Current Electricity Test

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Current Electricity Test - 79

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Question 1
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In the arrangement shown in figure, the current through $$5 \Omega$$ resistor is

A
$$2 A$$

B
Zero

C
$$\dfrac{12}{7}A$$

D
$$1 A$$

Solution

Using the Kirchhoff's voltage law
for loop ABCDA, $$2i_1+5(i_1+i_2)=12$$ or $$ 7i_1+5i_2=12 ...(1)$$
and for loop EBCFE, $$2i_2+5(i_1+i_2)=12$$ or $$ 5i_1+7i_2=12 ...(2)$$
Now, $$(1)\times 7 , (2)\times 5 \Rightarrow 49i_1+35i_2=84 ...(3)$$ and
$$25i_1+35i_2=60 ...(4)$$
$$(3)-(4) \Rightarrow 24i_1=24$$ or $$i_1=1 A$$
Substituting the value of $$i_1$$ in $$(1)$$ we get, $$i_2=\dfrac{12-7}{5}=1 A$$
Thus the current through $$5 \Omega=i_1+i_2=1+1=2A$$
Question 2
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Electrical resistance of certain materials, known as superconductors, changes abruptly from a non zero value to zero as their temperature is lowered below a critical temperature $$T_C(0)$$. An interesting property of superconductors is that their critical temperature becomes smaller than $$T_C(0)$$ if they are placed in a magnetic field, i.e, the critical temperature $$T_c(B)$$ is a function of the magnetic field strength B. The dependence of $$T_C(B)$$ on B is shown in the figure.
In the graphs given, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields $$B_1$$(solid line) and $$B_2$$(dashed line). if $$B_2$$ is larger than $$B_1$$ which of the following graphs shows the correct variation of R with T in these fields?

A

B

C

D
Question 3
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A cell of emf $$E$$ and internal resistance $$r$$ is connected in series with an external resistance $$nr$$. Then, the ratio fo the terminal potential difference to E.M.F is

A
$$1/n$$

B
$$\cfrac { 1 }{ n+1 } \quad $$

C
$$\quad \cfrac { n }{ n+1 } $$

D
$$\cfrac { n+1 }{ n } $$

Solution

Applying voltage division rule, to find terminal voltage(v);
Voltage drop $$nR=E\cdot\cfrac{nR}{nR+R}=\cfrac{E\cdot nR}{R[n+1]}$$
$$\Rightarrow \cfrac{V}{E}=\cfrac{n}{[n+1]}$$
Question 4
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Directions For Questions

The figure shows the circuit of a potentiometer. The length of the potentiometer wire $$AB$$ is $$50cm$$. The emf of the battery $${E}_{1}$$ is $$4$$ volt, having negligible internal resistance. Values of resistance $${R}_{1}$$ and $${R}_{2}$$ are $$15$$ ohm and $$5$$ ohm respectively. When both the keys are open, the null point is obtained at a distance of $$31.25cm$$ from end A but when both the keys are closed, the balance length reduces to $$5cm$$ only. Given $${R}_{AB}=10\Omega$$

...view full instructions

The emf of the cell $${E}_{2}$$ is:

A
$$1$$ volt

B
$$2$$ volt

C
$$3$$ volt

D
$$4$$ volt

Solution
Question 5
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The Kirchhoff's first law $$(\displaystyle\sum i=0)$$ and second law $$\left(\displaystyle \sum iR=\displaystyle\sum E\right)$$, where the symbols have their usual meanings, are respectively based on

A
Conservation of charge, conservation of energy

B
Conservation of charge, conservation of momentum

C
Conservation of energy, conservation of charge

D
Conservation of momentum, conservation of charge

Solution

$$\sum i=0$$ (Conservation of charge).
Charge can neither be created or destroyed.
$$iR=\sum E$$ (Conservation of energy).
The total potential around a loop$$=0$$ or
Drop-in Potential energy=Energy dissipated.
According to Kirchhoff's first law, a junction can act neither as a source of charge nor as a sink of charge. This supports the law of conservation of charge. According to Kirchhoff's second law, the energy per unit charge transferred to the moving charges is equal to the energy per unit charge transferred from them. This supports the law of conservation of energy.
Question 6
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Which of the following relation is wrong?

A
1 ampere $$ \times $$ 1 ohm = 1 Volt

B
1 watt$$ \times $$ 1 sec = 1 Joule

C
1 newton per coulomb = 1 Volt per meter

D
1 coulomb $$ \times $$ 1 volt = 1 watt

Solution
Question 7
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Directions For Questions

The figure shows the circuit of a potentiometer. The length of the potentiometer wire $$AB$$ is $$50cm$$. The emf of the battery $${E}_{1}$$ is $$4$$ volt, having negligible internal resistance. Values of resistance $${R}_{1}$$ and $${R}_{2}$$ are $$15$$ ohm and $$5$$ ohm respectively. When both the keys are open, the null point is obtained at a distance of $$31.25cm$$ from end A but when both the keys are closed, the balance length reduces to $$5cm$$ only. Given $${R}_{AB}=10\Omega$$

...view full instructions

The internal resistance of the cell $${E}_{2}$$ is:

A
$$4.5\Omega$$

B
$$5.5\Omega$$

C
$$6.5\Omega$$

D
$$7.5\Omega$$

Solution
Question 8
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If an alternate current main supply is given to be $$220 \ V$$. What would be the average electro motive force during a positive half cycle:

A
$$198 \ V$$

B
$$386 \ V$$

C
$$256 \ V$$

D
None of these

Solution

$$Vav= \dfrac{2}{\pi} Vo= \dfrac{2}{\pi} Vrms \times \sqrt{2}$$
$$= \dfrac{2\sqrt{2}}{\pi} Vrms$$
$$= \dfrac{2\sqrt{2}}{\pi} \times 220$$
$$198\ V$$
Question 9
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A potentiometer wire has length $$4$$m and resistance $$8\Omega$$.What should be the resistance that must be connected in series with the wire and an accumulator of emf $$2$$V, so as to get a potential gradient $$1$$mV per cm on the wire?

A
$$ 32\Omega$$

B
$$36\Omega$$

C
$$ 40\Omega$$

D
$$ 30\Omega$$

Solution

Given,$$l=4$$m
$$R=$$Potentiometer wire resistance $$=8\Omega$$
Potential gradient$$=\dfrac{dV}{dr}=1$$mVper cm
So, for $$400$$cm, $$\Delta V=400\times 1\times {10}^{-3}=0.4$$V
Let a resistor $${R}_{s}=$$connected in series, so as
$$\Delta V=\dfrac{V}{R+{R}_{s}}\times R$$
$$0.4=\dfrac{2}{8+R}\times 8$$
$$\Rightarrow 8+R=\dfrac{16}{0.4}=40$$
$$\Rightarrow R=32\Omega$$
Question 10
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Two heater coils separately take $$10$$ minute and $$5$$ minute to boil certain amount of water. If both the coils are connected in series, the time taken will be

A
$$15$$ min

B
$$7.5$$ min

C
$$3.33$$ min

D
$$2.5$$ min

Solution