Current Electricity Test

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Current Electricity Test - 50

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Question 1
1 / -0

For the following circuit the potential difference between x and y in volt is :-

A
10

B
50

C
100

D
0

Solution

Apply Kirchhoff's law
for loop pxqp: $$I_x(100+100)=200 \Rightarrow I_x=1 A$$
for loop pyqp: $$I_y(100+100)=200 \Rightarrow I_y=1 A$$
thus, $$V_p-V_x=100I_x=100 V $$ and $$V_p-V_y=100I_y=100 V$$
so, $$V_x-V_y=0$$
Question 2
1 / -0

The specific resistance of a wire

A
varies with its length.

B
varies with its cross-section.

C
varies with its mass.

D
does not depend upon its length, cross-section and mass.

Solution

The specific resistance or resistivity of a wire is the property of the material of it. It does not depend on mass, length or cross-section of wire. It may change with temperature.
Question 3
1 / -0

In the experiment of half deflection method the resistance R should be

A
> S

B
$$\approx$$ S

C
< S

D
$$\approx$$ G

Solution

In half deflection method, Rg = (R)*(S)/(R-S) .
Here , Rg= Galvanometer resistance, R = high resistance in series with the galvanometer, S = low resistance in parallel with the galvanometer. We want to approximate Rg in terms of S which we are able to vary in the experiment . If R>>S then Rg = S.
Question 4
1 / -0

A group of N cells whose emf varies directly with the internal resistance as per the equation $$E_{N}=1.5r_{N}$$ are connected as shown in the figure. The current I in the circuit is :-

A
5.1 A

B
0.51 A

C
1.5 A

D
0.15 A

Solution

If the current I flows in anticlockwise direction,

$$I=\frac{E_1+E_2+.......+E_N}{r_1+r_2+......+r_N}=\frac{1.5r_1+1.5r_2+.....+1.5r_N}{r_1+r_2+......+r_N}=1.5 A$$
Question 5
1 / -0

The resistance of a coil in a platinum resistance thermometer at $$0^{\circ}C$$ is 5 ohm and at $$100^{\circ}C$$ it is 5.75 ohm. Its resistance at an unknown temperature is 5.15 ohm. Then the unknown temperature will be :-

A
$$40^{\circ}C$$

B
$$10^{\circ}C$$

C
$$15^{\circ}C$$

D
$$20^{\circ}C$$

Solution

We know that $$R=R_0(1+\alpha. T)$$

Thus, temperature coeff
Question 6
1 / -0

The current voltage graph for a given metallic conductor at two different temperatures $$T_{1}$$ and $$T_{2}$$ are as shown in the figure. Then :-

A
$$T_{1} >T_{2}$$

B
$$T_{1} =T_{2}$$

C
nothing can be said about $$T_{1}$$ and $$T_{2}$$

D
$$T_{1}< T_{2}$$
Question 7
1 / -0

Two resistance $$R_{1}$$ and $$R_{2}$$ are made of different materials. The temperature coefficient of the material of $$R_{1}$$ is $$\alpha $$ and of the material of $$R_{2}$$ is $$-\beta $$. The resistance of the series combination of $$R_{1}$$ and $$R_{2}$$ will not change with temp., then ratio of resistance of two wire will be :-

A
$$\displaystyle \frac{\alpha }{\beta }$$

B
$$\displaystyle \frac{\alpha +\beta }{\alpha -\beta }$$

C
$$\displaystyle \frac{\alpha ^{2}+\beta ^{2}}{\alpha \beta }$$

D
$$\displaystyle \frac{\beta }{\alpha}$$

Solution

Here, $$R_1+R_2=$$ constant
When $$R_1$$ will increase, $$R_2$$ will decrease.
$$dR_1+dR_2=0$$
or $$\displaystyle R_1\alpha \Delta T-R_2\beta \Delta T=0 $$ as $$[R=R_0(1+\alpha T)]$$
or $$\displaystyle \frac{R_1}{R_2}=\frac{\beta}{\alpha}$$
Question 8
1 / -0

In a typical Wheatstone network the resistance in cyclic order are $$A=10\:\Omega ,\:B=5\:\Omega ,\:C=4\:
\Omega $$ and $$D=4\:\Omega $$ for the bridge to be balanced :-

A
$$10\:\Omega $$ should be connected in parallel $$A$$

B
$$10\:\Omega $$ should be connected in series with $$A$$

C
$$5\:\Omega $$ should be connected in series with $$B$$

D
$$5\:\Omega $$ should be connected in parallel with $$B$$

Solution

a) here, $$A=10||10=10/2=5, B=5,C=4,D=4$$

and $$A/B=1, C/D=1$$ so, $$A/B=C/D$$

b) here, $$A=10+10=20, B=5,C=4,D=4$$

and $$A/B=4,C/D=1$$ so$$, A/B \neq C/D$$

c) here,$$A=10,B=5+5=10,C=4,D=4$$

and $$A/B=1,C/D=1$$ so, $$A/B=C/D$$

d)here, $$A=10,B=5||5=5/2=2.5,C=4,D=4$$

And $$A/B=4,C/D=1$$ so,$$A/B \neq C/D$$
Question 9
1 / -0

At what temperature will the resistance of a copper wire become three times its value at $$0^{\circ}C$$ [Temperature coefficient of resistance for copper $$=4\times 10^{-3}per^{\circ}C$$] :-

A
$$400^{\circ}C$$

B
$$450^{\circ}C$$

C
$$500^{\circ}C$$

D
$$550^{\circ}C$$

Solution

As we know,
$$R=R_0(1+\alpha \Delta T)$$
give, $$R=3R_0$$ and $$\alpha =4\times 10^{-3}per^0C$$
putting Values,
$$3R_0=R_0(1+4\times 10^{-3}(T_2-T_1))$$
$$2=4\times 10^{-3}(T_2-0)$$ as $$T_1=0^0C$$
therefore,
$$T_2=500^0C$$
Question 10
1 / -0

Which of the statement is wrong :-

A
when all resistance are equal, then the sensitivity of wheatstone bridge is maximum.

B
when the galvanometer and the cell are interchanged, then the balancing of wheat stone bridge will be effected.

C
Kirchoff's first law for the currents meeting at the Junctions in an electric circuit shows the conservation of charge.

D
Rheostat can be used as potential divider.

Solution

Before interchanging cell and galvanometer (fig-a): the null condition is $$\frac{P}{Q}=\frac{S}{R}$$
After interchanging cell and galvanometer (fig-b): the null condition is $$\frac{P}{S}=\frac{Q}{R} \Rightarrow \frac{P}{Q}=\frac{S}{R}$$
Thus, in both case balanced condition is same so it will not be effected after interchanging cell and galvanometer.