Current Electricity Test

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Current Electricity Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

Electromotive force of a cell is basically a

A
force

B
power

C
work

D
current capacity

Solution

EMF$$=$$ work done per unit charge $$= \dfrac{Fd}{q}=\dfrac{qEd}{q}=Ed=V$$
Question 2
1 / -0

Kirchhoff's first and second law shows the conservation of :-

A
linear momentum and angular momentum.

B
charge and energy

C
mass and energy.

D
charge and linear momentum.

Solution

Kirchhoff's first law is the consequence of conservation of charge and second law is the conservation of energy.
Question 3
1 / -0

Eight identical cells each of potential E and internal resistance r are connected in series to form a closed circuit. An ideal voltmeter connected across 2 cells will read :-

A
13 E

B
zero

C
2 E

D
10 E

Solution

Applyin Kirchoff's voltage law on the closed loop gives,
$$8E-8(ir)=0$$
Hence, $$i=\dfrac{E}{r}$$
Hence potential difference across A and B= $$2E-2(ir)=0$$
Question 4
1 / -0

Three unequal resistors in parallel are equivalent to a resistant 1 ohm. If two of them are in the ratio 1 :2 and if no resistance value is fractional, the largest of the three resistance in ohm is :-

A
4

B
6

C
8

D
12

Solution

$$1={\dfrac {1}{R_1}}+{\dfrac {1}{R_2}}+{\dfrac {1}{R_3}}$$
$${\dfrac {R_1}{R_2}}={\dfrac {1}{2}}$$ (Given)
$$R_2=2R_1$$
$$1={\dfrac {2}{R_2}}+{\dfrac {1}{R_2}}+{\dfrac {1}{R_3}}$$
$$1={\dfrac {3}{R_2}}+{\dfrac {1}{R_3}}$$
$$1-{\dfrac {3}{R_2}}={\dfrac {1}{R_3}}$$
$${\dfrac {{R_2}-3}{R_2}}={\dfrac {1}{R_3}}$$
$${R_3}={\dfrac {R_2}{{R_2}-3}}$$
Now only $$R_2=6 \Omega$$ satisfies this above equation $$\Rightarrow R_3=2 \Omega$$ & $$R_1=3 \Omega$$
Question 5
1 / -0

In how many parts (equal) a wire of $$100\Omega$$ be cut so that a resistance of $$1\Omega$$ is obtained by connecting them in parallel?

A
$$10$$

B
$$5$$

C
$$100$$

D
$$50$$

Solution

The resistance of the conductor is proportional to length of the conductor.
That is, $$R=\rho \dfrac { l }{ A } $$.
The resistance of one wire of length l, R1 = 100 ohms.
If we cut it in a half: $$\dfrac { { R }_{ 1 } }{ l } =\dfrac { { R }_{ 2 } }{ \dfrac { l }{ 2 } } \quad \rightarrow \quad { R }_{ 2 }=\dfrac { { R }_{ 1 } }{ 2 } $$.
If we cut it in n parts the resistance will be $$R=\dfrac { { R }_{ 1 } }{ n } $$.
Because we shorten the length n times so it became $$\dfrac { l }{ n } $$.
When we connect these parts in parallel the output resistance must be 1 ohm.
So, $$1=\dfrac { 1 }{ R } +\dfrac { 1 }{ R } +...=n\dfrac { 1 }{ R } =n\dfrac { 1 }{ \dfrac { { R }_{ 1 } }{ n } } =\dfrac { { n }^{ 2 } }{ { R }_{ 1 } } =\dfrac { { n }^{ 2 } }{ 100 } \rightarrow n=10$$.
Hence, a resistance of 1 ohm is obtained by connecting 10 equal parts of the wire and connecting them in parallel.
Question 6
1 / -0

Referring to the circuit, $$R$$ is the resistance of a potentiometer. As the sliding contact is moved from $$a$$ to $$b$$ the reading in the ideal ammeter will :

A
increase

B
decrease

C
initial decrease and then increase

D
initially increase and then decrease

Solution

$$R_{ac}$$ increases as x increases therefore current through ammeter decreases.
$$R_{ac}=$$Resistance from point a to c
Question 7
1 / -0

Two long straight cylindrical conductors with resistivities $$\rho_1$$ and $$\rho_ 2$$ respectively are joined together as shown in figure. If current I flows through the conductors, the magnitude of the total free charge at the interface of the two conductors is

A
zero

B
$$\displaystyle \frac{(\rho_1 - \rho_2) I \varepsilon_0}{2}$$

C
$$\varepsilon_0 I |\rho_1 - \rho|$$

D
$$\varepsilon_0 I |\rho_1 + \rho|$$

Solution

Apply Gauss's law : $$\displaystyle \frac{q_{in}}{\varepsilon_0} = $$ (outgoing flux - incoming flux) $$= \Delta (EA)$$
$$\displaystyle V = IR = \frac{I \rho l}{A} \Rightarrow \frac{V}{l} A = I \rho \Rightarrow EA = I \rho$$
$$\Rightarrow \displaystyle \frac{q_{in}}{\varepsilon_0} = I |\rho_1 - \rho_2| \Rightarrow q_{in} = I \varepsilon_0 |\rho_1 - \rho_2|$$
Question 8
1 / -0

In the electrolysis of NaCl

A
$$Cl^-$$ is oxidised at anode

B
$$Cl^-$$ is reduced at anode

C
$$Cl^-$$ is reduced at cathode

D
$$Cl^-$$ is neither reduced nor oxidised

Solution

Answer is A.
Electrolysis of an aqueous solution of table salt (NaCl, or sodium chloride) produces aqueous sodium hydroxide and chlorine, although usually only in minute amounts. NaCl (aq) can be reliably electrolysed to produce hydrogen. Hydrogen gas will be seen to bubble up at the cathode, and chlorine gas will bubble at the anode.
As the electricity from the battery passes through and between the electrodes, the water splits into hydrogen and chlorine gas, which collect as very tiny bubbles around the electrode tips. Hydrogen collects around the cathode and chlorine gas collects around the anode.
Hence, in the electrolysis of NaCl, $$Cl^-$$ is oxidized at the anode.
Question 9
1 / -0

Eels are able to generate current with biological cells called electroplaques. The electroplaques in an Eel are arranged in 100 rows, each row stretching horizontally along the body of the fish containing 5000 electroplaques. The arrangement is suggestively shown below. Each electroplaques has an emf of 0.15 V and internal resistance of $$0.25\Omega $$. The water surrounding the Eel completes a circuit between the head and its tail. If the water surrounding it has a resistance of $$500\Omega $$, the current an Eel can produce in water is about :-

A
1.5 A

B
3.0 A

C
15 A

D
30 A

Solution

Let the current in each row be $$i$$.
Hence current in water is contributed by each row and total current in water becomes $$100i$$.
Applying Kirchoff's Voltage Law to row and water,
$$5000E-(5000r)i-500(100i)=0$$
$$\implies i=\dfrac{5000E}{5000r+100R}$$
$$=0.015A$$
Hence current in water is $$100i=1.5A$$
Question 10
1 / -0

Two wires of resistance $$R_{1}$$ and $$R_{2}$$ at $$0^{o}C$$ have temperature coefficient of resistance $$\alpha _{1}$$ and $$\alpha _{2}$$, respectively. These are joined in series. The effective temperature coefficient of resistance is :-

A
$$\displaystyle \frac{\alpha _{1}+\alpha _{2}}{2}$$

B
$$\displaystyle \sqrt{\alpha _{1}\alpha _{2}}$$

C
$$\displaystyle \frac{\alpha _{1}R_{1}+\alpha _{2}R_{2}}{R_{1}+R_{2}}$$

D
$$\displaystyle \frac{\sqrt{R_{2}R_{1}\alpha _{1}\alpha _{2}}}{\sqrt{R_{1}^{2}+R_{2}^{2}}}$$

Solution

The thermal coefficient of resistance is $$\Rightarrow \alpha={\frac {1}{R}}{\frac {dR}{dt}}$$
Equivalent resistance in series= $${R_1}+{R_2}={R_{eq.}}$$
$${\frac {dR_{eq.}}{dt}}={\frac {dR_{1}}{dt}}+{\frac {dR_{2}}{dt}}$$
$${\alpha_{eq.}}{R_{eq.}}={\alpha_{1}}{R_{1}}+{\alpha_{2}}{R_{2}}$$
$${ \alpha }_{ eq. }=\frac { { \alpha }_{ 1 }{ R }_{ 1 }+{ \alpha }_{ 2 }{ R }_{ 2 } }{ { R }_{ 1 }+{ R }_{ 2 } }$$