Current Electricity Test

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Current Electricity Test - 44

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Question 1
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Directions For Questions

The emf of the driver cell of a potentiometer is 2 V, and its internal resistance is negligible. The length of the potentiometer wire is 100 cm, and resistance is $$5 \Omega$$.

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How much resistance is to be connected in series with the potentiometer wire to have a potential gradient of $$0.05$$ mV cm$$^{-1}$$?

A
$$1990\Omega$$

B
$$2000\Omega$$

C
$$1995\Omega$$

D
$$none$$ $$of$$ $$these$$

Solution

Here,$$k=0.05$$ mV cm$$^{-1}=5$$ mV m$$^{-1}$$
Let $$R$$ be the resistance is to be connected in series with the potentiometer wire, the potential of driver cell is
$$V = I(R+5)$$.................................................(1)
The current through the potentiometer wire is,
$$I=\dfrac {kl}{r}=\dfrac {5\times 10^{-3}\times 1}{5}=10^{-3}A$$
Hence, from(1)
$$2=I(R+r)=10^{-3}(R+5)$$
$$R= 2000-5$$
$$R=1995\Omega$$
Question 2
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A $$6V$$ battery of negligible internal resistance is connected across a uniform wire of length $$1 m$$. The positive terminal of another battery of e.m.f. $$4V$$ and internal resistance $$1\Omega$$ is joined to the point $$A$$ as shown in the figure. The ammeter shows zero deflection when the jockey touches the wire at the point $$C$$.
The length AC is equal to

A
2/3 m

B
1/3 m

C
3/5 m

D
1/2 m

Solution

we have,

$$E = \dfrac {Irl}{L}$$

$$l = \dfrac {EL}{Ir}$$

$$l = \dfrac {(4)(1)}{(6)(1)}$$.......$$\because I=\dfrac {6}{1}=6A$$

$$l = \dfrac {2}{3}$$
Question 3
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In the figure, the potentiometer wire $$AB$$ of length $$L$$ and resistance $$9r$$ is joined to the cell $$D$$ of emf $$\varepsilon$$ and internal resistance $$r$$. The cell $$C's$$ emf is $$\varepsilon/2$$ and its internal resistance is $$2r$$. The galvanomater $$G$$ will show no deflection when the length $$AJ$$ is

A
$$\displaystyle\frac{4L}{9}$$

B
$$\displaystyle\frac{5L}{9}$$

C
$$\displaystyle\frac{7L}{18}$$

D
$$\displaystyle\frac{11L}{18}$$

Solution

voltage drop on the balancing length of the wire $$\Rightarrow$$$$i \ {\rho} \ x$$
where $$ i$$= current flowing in primary circuit
$$\rho$$ = resistance per unit length of the wire
x = balancing length of wire AB when deflection in galvanometer is zero
When there is no any deflection in the galvanometer $$\Rightarrow$$ voltage of secondary circuit $$=$$ voltage drop across the wire AJ

$$\Rightarrow {\dfrac {E}{2}}=(\dfrac {E}{10r})(\dfrac {9r}{L}){x}$$ where $$x$$ is the length of wire AJ

$${x}={\dfrac {5L}{9}}$$
Question 4
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A brass disc and a carbon disc of same radius are assembled alternatively to make a cylindrical conductor. The resistance of the cylinder is independent of the temperature. The ratio of thickness of the brass disc to that of the carbon disc is $$\underline{\hspace{0.5in}}$$ .$$\alpha$$ is temperature coefficient of resistance & Neglect linear expansion

A
$$\displaystyle\frac{\alpha_C\rho_C}{\alpha_B\rho_B}$$

B
$$\displaystyle\frac{\alpha_C\rho_B}{\alpha_B\rho_C}$$

C
$$\displaystyle\frac{\alpha_B\rho_C}{\alpha_C\rho_B}$$

D
$$\displaystyle\frac{\alpha_B\rho_B}{\alpha_C\rho_C}$$

Solution

Given the resistance is independent of temperature .
Hence ,

$$\dfrac{\rho _B l _B}{A} ( \alpha _ B \Delta T ) = \dfrac{\rho _C l _C}{A} ( \alpha _ C \Delta T ) $$

$$\therefore \dfrac{l_B}{l_C} = \dfrac{\rho _C \alpha _C}{\rho _B \alpha _B} $$
Question 5
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Calculate the value of current $$I_2$$ in the section of networks shown in figure.

A
13 A

B
7 A

C
10 A

D
3 A

Solution

Applying KCL to the junction at point A, we get
$$15 -8-I_2 = 0$$
$$\therefore I_2 = 7A$$
Question 6
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Calculate the value of current $$I_4$$ in the section of networks shown in figure.

A
13 A

B
7 A

C
10 A

D
3 A

Solution
Question 7
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A wire of length $$L$$ and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised at $$\triangle T$$ in time $$t$$. $$N$$ number of similar cells is now connected in series with a wire of the same material and cross-section but of length $$2L$$. The temperature of the wire is raised by the same amount $$\triangle T$$ in the same time $$t$$. The value of $$N$$ is:

A
4

B
6

C
8

D
9

Solution

We know,

$$(\dfrac{V^2}{R}) = m \Delta T$$

where variables have their usual meanings.

For given data, $$V^2 = Rm$$

Now,

$$(3V^2) \propto Rm$$

$$(NV^2) \propto (2R)(2m)$$

From above equations we can write,

$$ \dfrac {N^2}{9} = 4$$

$$\therefore {N^2} = 36$$

$$\therefore {N} = 6$$
Question 8
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In a balanced wheat stone bridge, current in the galvanometer is zero. It remains zero when
$$[1]$$. battery emf is increased
$$[2]$$. all resistances are increased by 10 ohms
$$[3]$$. all resistances are made five times
$$[4]$$. the battery and the galvanometer are interchanged

A
only $$[1]$$ is correct

B
$$[1], [2]$$ and $$[3]$$ are correct

C
$$[1], [3]$$ and $$[4]$$ are correct

D
$$[1]$$ and $$[3]$$ are correct

Solution

The balancing condition for Wheatstone's bridge is, $$\dfrac {R_1}{R_2} = \dfrac {R_3}{R_4}$$
This condition is independent of emf of cell, hence if battery emf is increased the bridge will remain balanced means no current will flow through the galvanometer. Also,if resistances are increased five times, we get
$$\dfrac {5R_1}{5R_2} = \dfrac {5R_3}{5R_4}$$

$$\dfrac {R_1}{R_2} = \dfrac {R_3}{R_4}$$

Hence condition remains the same.
If galvanometer and cell are interchanged,

$$\dfrac {R_3}{R_4} = \dfrac {R_1}{R_2}$$
$$\dfrac {R_1}{R_2} = \dfrac {R_3}{R_4}$$

Hence condition remains the same.
If resistances are increased by 10 ohms, we get

$$\dfrac {R_1+10}{R_2+10} = \dfrac {R_3+10}{R_4+10}$$

$$( R_1+10)(R_4+10) = ( R_3+10)(R_2+10)$$

$$R_1R_4 +10(R_1+R_4) +100 = R_2R_3 +10(R_2+R_3) +100$$

This will not yield the balanced condition for bridge.
Question 9
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Calculate the value of current $$I_3$$ in the section of networks shown in figure.

A
13 A

B
7 A

C
10 A

D
3 A

Solution
Question 10
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To measure a small resistance $$\sim 10^{-5}\Omega$$, one should used

A
Wheatstone bridge

B
Postoffice box

C
Wein's bridge

D
Carrey Foster bridge

Solution

A Wheatstone bridge is an electrical circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit.
A Post Office Box can also be used to measure an unknown resistance.
Wien's bridge is used for precison measurement of capacitance in terms of resistance and frequency.
Carrey Foster bridge is used to measure very small resistances (of the order of micro ohm). Also, are used to measure very small resistance difference between large resistances.