in series. The equivalent resistivity of the combination is:

A
$${\rho _1} + {\rho_2}$$

B
$$\frac{1}{2}\left( {\,{\rho _1} + {\rho_2}} \right)$$

C
$$\sqrt {{\rho _1}{\rho _2}} $$

D
$$2\left( {\,{\rho _1} + {\rho_2}} \right)$$

Solution

$$\textbf{Step 1: Calculation of resistance of both wires}$$
We know that Resistance, $$R=\rho$$ $$l/A$$

As, the physical dimensions are same then Length $$l$$ and Area $$A$$ are constant
So, $$R_{1}=\rho_{1}$$ $$l/A$$
$$ \&$$ $$R_{2}=\rho_{2}$$ $$l/A$$

$$\textbf{Step 2: Equivalent resistance and resistivity}$$
For series combination $$R_{eq} = R_{1}+R_{2}=(\rho_{1}+ \rho_{2})l/A$$
Compare this with $$R_{eq} = (\rho_{eq} )l/A$$
So, $$\rho_{eq} =\rho_1+\rho_{2}$$

Hence, Option A is correct