Current Electricity Test

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Current Electricity Test - 73

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Question 1
1 / -0

When battery and galvanometer are interchanged in the case of Wheatstone bridge, then?

A
If the bride was in adjustment before interchange, it will not be in adjustment after interchange

B
If the bridge was in adjustment before interchange, it will be in adjustment after interchange

C
If the bridge was in adjustment before interchange, it may or may not be in adjustment after interchange

D
All the above

Solution
Question 2
1 / -0

In the circuit shown in the figure, the voltage across $$15\Omega$$ resistor is $$30$$V having the polarity as indicated. What is the value of R?

A
$$10\Omega$$

B
$$35\Omega$$

C
$$17.5\Omega$$

D
$$37.5\Omega$$

Solution
Question 3
1 / -0

When the switch is closed, the initial current through the $$1\Omega$$ resistor is?

A
$$12$$A

B
$$4$$A

C
$$3$$A

D
$$\dfrac{10}{7}$$A

Solution
Question 4
1 / -0

Two cells of e.m./f.s $$ E_{1} $$ and $$ E_{2} $$ $$ (E_{1}> E_{2}) $$ are connected as shown Fig. 6.45
When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A and C, the balancing length is 100 cm. The ratio $$ E_{1}/E_{2} $$ is

A
3:1

B
1:3

C
2:3

D
3:2

Solution
Question 5
1 / -0

Directions For Questions

A network of resistance is constructed with $$R_{1}$$ and $$R_{2}$$ as shown in Fig $$5.243$$. The potential at the points $$1,2,3 ... N$$ are $$V_{1}, V_{2}, V_{3},.... V_{n}$$, respectively, each having a potential $$k$$ times smaller than the previous one.

...view full instructions

The ratio $$R_{2}/R_{3}$$ is

A
$$\dfrac{(k-1)^{2}}{k}$$

B
$$k^{2}-\dfrac{1}{k}$$

C
$$\dfrac{k}{k-1}$$

D
$$k-\dfrac{1}{k^{2}}$$

Solution

Current in $$R_1$$ and $$R_2$$ will be the same:

$$\dfrac {V_{n-1}-V_n}{R_1}=\dfrac {V_n}{R_3}$$

$$\dfrac {V_{n-1}-\dfrac {V_{n-1}}{k}}{R_1}=\dfrac {V_{n-1}}{k\ R_3}$$

$$\Rightarrow \ R_1 =R_3 (k-1)$$

Put the value of $$R_1$$ in the above expression, we get:
$$\dfrac {R_2}{R_3}=\dfrac {k}{k-1}$$
Question 6
1 / -0

In a meter bridge experiment, null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If $$ X< Y $$, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4 X against Y

A
50 cm

B
80 cm

C
40 cm

D
70 cm

Solution

Case: 1
$$\dfrac{X}{Y}=\dfrac{20}{80}=\dfrac{1}{4}$$
or $$Y=4 X$$
Case: 2
When X become 4X
$$\dfrac{4X}{Y}=\dfrac{l}{100-l}$$

or $$\dfrac{4X}{4X}=\dfrac{l}{100-l}$$

or $$l=50$$ cm
Question 7
1 / -0

When temperature of the conductor is increased, then the collision frequency between current carries ( electrons ) and atoms increases. This results in

A
decreased in the resistance of the conductor

B
increase in the resistance of the conductor

C
no change in the resistance

D
decrease or increase in the resistance depending upon the type of the conductor

Solution

When temperature increases,collision frequency increases, it means more number of collisions per time. Time will decrease the relaxation time. But $$R \propto \dfrac{1}{\tau }$$, so $$R$$ increases.
Question 8
1 / -0

The equivalent resistance between $$A$$ and $$B$$ in the network in figure is

A
$$\dfrac{4}{3}\Omega$$

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

C
$$3\Omega$$

D
$$2\Omega$$

Solution

Above circuit can be reduced as shown ,
Now,
The equivalent of the network is given in figure.
The equivalent of the above network is a parallel combination of $$3\Omega, 4\Omega $$ and $$6\Omega$$, i.e,
$$\dfrac{1}{R}=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{6}\Rightarrow R=\dfrac{4}{3}\Omega$$
Question 9
1 / -0

Directions For Questions

A network of resistance is constructed with $$R_{1}$$ and $$R_{2}$$ as shown in Fig $$5.243$$. The potential at the points $$1,2,3 ... N$$ are $$V_{1}, V_{2}, V_{3},.... V_{n}$$, respectively, each having a potential $$k$$ times smaller than the previous one.

...view full instructions

The ratio $$R_{1}/R_{2}$$ is

A
$$k^{2}-\dfrac{1}{k}$$

B
$$\dfrac{k}{k-1}$$

C
$$k-\dfrac{1}{k^{2}}$$

D
$$\dfrac{(k-1)^{2}}{k}$$

Solution

Given ,

$$V_1 =\dfrac {V_0}{k}, V_2 =\dfrac {V_1}{k}, V_3 =\dfrac {V_2}{k},$$

$$\ I=I_1 +I_2$$

$$\dfrac {V_0-V_0 /k}{R_1}=\dfrac {V_0/ k-V_0 /k^2}{R_1}+\dfrac {V_0 /k}{R_2}$$

$$\dfrac {R_1}{R_2}=\dfrac {(k-1)^2}{k}$$
Question 10
1 / -0

In Fig. when an ideal voltmeter is connected across 4000$$\Omega $$ resistance, it reads 30 V. If the voltmeter is connected across 3000$$\Omega $$ resistance, it will read

A
$$20 V$$

B
$$22.5 V$$

C
$$35 V$$

D
$$40 V$$

Solution

Let $$I$$ be the current in the circuit then, $$I\times R=V$$
$$\Rightarrow I\times 4000=30 V$$
$$\Rightarrow I=\dfrac{30}{4000}$$
Reading of voltmeter across $$3000\Omega$$
$$V=I\times 3000=\dfrac{30}{4000}\times 3000=22.5 V$$