Current Electricity Test

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Current Electricity Test - 53

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Question 1
1 / -0

What is the effective resistance between points P and Q?

A
$$\displaystyle \frac{r}{2}$$

B
$$\displaystyle \frac{r}{3}$$

C
$$\displaystyle \frac{2r}{3}$$

D
$$\displaystyle \frac{4r}{3}$$

Solution

$$\textbf{Hint: Resistances are connected in parallel}$$
Step 1: Find the the type of connection for each resistance
After moving from first 2r resistance, it has two paths one is with one more 2r resistance and other is pass through the conductor. So, these two 2r resistances are in parallel connection. Similarly, there is one more way that goes from above and connects directly to r which is other option with these 2r path. So, it is also in parallel connection. Hence, all resistance are parallel to each other.
Step2: Find the net resistance of the circuit
Assume that net resistance is R_{net}. So, from parallel resistance formula,
$$\dfrac{1}{R_{net}} = \dfrac{1}{2r} + \dfrac{1}{2r} + \dfrac{1}{r} = \dfrac{1}{r} (\dfrac{1}{2} + \dfrac{1}{2} + 1) = \dfrac{2}{r}$$

$$\Rightarrow R_{net} = \dfrac{r}{2}$$
$$Answer:$$
Hence, option A is the correct answer.
Question 2
1 / -0

What is the effective resistance between points P and Q?

A
$$\dfrac {r}{2}$$

B
$$\dfrac {r}{3}$$

C
$$\dfrac {2r}{3}$$

D
$$\dfrac {4r}{3}$$

Solution

In the given combination, point $$B$$ is equivalent to point $$P$$ as they are connected by a conducting wire. Similarly, point $$A$$ is equivalent to point $$Q$$ for the same reason.
So, the first resistance ($$2r$$) is connected between $$P$$ and $$A$$, i.e. $$P$$ and $$Q$$.
The second resistance ($$2r$$) is connected between $$A$$ and $$B$$, i.e. $$P$$ and $$Q$$.
the third resistance ($$r$$) is connected between $$B$$ and $$Q$$, i.e. $$P$$ and $$Q$$.
Therefore, $$\text{the three resistors are connected in parallel to each other.}$$

$$\therefore$$ Equivalent resistance between P and Q $$\dfrac{1}{R_{eq}} = \dfrac{1}{2r}+\dfrac{1}{2r}+\dfrac{1}{r}$$
$$\implies$$ $$R_{eq} = \dfrac{r}{2}$$
Question 3
1 / -0

A man has five resistors each of value $$\frac {1}{5}\Omega $$. What is the minimum resistance he can obtain by connecting them ?

A
$$\dfrac {1}{10}\Omega $$

B
$$\dfrac {1}{5}\Omega $$

C
$$\dfrac {1}{50}\Omega $$

D
$$\dfrac {1}{25}\Omega $$

Solution

Minimum resistance is obtained when all the resistance are connected in parallel.
Effective resistance in parallel connection $$1\dfrac{1}{R_{min}} = \dfrac{1}{R}+\dfrac{1}{R}+....$$ 5 terms
$$\implies$$ Minium resistance $$R_{min} = \dfrac{R}{5} =\dfrac{1/5}{5} =\dfrac{1}{25}\Omega$$
Question 4
1 / -0

The specific resistance of a rod of copper as compared to that of thin wire of copper is

A
more

B
less

C
same

D
depends upon the length and area of wire

Solution

Specific resistance is a property of material and its temperature and is independent of the dimension .

Hence, specific resistance of both copper rod and copper wire is same.

Answer-(C)
Question 5
1 / -0

A 4$$\Omega$$ resistance is bent through 180$$^o$$ at its midpoint and the two halves are twisted together. Then the resistance is

A
1$$\Omega$$

B
2$$\Omega$$

C
5$$\Omega$$

D
8$$\Omega$$

Solution

As the $$4\Omega$$ resistance is folded at its midpoint, thus the resistance of each segment which are now connected in parallel is $$2\Omega$$.
Equivalent resistance between A and B $$\dfrac{1}{R_{AB}} = \dfrac{1}{2}+\dfrac{1}{2}$$
$$\implies$$ $$R_{AB} = 1\Omega$$
Question 6
1 / -0

The resistance of carbon _______ with rise in the temperature.

A
Increases

B
Decreases

C
Remains the same

D
First increases then decreases

Solution

Since carbon is a semiconductor its temperature coefficient of resistance$$(\alpha)$$ is negative.

$$R=R_o[1+\alpha(T-T_o)]$$

Since $$\alpha$$ is negative, hence resistance $$R$$ decreases with increasing value of $$T$$.

Answer_(B)
Question 7
1 / -0

The specific resistance of a rod of copper as compared to that of thin wire of copper is

A
less

B
more

C
same

D
depends upon the length and area of crosssection of wire

Solution

Specific resistance also called resistivity is a property of material and not that of dimension.
Hence, specific resistance of copper rod and wire are same.

Answer-(C)
Question 8
1 / -0

The smallest resistance which can be obtained with ten 0.1 ohm resistors is

A
1 ohm

B
0.1 ohm

C
0.01 ohm

D
0.001 ohm

Solution

When the resistances are connected in parallel, the effective resistance becomes less than the smallest individual resistance. Thus smallest resistance is obtained when ten $$0.1\Omega$$ resistors are connected in parallel.
$$\therefore$$ Effective resistance $$R_p = \dfrac{r}{10} =\dfrac{0.1}{10} = 0.01\Omega$$
Question 9
1 / -0

We have n resistors each of resistance R. The ratio of the combination for maximum and minimum values is

A
n

B
$$n^2$$

C
$$n^3$$

D
$$\displaystyle \frac{1}{n}$$

Solution

The maximum resistance is obtained when all the $$n$$ resistors are connected in series.
Thus maximum resistance $$R_{max} = nR$$
Minimum resistance is obtained when all the $$n$$ resistors are connected parallel to each other.
$$\therefore$$ Minimum resistance $$R_{min} = \dfrac{R}{n}$$
$$\implies$$ $$\dfrac{R_{max}}{R_{min}} = \dfrac{nR}{\frac{R}{n}} = n^2$$
Question 10
1 / -0

A wire had a resistance of $$12\ \Omega$$. It is bent in the form of a circle. The effective resistance between two points on any diameter is:

A
$$3\ \Omega$$

B
$$6\ \Omega$$

C
$$12\ \Omega$$

D
$$24\ \Omega$$

Solution

Resistance of the wire, $$R=12\ \Omega$$
As the length of each segment is half of the complete wire, thus the resistance of each segment, $$R' =\dfrac{12}{2} =6\Omega$$
Effective resistance between P and Q ($$R'$$ in parallel), $$R_{eq}$$
$$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$$

$$\dfrac{1}{R_{eq}}=\dfrac{1}{R'}+\dfrac{1}{R'}$$

$$\dfrac{1}{R_{eq}}=\dfrac{2}{R'}$$

$$R_{eq}= \dfrac62$$

$$R_{eq} =3\ \Omega$$