Electricity Test

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Electricity Test - 56

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Question 1
1 / -0

Two resistance are connected in series as shown in diagram. What is the value of $$R$$?

A
$$4 \Omega$$

B
$$6 \Omega$$

C
$$5 \Omega$$

D
$$3 \Omega$$

Solution

Voltage across $$5\Omega$$ resistor is $$V=10V$$

Current, $$I=\dfrac{V}{R}=\dfrac{10}{5}$$

$$\implies I=2A$$

Since resistance $$5\Omega$$ and $$R$$ are in series and hence current through both resistance is same.
Hence current through $$R$$ is also $$2A$$.

Now for $$R$$, potential difference is $$V=6V$$ and $$I=2A$$.

Hence , $$R=\dfrac{V}{I}=\dfrac{6}{2}$$

$$\implies R=3\Omega$$

Answer-(D)
Question 2
1 / -0

In a circuit two or more cells of the same e.m.f. are connected in parallel in order to

A
Increase the P.D across a resistance in the circuit

B
Decrease the P.D across a resistance in the circuit

C
Facilitate drawing more current from the battery system

D
Change the e.m.f. across the system of batteries

Solution

$$E_{eff}=\dfrac{E_1}{r_1}+\dfrac{E_2}{r_2}$$ where r is internal resistancE
It will increase the current ,$$I=\dfrac{E_{eff}}{R}$$
Question 3
1 / -0

Five resistors each of $$1\ ohm$$ are connected in parallel. The resultant resistance of the combination is

A
$$5\Omega$$

B
$$1\Omega$$

C
$$0.2\Omega$$

D
$$2\Omega$$

Solution

All resistors are connected in parallel
$$\dfrac{1}{R_e}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}+\dfrac{1}{R_5}$$
$$\dfrac{1}{R_e}=\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1} = \dfrac{5}{1}$$
$$R_e=\dfrac{1}{5}=0.2\Omega$$
Question 4
1 / -0

Two resistances are connected in series as shown in the diagram. What is the current through $$R$$?

A
$$1 A$$

B
$$2 A$$

C
$$3 A$$

D
$$4 A$$

Solution

The voltage across $$5\Omega$$ resistor is $$V=10V$$

Current,
$$I=\dfrac{V}{R}$$; according to ohm' law

$$\Rightarrow I=\dfrac{10}{5}$$

$$\Rightarrow I=2A$$

Since resistance $$5\Omega$$ and $$R$$ are in series and hence current through both resistance is the same.
Hence current through $$R$$ is also $$2A$$.
Question 5
1 / -0

Resistance of $$2\ \Omega$$ and $$3\ \Omega$$ are connected in series. If the potential difference across the $$2\ \Omega$$ resistor is $$3\ V$$, the potential difference across $$3\Omega$$ is:

A
$$4.5\ V$$

B
$$9\ V$$

C
$$3\ V$$

D
$$2\ V$$

Solution

Given,
Resistors of resistances, $$R_1=2\ \Omega$$ and $$R_2=3\ \Omega$$ are connected in series.
Potential difference across $$2\ \Omega$$ resistor, $$V_1=3\ V$$
Potential difference across $$2\ \Omega$$ resistor, $$V_2$$

According to Ohm's law,
$$V=IR$$
$$\implies I=\dfrac VR$$

$$\implies I=\dfrac {V_1}{R_1}$$

$$I=\dfrac {3}{2}$$

$$I=1.5\ A$$

We know, current in a series circuit remains constant.
$$V_2=IR_2$$
$$V_2=1.5\times 3$$
$$V_2=4.5\ V$$
Question 6
1 / -0

If the length of a material changes 4 times and area changes 2 times. Then the resistivity of the material will be changing

A
Two times

B
Four times

C
Will not change

D
Continuously change

Solution

Resistivity is an intrinsic property of a material that signifies how efficiently a given material opposes the flow of electric current. It does not depend on the dimensions of a resistor. Resistivity is influenced by temperature - for most materials, the resistivity increases with temperature.
Hence, here in the question resistivity remains the same irrespective of the change in dimensions of the conductor.

The correct option is C.
Question 7
1 / -0

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of $$4/3$$ and $$2/3$$, then the ratio of the currents passing through the wire will be

A
$$3$$

B
$$\dfrac{1}{3}$$

C
$$\dfrac{8}{9}$$

D
$$2$$

Solution

$$R=\rho\dfrac{l}{A}$$

$$\implies R=\rho\dfrac{l}{\pi r^2}$$

$$\implies R\propto \dfrac{l}{r^2}$$

Hence, $$\dfrac{R'}{R}=\dfrac{l'}{r'^2}\dfrac{r^2}{l}$$

$$\dfrac{R'}{R}=\dfrac{l'}{l}\dfrac{r^2}{(r')^2}$$

$$\implies \dfrac{R'}{R}=\dfrac{4/3}{(2/3)^2}$$

$$\implies \dfrac{R'}{R}=3$$

Since, $$I\propto \dfrac{1}{R}$$

Hence, $$\dfrac{I'}{I}=\dfrac{1}{3}$$

Answer-(B)
Question 8
1 / -0

A resistor of $$6 k\Omega $$ with tolerance $$10$$% and another of $$4k\Omega $$ with tolerance $$10$$% are connected in series. The tolerance of combination is about

A
$$5$$%

B
$$10$$%

C
$$12$$%

D
$$15$$%

Solution

Let the resistances are $$R_1= 6\ k\Omega$$ and $$R_2=4\ k\Omega$$.
Tolerence in the resistances : $$\Delta R_1 = 600\Omega$$ $$\Delta R_2 = 400\Omega$$
Total resistance of the combination $$R_{eq} = R_1+R_2 =6+4 =10k\Omega$$
Total tolerance in the equivalent resistance: $$\Delta R_{eq} =\Delta R_1+\Delta R_2 =600+400 = 1k\Omega $$
Hence percentage tolerance of combination $$ = \dfrac{1k\Omega}{10k\Omega} \times 100 =10$$ %
Question 9
1 / -0

The resistance of the series combination of two resistances is S. When they are joined in parallel the total resistance is P. If S = nP, then the minimum possible value of n is

A
4

B
3

C
2

D
1

Solution

Let two resistor be $$R_{1} \ and \ R_{2}$$

In series combination the equivalent resistance is given by $$ S = R_{1} + R_{2}$$

In parallel combination the equivalent resistance is given by $$ P = \dfrac{R_{1} \times R_{2}}{R_{1} + R_{2}}$$

By relation S = nP
We can write it $$R_{1} + R_{2} = n\dfrac{R_{1} \times R_{2}}{R_{1} + R_{2}}$$
Rearranging the term $$(R_{1} + R_{2})^2 = n R_{1} \times R_{2}$$

$$R_1^2 + R_2^2+ 2R_1R_2=nR_1R_2$$

$$R_1^2 + R_2^2 + (2-n)R_1R_2 = 0$$

Dividing the whole equation with $$R_2^2$$, we will get

$$(\dfrac{R_1}{R_2})^2 + 1 + (2-n) \dfrac{R_1}{R_2} = 0$$

Put $$\dfrac{R_1}{R_2} = k$$

$$k^2 + (2-n)k +1 = 0$$

For real value of k, $$D \geq 0$$
$$(2-n)^2 - 4 \geq 0$$

For minimum value of n,
$$(2-n)^2 - 4 = 0$$
$$4 + n^2 - 4n-4=0$$
$$n^2 - 4n = 0$$
$$n(n-4)=0$$
$$\implies n=0 \ or \ 4$$

Since $$n$$ can't be 0,
So $$ n = 4$$
Question 10
1 / -0

The resistance of one ohm is very approximately equal to that of a column of mercury $$1.06 m$$ long and of uniform cross-section of one hundredth of $$1\ cm^2$$. Find the resistivity of mercury.

A
$$6.4 \times 10^{-7} \Omega m$$

B
$$7.4 \times 10^{-7} \Omega m$$

C
$$9.4 \times 10^{-7} \Omega m$$

D
$$2.4 \times 10^{-7} \Omega m$$

Solution

Resistance, $$(R) =$$ $$\rho\dfrac{l}{A}$$ where $$\rho$$ is resistivity, $$l$$ is the length and $$A$$ is the cross sectional area.
Now $$1$$ ohm is equal to very approximately that of a column of mercury $$1.06\ m$$ long and of uniform cross-section of one hundredth of $$1 cm^2$$=$$10^{-6} m^2$$Putting the values of $$R, l $$ and $$A$$
$$\rho=\dfrac{10^{-6}}{1.06}=9.43\times 10^{-7} $$

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Electricity Test - 56

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Electricity Test - 56

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Question 1

$$4 \Omega$$

$$6 \Omega$$

$$5 \Omega$$

$$3 \Omega$$

Solution

Question 2

Increase the P.D across a resistance in the circuit

Decrease the P.D across a resistance in the circuit

Facilitate drawing more current from the battery system

Change the e.m.f. across the system of batteries

Solution

Question 3

$$5\Omega$$

$$1\Omega$$

$$0.2\Omega$$

$$2\Omega$$

Solution

Question 4

$$1 A$$

$$2 A$$

$$3 A$$

$$4 A$$

Solution

Question 5

$$4.5\ V$$

$$9\ V$$

$$3\ V$$

$$2\ V$$

Solution

Question 6

Two times

Four times

Will not change

Continuously change

Solution

Question 7

$$3$$

$$\dfrac{1}{3}$$

$$\dfrac{8}{9}$$

$$2$$

Solution

Question 8

$$5$$%

$$10$$%

$$12$$%

$$15$$%

Solution

Question 9

4

3

2

1

Solution

Question 10

$$6.4 \times 10^{-7} \Omega m$$

$$7.4 \times 10^{-7} \Omega m$$

$$9.4 \times 10^{-7} \Omega m$$

$$2.4 \times 10^{-7} \Omega m$$

Solution

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