Current Electricity Test

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Current Electricity Test - 57

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Weekly Quiz Competition

Question 1
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Copper and Carbon wires are connected in series and the combined resistor is kept at $$\displaystyle { 0 }^{ \circ }C$$. Assuming the combined resistance does not vary with temperature, the ratio of the resistances of Carbon and Copper wires at $$\displaystyle { 0 }^{ \circ }C$$ is:
(Temperature coefficients of resistivity of Copper and Carbon respectively are $$\displaystyle 4\times { { 10 }^{ -3 } }/{ ^{ \circ }{ C } }$$ and $$\displaystyle -0.5\times { { 10 }^{ -3 } }/{ ^{ \circ }{ C } }$$)

A
$$8$$

B
$$6$$

C
$$2$$

D
$$4$$

Solution

It is given that combined series resistance does not change with temperature.
Thus, $$R_{Cu}+R_{C}=R_{Cu}(1+\alpha_{Cu}\Delta T)+R_C(1+\alpha_C\Delta T)$$
for all values of $$\Delta T$$
Thus, $$R_{Cu}\alpha_{Cu}=-R_{C}\alpha_{C}$$
$$\implies \dfrac{R_C}{R_{Cu}}=-\dfrac{\alpha_{Cu}}{\alpha_C}$$
$$=-\dfrac{4\times 10^{-3}}{-0.5\times 10^{-3}}=8$$
Question 2
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The electric current $$i$$ in the circuit shown is:

A
$$6\ A$$

B
$$2\ A$$

C
$$3\ A$$

D
$$4\ A$$

Solution

Using Kirchhoff's junction law at A : $$3 + 2 = i_1$$
$$\implies i_1 = 5$$ A

Using Kirchhoff's junction law at B : $$ i_1 = 2 + i_2$$
$$\therefore$$ $$ 5 = 2 + i_2$$ $$\implies i_2 = 3$$ A

Using Kirchhoff's junction law at C : $$ i_2 + 1 = i$$
$$\therefore$$ $$ 3 + 1 = i$$ $$\implies i = 4$$ A
Question 3
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A group of $$N$$ cells whose emf varies directly with the internal resistance as per the equation $$\displaystyle { E }_{ N }=1.5{ r }_{ N }$$ are connected as shown in the figure above. The current $$I$$ in the circuit is:

A
$$5.1\ A$$

B
$$0.51\ A$$

C
$$1.5\ A$$

D
$$0.15\ A$$

Solution

Current, $$I=\dfrac{\Sigma E}{\Sigma r}$$

$$=\dfrac{1.5r_1+1.5r_2+........+1.5r_N}{r_1+r_2+......+r_N}$$

$$=1.5 \ A$$
Question 4
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A potentiometer wire is $$100\ $$cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $$50$$ cm and $$10$$ cm from the positive end of the wire in the two cases. The ratio of emf's is:

A
$$5:1$$

B
$$5:4$$

C
$$3:4$$

D
$$3:2$$

Solution

If the emfs of the two cells are taken as $$E_1$$ and $$E_2$$ respectively.

So, $$\dfrac{E_1+E_2}{E_1-E_2}=\dfrac{50}{10}$$

or, $$\dfrac{2E_1}{2E_2}=\dfrac{50+10}{50-10}$$

or, $$\dfrac{E_1}{E_2}=\dfrac{3}{2}$$
Question 5
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The resistance across $$A$$ and $$B$$ in the given figure will be:

A
$$3R$$

B
$$R$$

C
$$\cfrac{R}{3}$$

D
None of the above

Solution

Given: $$R_1 = R_2 = R_3 = R$$

The three resistances are connected in parallel between the points A and B.
Equivalent resistance for parallel combination:
$$\Rightarrow \dfrac{1}{R_{AB}} =\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$$

$$\Rightarrow \dfrac{1}{R_{AB}} =\dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} = \dfrac{3}{R}$$

$$\Rightarrow $$ $$R_{AB} = \dfrac{R}{3}$$
Question 6
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In shown diagram, if voltage of battery is doubled and resistance kept constant then what is the new Power dissipated at resistor(Consider $$P$$ was the initial power dissipation) ?

A
$$\dfrac{P}{4}$$

B
$$\dfrac{P}{2}$$

C
P

D
2P

E
4P

Solution

Power, $$P=\dfrac{V^2}{R}$$
As R is kept constant so $$P \propto V^2$$
$$P'$$is new power dissipation.
So, $$\dfrac{P'}{P}=\dfrac{(2V)^2}{V^2}=\dfrac{4V^2}{V^2}$$ or $$P'=4P$$
Thus, option E is correct.
Question 7
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A potentiometer has uniform potential gradient. The specific resistance of the material of the potentiometer wire is $$\displaystyle { 10 }^{ -7 }$$ ohm-metre and the current passing through it is $$0.1$$ ampere, cross-section of the wire is $$\displaystyle { 10 }^{ -6 }{ m }^{ 2 }$$. The potential gradient along the potentiometer wire is:

A
$$\displaystyle { 10 }^{ -6 }{ V }/{ m }$$

B
$$\displaystyle { 10 }^{ -4 }{ V }/{ m }$$

C
$$\displaystyle { 10 }^{ -8 }{ V }/{ m }$$

D
$$\displaystyle { 10 }^{ -2 }{ V }/{ m }$$

Solution

Given : Resistivity $$\rho = 10^{-7} \Omega m$$
Cross-section area $$A =10^{-6} m^2$$
Current in the wire $$i = 0.1$$ A
Resistance of the wire $$R = \rho \dfrac{l}{A}$$

Thus resistance gradient $$R' = \dfrac{R}{l} = \dfrac{\rho}{A} = \dfrac{10^{-7}}{10^{-6}} = 0.1$$

$$\therefore$$ Potential gradient $$V'= i R' = 0.1 \times 0.1 = 10^{-2}$$ V/m
Question 8
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If a third identical resistor is added in parallel in a circuit of two identical parallel resistors. Calculate the ratio of the new equivalent resistance to the old?

A
$$\dfrac{4}{9}$$

B
$$\dfrac{2}{3}$$

C
$$1$$

D
$$\dfrac{3}{2}$$

E
$$\dfrac{9}{4}$$

Solution

Let the resistance of each resistor be $$R$$.
Equivalent resistance of two resistors in parallel, $$\dfrac{1}{R_{eq}} = \dfrac{1}{R} + \dfrac{1}{R}$$
$$\implies R_{eq} = \dfrac{R}{2}$$

Equivalent resistance of three resstors in parallel, $$\dfrac{1}{R'_{eq}} = \dfrac{1}{R} + \dfrac{1}{R}+ \dfrac{1}{R}$$
$$\implies R'_{eq} = \dfrac{R}{3}$$

$$\therefore\ \dfrac{R'_{eq}}{R_{eq}} = \dfrac{R/3}{R/2} = \dfrac{2}{3}$$
Question 9
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Three resistor of $$10 \Omega$$, $$20 \Omega$$, and $$30 \Omega$$ are connected in parallel in a circuit. Calculate the equivalent resistance.

A
$$6000 \ \Omega$$

B
$$0.017 \ \Omega$$

C
$$1 \ \Omega$$

D
$$60 \ \Omega$$

E
$$5.45 \ \Omega$$

Solution

Given : $$R_1 = 10\Omega$$ $$R_2= 20\Omega$$ $$R_3 = 30\Omega$$
Equivalent resistance of parallel combination $$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} +\dfrac{1}{R_2} + \dfrac{1}{R_3}$$

$$\therefore$$ $$\dfrac{1}{R_{eq}} = \dfrac{1}{10} +\dfrac{1}{20} + \dfrac{1}{30}$$ $$\implies R_{eq} = 5.45$$ $$\Omega$$
Question 10
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This question relates to the DC circuit shown below.
The values for current, voltage, and resistance in the circuit are graphed from point A through point G in the graphs directly below.
Which of the graphs shows the voltage from point A to point G?

A
A

B
B

C
C

D
D

E
E

Solution

Here the current through entire circuit is constant and the resistance is increasing from A to G. So by Ohm's law, the voltages throughout the circuit will decrease as the distance from point A increases. Thus, each increase in the resistance produces a corresponding decrease in voltage. So graph A will be the correct.