Current Electricity Test

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Current Electricity Test - 45

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Question 1
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The resistance of a carbon filament at $$0^oC$$ is 104$$\Omega$$. It is connected in series with an iron wire. The temperature coefficients of resistivity of carbon and iron are $$- 0.0003$$ and $$0.0052$$ per $$^oC$$ respectively. What must be the resistance of iron wire so that the combined resistance does not change with temperature?

A
1.8 $$\Omega$$

B
3.7 $$\Omega$$

C
6.0 $$\Omega$$

D
15 $$\Omega$$

Solution

Let, $$R_c$$ and $$R_i$$ be the resistances of copper and iron wires

respectively. $$\alpha_c$$ and $$\alpha_i$$ be the temperature coefficients of resistivity for copper and iron wires respectively.

The change in resistance with temperature is given by

$$R = R_0 + R_0 \alpha \Delta T$$

where, $$\Delta T $$ is the temperature difference. and $$R_0 = R_c + R_i$$ at $$0^\circ C$$

the combined resistance does not change with temperature hence, the decrease in resistance of copper wire is equal to the increase in resistance of iron wire.

Therefore,

$$- R_c \alpha_c \Delta T = R_i \alpha_i \Delta T$$

$$- 104 \times (-0.0003) = R_i \times 0.0052$$

Negative sign indicates the decrement in resistance.

$$0.0312 = R_i \times 0.0052$$

$$R_i = 6 \Omega$$
Question 2
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A battery of emf $$E_0 = 12 \space V$$ is connected across a $$4m$$ long uniform wire having resistance $$4 \space\Omega/m$$. The cells of small emfs $$\epsilon_1 = 2V$$ and $$\epsilon_2 = 4V$$ having internal resistance $$2\space\Omega$$ and $$6\space\Omega$$ respectively, are connected as shown in the figure. If galvanometer shows no deflection at the point $$N$$, the distance of point $$N$$ from the point $$A$$ is equal to

A
$$\displaystyle\frac{1}{6}m$$

B
$$\displaystyle\frac{1}{3}m$$

C
$$25 \space \text{cm}$$

D
$$50 \space \text{m}$$

Solution

$$\text{Voltage drop on the balancing length of the wire} \Rightarrow i{\rho}x$$

$$\text{Here}\ i= \text{Current flowing in primary circuit}$$

$$\rho=\text{Resistance per unit length of the wire}$$

$$x = \text{Balancing length of wire AB when deflection in galvanometer is zero}$$

$${\dfrac {1}{R_{eq.}}} \Rightarrow ({\dfrac {1}{2}}+{\dfrac {1}{6}})={\dfrac {4}{6}}$$

$${ E }_{ eq. }=\dfrac { \left( \dfrac { { E }_{ 1 } }{ { r }_{ 1 } } -\dfrac { { E }_{ 2 } }{ { r }_{ 2 } } \right) }{ \dfrac {1}{ R }_{ eq. } } =\left( \dfrac { \dfrac { 2 }{ 2 } -\dfrac { 4 }{ 6 } }{ \dfrac { 4 }{ 6 } } \right) =\dfrac { 1 }{ 2 }\ \text{volt}$$

$$\text{when deflection is zero} \Rightarrow\ E\ \text{of secondary circuit}=\text{voltage drop across the wire AN}$$

$${\dfrac {1}{2}}={\dfrac {12}{8+16}}{\times}{4}{\times}{x}$$

$${\Rightarrow} x={\dfrac {1}{4}}\ \text{m}.$$$${\Rightarrow} x=25\ \text{cm}$$
Question 3
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In Fig. A B is 300 cm long wire having resistance $$10\Omega$$ per meter. Rheostate is set at $$20\Omega$$. The balance point will be attained at

A
$$1.0 m$$

B
$$1.25 m$$

C
$$1.5m$$

D
$$\text{cannot be determined}$$

Solution

$$V_{AB}=\dfrac {6\times 30}{50}=3.6V$$. Terminal voltage of cell $$=\dfrac {2\times 1.5}{2}=1.5 V$$

Using $$V=kl\Rightarrow 1.5=\dfrac {3.6}{300}l$$ or $$l=125 cm$$
Question 4
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Which of the V-I graph obeys Ohm's law?

A

B

C

D

Solution

Ohm's law is: The current flowing through a conductor is directly proportional to the voltage applied across the conductor
means $$ I\propto V $$ or $$I=\dfrac { V }{ R } $$.....eq.( 1 )
from equation 1 we can see , the slope of $$ I - V$$ graph is $$ 1/R$$ which is a constant. graphically it can be shown as shown in figure of option B. Hence option B is correct.
Question 5
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In the circuit shown in the fig the heat produced in resistance $$R_1$$ can be measured by

A
connecting both voltmeter and an ammeter in parallel to $$R_1$$.

B
connecting ammeter in series with $$R_1$$ and voltmeter in parallel to both $$R_1$$ and $$R_2$$.

C
connecting ammeter in parallel to $$R_1$$ and voltmeter in series with $$R_1$$.

D
connecting voltmeter in parallel to $$R_1$$ and ammeter in series with $$R_1$$.

Solution

Voltmeter have nearly infinite resistance and ammeter have negligible resistance. Hence, by connecting voltmeter in parallel to the resistance $$($$ here $$R_1 )$$ the exact potential can be measured and by connecting ammeter in series with the resistance $$($$ here $$R_1 )$$ the total current can be measured through the resistance.
Question 6
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Two cells of same emf are connected in series. Their internal resistances are $$r_1$$ and $$r_2$$ respectively and $$r_1 > r_2$$. When this combination is connected to an external resistance R then the potential difference between the terminals of first cell becomes zero. In this condition the value of R will bw

A
$$\frac {r_1-r_2}{2}$$

B
$$\frac {r_1+r_2}{2}$$

C
$$r_1-r_2$$

D
$$r_1+r_2$$

Solution

Let emf of each cell is $$E$$.
As they are connected in series so current in circuit is $$I=\dfrac{E+E}{r_1+r_2+R}=\dfrac{2E}{r_1+r_2+R}$$

Potential across terminal of first cell is $$V_1=E-Ir_1=E-\dfrac{2Er_1}{r_1+r_2+R}$$

As $$V_1=0 \Rightarrow E-\dfrac{2Er_1}{r_1+r_2+R}=0$$

$$r_1+r_2+R-2r_1=0$$

$$R=r_1-r_2$$
Question 7
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Three voltmeters, all having different resistances, are joined as shown. When some potential difference is applied across A and B, their readings are $$V_1, V_2, V_3$$

A
$$V_1=V_2$$

B
$$V_1\neq V_2$$

C
$$V_1+V_2=V_3$$

D
$$V_1+V_2 > V_3$$

Solution

Here, three voltmeters are joined as shown in given figure.
We know,
$$V = IR$$
Also, all voltmeters have different resistances, then for same current $$V \propto R$$
Hence, $$V_1 \propto R_1$$
similarly, $$V_2 \propto R_2$$ and $$V_3 \propto R_3$$
$$\Rightarrow V_1 \neq V_2$$
When some potential difference is applied the current will divide into two branches i.e. say $$I_1$$ in one having $$V_1$$ and $$V_2$$ in series and say $$I_2$$ in other having $$V_3$$ parallel to both of them.
Now, applying KVL to the loop
$$I_1R_1 +I_1R_2- I_2V_3 = 0$$
$$V_1 + V_2 - V_3 = 0$$
$$V_1 + V_2 = V_3$$
Question 8
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The length of a potentiometer wire is 10m. The distance between the null points on its wire corresponding to two cells comes out to be 60 cm. If the difference of emf's of the cells is 0.4 volt then the potential gradient on potentiometer wire will be

A
0.67 V/m

B
0.5 V/m

C
2.5 V/m

D
0 V/m

Solution

Let, $$E_1$$ and $$E_2$$ be the emf's of two cells, $$l_1$$ and $$l_2$$ be the balancing lengths obtained for two cells respectively.
Given that,

$$E_1 - E_2 = 0.4 V$$
and

$$l_1 - l_2 = 60 cm = 0.6 m$$

Now, the potential is the fall of potential per unit length of potentiometer wire,
hence

$$k = \dfrac{E_1 - E_2}{l_1 - l_2}$$

$$k = \dfrac{0.4}{0.6}$$

$$k = 0.67 Vm^{-1}$$
Question 9
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On comparing the emf's $$E_1$$ and $$E_2(E_1 > E_2)$$ of two cells by a potentiometer, the balancing lengths come out to be $$l_1$$ and $$l_2$$ respectively, then

A
$$l_1 < l_2$$

B
$$l_1 - l_2$$

C
$$l_1 > l_2$$

D
none of the above

Solution

For potentiometer when null deflection is obtained in galvanometer, the emf of battery is equal to the potential difference across the balancing length hence,
$$E = k l$$
where, E is the emf of battery, k is potential gradient and l is the balancing length.
Therefore,
$$E_1 = k l_1 \Rightarrow E_1 \propto l_1$$
$$E_2 = k l_2 \Rightarrow E_2 \propto l_2$$
$$ if \ E_1 > E_2 \Rightarrow l_1 > l_2$$
Question 10
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The potentiometer is more appropriate for measuring potential difference than a voltmeter because

A
the resistance of voltmeter is high

B
the sensitivity of a potentiometer is higher than that of voltmeter

C
the resistance of potentiometer wire is very low

D
the potentiometer does not draw any current from the unknown source of emf

Solution

A voltmeter, when connected across two points, develops little current through itself due to the potential difference between the two points. This changes the actual circuit and hence an accurate reading is not obtained.

A potentiometer reading is based on finding an equipotential point such that no current flows. The circuit is not affected in such case and a more appropriate reading is obtained.