[3 Mark Questions]

Question 13 Marks

State three factors on which the heat produced by an electric current depends. How does it depend on these factors?

Answer

Heat produced by an electric current depends on the following factors:

Heat produced is directly proportional to square of current.
Heat produced is directly proportional to resistance.
Heat produced is directly proportional to the time for which current flows.

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Question 23 Marks

Consider the circuit given below where A, B and C are three identical light bulbs of constant resistance.

List the bulbs in order of increasing brightness.
If C burns out, what will be the brightness of A now compared with before?
If B burns out instead, what will be the brightness of A and C compared with before?

Answer

C will be the brightest. Voltage will be distributed equally between A and B, so they will have equal brightness but lesser than that of C.
A gets the same voltage as before, so its brightness remains the same.
If B burns put, A will also stop glowing because it is connected in series with B. However, brightness of C remains the same.

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Question 33 Marks

The electrical resistivities of four materials P, Q, R and S are given below:
$\begin{matrix}\text{P}&6.84\times10^{-8}\Omega\text{m}\\\text{Q}&1.70\times10^{-8}\Omega\text{m}\\\text{R}&1.0\times10^{15}\Omega\text{m}\\\text{S}&11.0\times10^{-7}\Omega\text{m}\end{matrix}$
Which material will you use for making:

Heating element of electric iron.
Connecting wires of electric iron.
Covering of connecting wires?

Give reason for your choice in each case.

Answer

S; because it has high resistivity of $\frac{11}{10000000}\text{ohm\ m}$ (it is actually nichrome).
Q; because it has very low resistivity of $\frac{1.7}{100000000}\text{ohm\ m}$ (it is actually copper).
R; because it has very very high resistivity of$1.0\times100000000000000\ \text{ohm\ m}$ (it is actually rubber).

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Question 43 Marks

Calculate the resistance of a copper wire 1.0km long and 0.50mm diameter if the resistivity of copper is $1.7\times10\Omega\ \text{m}.$

Answer

I = 1km = 1000m
$\text{r}=\frac{\text{d}}{2}=\frac{0.5}{2}\text{mm}=0.25\text{mm}=0.25\times10^{-3}\text{m}$
$\rho=1.7\times10^{-8}\Omega\text{m}$
$\text{R}=\rho\frac{\text{I}}{\text{A}}=\rho\frac{\text{l}}{\pi\text{r}^2}$
$\text{R}=1.7\times10^{-8}\times\frac{1000}{3.14\times(0.25\times10^{-3})^2}=86.6\Omega$

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Question 53 Marks

A $4 Ω$ coil and a $2 Ω$ coil are connected in parallel. What is their combined resistance? A total current of 3A passes through the coils. What current passes through the $2 Ω$ coil?

Answer

$4\Omega$ and $2\Omega$ coil are connected in parallel.
Combined resistance is R
$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$
$\text{R}=\frac{4}{3}\Omega$
Total current $\text{I}=\frac{\text{V}}{\text{R}}=3\Omega$
$\frac{\text{V}}{\frac{4}{3}}=3$
$\text{V}=3\times\frac{4}{3}=4\text{V}$
Current through $2\Omega$ coil $=\frac{\text{V}}{2}=\frac{4}{2}=2\text{A}$

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Question 63 Marks

When a current of 4.0A passes through a certain resistor for 10 minutes, 2.88 × 10J of heat are produced. Calculate:

The power of the resistor.
The voltage across the resistor.

Answer

Given: $I =4 amp , t =10 min=10 \times 60=600 sec , H =2.88 \times 10^4 J$
a. We have
$K=I^2 R T$
$28800=4^2 \times R \times 600$
$R=3 ohm$
We know that
$P=I^2 \times R$
$=4^2 \times 3$
$P=48 W$
b. $V=$ ?
We know that
$V=I R$
$V=4 \times 3$
$V=12 V$

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Question 73 Marks

A wire is 1.0m long, 0.2mm in diameter and has a resistance of $10Ω.$ Calculate the resistivity of its material?

Answer

l = 1m
$\text{r}=\frac{\text{d}}{2}=\frac{0.2}{2}\text{mm}=0.1\text{mm}=0.0001\text{m}$
R = 10 ohm
We know that,
$\text{R}=\text{P}\frac{\text{I}}{\text{A}}$
$\text{P}=\frac{\text{RA}}{\text{I}}$
$=\frac{10\times\pi\times(0.0001)^2}{1}$
$=31.4\times10^{-8}\Omega\text{m}$

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Question 83 Marks

Two resistances when connected in parallel give resultant value of 2 ohm; when connected in series the value becomes 9 ohm. Calculate the value of each resistance.

Answer

Two resistance when connected in series, resultant value is 9 ohms.
Two resistance when connected in parallel, resultant values is 2 ohms.
Let the two resistance be $R _1$ and $R _2$
If connected in series, then
$9=R_1+R_2$
$R_1=9-R_2$
If connected in parallel, then
$\frac{1}{2}=\frac{1}{R_1}+\frac{1}{R_2}$
From aboves equations we get that
$\frac{1}{2}=\frac{\left(R_1+R_2\right)}{R_1 R_2}$
$\frac{1}{2}=\frac{9}{\left(9-R_2\right) R_2}$
$9 R_2-R_2{ }^2=18$
$R_2{ }^2-9 R_2+18=0$
$\left(R_2-6\right)\left(R_2-3\right)=0$
$R_2=6,3$
So if $R_2 6$ ohms, then $R_1=9-6=3$ ohms.
If $R_2=3$ ohms, then $R_1=9-3=6$ ohms.

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Question 93 Marks

A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination of resistances in half a minute.

Answer

Given: R1 $=40$ ohms, R2 $=60$ ohms (in series), $V=220 V, t =30 sec$ we know that
Total resistance, $R=40+60=100$ ohms
By ohm's law
$V=I$
$I=\frac{V}{R}$
$I=\frac{220}{100}=2.2 amp$
Putting the value of $I, R$ and $t$ in eq. $H=I^2 R T$
$H=2.2^2 \times 100 \times 30$
$H=14520 J$

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Question 103 Marks

How much energy is consumed when a current of 5 amperes flows through the filament (or element) of a heater having resistance of 100 ohms for two hours? Express it in joules.

Answer

I = 5 amp, R = 100 ohms, t = 2h
We know that
Electric energy consumed = P x t = I * I * Rt
= 25 × 100 × 2
= 5000Wh
= 5kwh
We know that 1kwh = 3.6 × 106J
Therefore, 5kwh = 5 × 3.6 × 106J = 18 × 106J.

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Question 113 Marks

A p.d. of 6V is applied to two resistors of $3 Ω$ and $6 Ω$ connected in parallel. Calculate:
The combined resistance.
The current flowing in the main circuit.
The current flowing in the $3 Ω$ resistor.

Answer

V = 6V,
$R _1=3$ ohm, $R _2=6$ ohm (in parallel)
Combined resistance, $\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$
$\frac{1}{\text{R}}=\frac{1}{3}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$
R = 2 ohm
Current flowing in the main circuit, $\text{I}=\frac{\text{V}}{\text{R}}=\frac{6}{2}=3\text{A}$
Current flowing in 3 ohm resistor $=\frac{\text{V}}{\text{R}_1}=\frac{6}{3}=2\text{A}$

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Question 123 Marks

Calculate the work done in moving a charge of 4 coulombs from a point at 220 volts to another point at 230 volts.

Answer

solution $V _1=220 V, V _2=230 V$, Charge moved $=4 C$
Thus, the potential difference $= V _2- V _1=230-220=10$.
We know that,
Work done $=$ Potential difference $\times$ Charge moved
$=10 \times 4$
Work done $=40$ joules.

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Question 133 Marks

Distinguish between resistance and resistivity.

Answer

a. Resistance is the property of the conductor, while resistivity is the property of the material of the conductor.
b. Resistance of a conductor is the opposition to the flow of electric current through it. resistivity of a substance is the opposition to the flow of electric current by a rod of that substance which is 1 m long and $1 m^2$ in cross section.
c. Resistivity of a substances depends on the nature of the substance and temperature.
d. Resistivity of a substance on the nature of the substance and its temper. it does not depend the length or thickness of the conductor.

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Question 143 Marks

Show how you would connect two 4 ohm resistors to produce a combined resistance of:

2 ohms.
8 ohms.

Answer

By connecting in parallel: Since equivalent resistance will be
$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$
Therefore. R = 2 ohm
By connecting in series: since equilaent resistance will be R = 4 ohm + 4 ohm = 8 ohm.

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Question 153 Marks

Calculate the resistance of an aluminium cable of length 10km and diameter 2.0mm if the resistivity of aluminium is $2.7\times10 Ω\ \text{m}.$

Answer

I = 10km = 10000m
d = 2mm
$r =1 mm=10^{-3} m$
$\rho=2.7\times10^{-8}\Omega\ \text{m}$
$\text{R}=\rho\frac{\text{I}}{\text{A}}$
$=2.7\times10^{-8}\times\frac{10000}{3.14\times(10^{-3})^2}$
$=0.859\times10^2\Omega$
$=86\Omega$

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Question 163 Marks

You are given one hundred $1 Ω $ resister. What is the smallest and largest resistance you can make in a circuit using these?

Answer

Given: $n =100, R =1 ohm$
For obtaining the smallest resistance, these resistance are connected in parallel:
Equivalent resistance $=\frac{1}{1}+\frac{1}{1}+\frac{1}{1} \ldots . .100$ times $=\frac{100}{1}$
$R_{eq}=\frac{1}{100}=0.01 ohm$
For obtaining the largest resistance, these are connected in series:
Equivalent reisistance $=1+1+1$......... 100 Times = 100
$R_{\text {eq }}=100$ ohm

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Question 173 Marks

Calculate the cost of operating a heater of 500W for 20hours at the rate of? ₹3.90 per unit.

Answer

Given P = 500W = 0.5KW, t = 20hr
We know that
Energy consumed = P × t = 0.5 × 20
= 10KWh
Total cost = 10 × cost per unit
cost per unit = Rs. 3.9 per unit
Therefore, total cost = 10 × 3.9 = Rs. 39

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Question 183 Marks

Give three reasons why different electrical appliances in a domestic circuit are connected in parallel.

Answer

Different electrical appliances in a domestic circuit are connected in parallel because of the following advantages:

If one electrical appliance stops working due to some defect, then all other appliances keep working properly.
Each electrical appliance has its own switch due to which it can be turned on or turned off independently, without affecting other appliances.
Each electrical appliance gets the same voltage as that of the power supply line.

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Question 193 Marks

Two exactly similar electric lamps are arranged (i) in parallel, and (ii) in series. If the parallel and series combination of lamps are connected to 220V supply line one by one, what will be the ratio of electric power consumed by them?

Answer

Let resistance of each lamp = R ohms.
Case1: Parellel connection
Resultant resistance $=\frac{1}{\frac{1}{\text{R}}+\frac{1}{\text{R}}}=\frac{\text{R}}{2}$
Electric power consumed $\text{P}_1=\frac{\text{V}_2}{\text{R}}=\frac{220^2}{\frac{\text{R}}{2}}=\frac{96800}{\text{R}}$
Case2: Series connection
Resultant resistance = R + R = 2R
Electric Power consumed $\text{P}_2=\frac{\text{V}^2}{2\text{R}}=\frac{24200}{\text{R}}$
$\therefore\frac{\text{P}_1}{\text{P}_2}=\frac{\frac{96800}{\text{R}}}{\frac{24200}{\text{R}}}=\frac{4}{1}$

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Question 203 Marks

A piece of wire of resistance $20 Ω$ is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.

Answer

We know that the resistance of a conductor is given by:
$\text{R}=\rho\frac{1}{\text{A}}=20\Omega$
Where $\rho=\text{resistivity}$
l = length of the conductor
A = area of cross-section of the conductor
Now the length is increased to twice the original length. Then let the new resistance be denoted by R'.
$\text{R}'=\rho\frac{2\text{I}}{\frac{\text{A}}{2}}=4\rho\frac{1}{\text{A}}$
$\text{R}'=4\text{R}=4\times20=80\Omega$
Thus, the new resistance will become four times.

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Question 213 Marks

The electrical resistivities of five substances A, B, C, D and E are given below:
$\begin{matrix}\text{B}&110\times10^{-8}\Omega\text{ m}\\\text{C}&2.60\times10^{-8}\Omega\text{ m}\\\text{D}&10.0\times10^{-8}\Omega\text{ m}\\\text{E}&1.70\times10^{-8}\Omega\text{ m}\end{matrix}$

Answer

E is best conductor of electricity due to its least electrical resistivity.
C, because its resistivity is lesser than that of A.
B, because it has the highest electrical resistivity.
C and E, because of their low electrical resistivities.

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Question 223 Marks

What is the resistance between A and B in the figure given below?

Answer

The three resistance of 20 ohm, 10 ohm and 20 ohm on the extreme right side are in series.
So, the resultant of these three resistance = 20 + 20 + 10 = 50 ohms.
This 50 ohms is in parallel with 30 ohms. so resultant of these two will be
$\frac{1}{\text{R}}=\frac{1}{30}+\frac{1}{50}$
$\frac{1}{\text{R}}=\frac{80}{1500}$
R = 18.75 ohms
Now, the resistance 10 ohms, 18.75 ohms and 10 ohms are in series.
Therefore, resultant resistance = 18.75 + 10 + 10 = 38.75 ohms.

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Question 233 Marks

What is Ohm’s law? Explain how it is used to define the unit of resistance.

Answer

Ohm’s law gives a relationship between current (I) and potential difference (V). According to ohm’s law: At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends. If I is the current flowing through a conductor and V is the p.d. across its ends, then according
to the ohm’s law:
$\text{I}\propto\text{V}$
or, $\text{I}\propto\text{V}$
or, V = RI
or, $\text{R}=\frac{\text{V}}{\text{I}}$
where, R is a constant called ''resistance'' of the conductor.
The unit of resistance is ohm.
If V = 1 volt and I = 1 amp, then $\text{R}=\frac{1}{1}=1\text{ohm}$
Thus, 1 ohm is the resistance of a conductor such that when a potential difference of 1 volt is applied to its ends, a current of 1 amp flows through it.

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Question 243 Marks

For the circuit shown in the diagram below:

What is the value of :
Current through $6 Ω$ resistor?
Potential difference across $12 Ω$ resistor?

Answer

As per the circuit
V = 4V
Total resistance in line $1=R_1=6+3=3$ ohm
Total resistance in line $2=R_2=12+3=15$ ohm
Current through $6\Omega$ resistor = current through line $1=\frac{\text{=V}}{\text{R}_1}=\frac{4}{9}=0.44\Omega$
p.d. across line 2 is 4V
current through line $2=\frac{\text{V}}{\text{R}_2}=\frac{4}{15}\Omega$
p.d. across $12\Omega$ resistor $=\frac{4}{15}\times12=3.2\text{V}$

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Question 253 Marks

In which of the following cases more electrical energy is consumed per hour?

A current of 1 ampere passed through a resistance of 300 ohms.
A current of 2 amperes passed through a resistance of 100 ohms.

Answer

$R =300 \text { ohm, } I=1 A, t=1 h$
$P =I^2 R=1^2 \times 300=300 W$
$E =P \times t=300 \times 1=300 Wh$
ii. $R=100$ ohm, $I=2 A, t =1 h$
$P=I^2 R=2^2 \times 100=400 W $
$E=P \times t=400 \times 1=400 W h$
Hence, in case (ii), the electrical energy consumed per hour is more.
Hence, in case (ii), the electrical energy consumed per hour is more.

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Question 263 Marks

An electric bulb is rated as 10W, 220V. How many of these bulbs can be connected in parallel across the two wires of 220V supply line if the maximum current which can be drawn is 5A?

Answer

P = 10W, V = 220V, I = 5A
We know that
P = VI
= 220 X 5
P = 1100W
Power of one bulb = 10W
Total no. of bulbs that can be connected $=\frac{1100}{10}=110$

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Question 273 Marks

Give three reasons why different electrical appliances in a domestic circuit are connected in parallel.

Answer

Different electrical appliances in a domestic circuit are connected in parallel because of the following advantages:

If one electrical appliance stops working due to some defect, then all other appliances keep working properly.
Each electrical appliance has its own switch due to which it can be turned on or turned off independently, without affecting other appliances.
Each electrical appliance gets the same voltage as that of the power supply line.

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Question 283 Marks

In the circuit shown below, the voltmeter reads 10V.

What is the combined resistance?
What current flows?
What is the p.d. across $ 2 Ω$ resistor?
What is the p.d. across $3 Ω$ resistor?

Answer

I = 6V
$\text{R}_1=2\Omega,\text{R}_2=3\Omega$
Combined resistance, $R _{\text {tot }}=2+3=5$
$I=\frac{V}{R_{\text {tot }}}=\frac{10}{5}=2 A$
$\text { p.d. across } 2 \Omega \text { resistor }=I \times R_1=2 \times 2=4 V \text {. }$
$\text { p.d. across } 3 \Omega \text { resistor }=I \times R_2=2 \times 3=6 V \text {. }$

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Question 293 Marks

How does the resistance of a wire change when:

Its length is tripled?
Its diameter is tripled?
Its material is changed to one whose resistivity is three times?

Answer

$\text{R}=\rho\frac{\text{I}}{\text{A}}$

I → 3I

$\text{R}'=\rho\frac{3\text{I}}{\text{A}}=3\text{R}$

Resistance gets tripled.

d → 3d

$\text{R}=\rho\frac{\text{I}}{\text{A}}=\text{R}=\rho\frac{\text{I}}{\pi\text{r}^2}=\rho\frac{\text{I}}{\pi\big(\frac{\text{d}}{2}\big)^2}$

$\text{R}'=\rho\frac{\text{I}}{\pi\big(\frac{3\text{d}}{2}\big)^2}=\frac{1}{9}\rho\frac{\text{I}}{\pi\big(\frac{\text{d}}{2}\big)^2}=\frac{\text{R}}9{}$

Resistance becomes $\frac{1}{9}\text{th}.$

$\text{R}=\rho\frac{\text{I}}{\text{A}}$

$\rho\rightarrow3\rho$

$\text{R}'=3\rho\frac{\text{I}}{\text{A}}=3\text{R}$

Resistance becomes 3 times.

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Question 303 Marks

Three 2V cells are connected in series and used as a battery in a circuit.

What is the p.d. at the terminals of the battery?
How many joules of electrical energy does 1C gain on passing through

One cell.
All three cells?

Answer

If three cells of 2 volt each are connected in series to make a battery, then the total potential difference between terminals of the battery will be 6V.

Given: p.d. = 2V, Charge moved = 1C

We know that

Work done = p.d × charge moved

= 2 × 1

Work done = 2 joules

Given: p.d = 6V, Charge moved = 1C

Work done = p.d × charge moved

= 6 × 1

Work done = 6 joule.

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Question 313 Marks

What will be the resistance of a metal wire of length 2 metres and area of cross-section 1.55 × 10m , if the resistivity of the metal be 2.8 × 10 m?

Answer

I = 2m
$A=1.55 \times 10^{-6} m^2$
$P=2.8 \times 10^{-8} m$
$\text{R}=\text{P}\frac{\text{I}}{\text{A}}$
$=2.8\times10^{-8}\times\frac{2}{1.55\times10^{-6}}$
$0.036\Omega$

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Question 323 Marks

Calculate the area of cross-section of a wire if its length is 1.0m, its resistance is $23 Ω$ and the resistivity of the material of the wire is $1.84\times10 Ω\ \text{m}.$

Answer

I = 1.0m
R = 23 ohm
$\rho=1.84\times10^{-6}\ \text{ohm-meter}$
We have
$\text{R}=\rho\frac{\text{I}}{\text{A}}$
$23=1.84\times10^{-6}\times\frac{1}{\text{A}}$
$\text{A}=\frac{1.84\times10^{-6}}{23}$
$=0.08\times10^{-6}\text{m}^2$
$=8\times10^{-8}\text{m}^2$

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Question 333 Marks

A resistor of 8 ohms is connected in parallel with another resistor X. The resultant resistance of the combination is 4.8 ohms. What is the value of the resistor X?

Answer

Given:
A resistor of 8ohm is connected in parallel with a resistor of x.
And resultant is 4.8.
Then X = ?
We know that for parallel case
$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{X}}$
$\frac{1}{4.8}=\frac{1.8}{}+\frac{1}{\text{x}}$
$\frac{1}{4.8}-\frac{1}{8}=\frac{1}{\text{x}}$
After solving we get that
X = 12 ohms.

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Question 343 Marks

The electrical resistivities of four materials A, B, C and D are given below:
$\text{A}\ -110\times10^{-8}\Omega\text{ m}$
$\text{B}-\ 1.0\times10^{10}\Omega\text{ m}$
$\text{C}-\ 10.0\times10^{-8}\Omega\text{ m}$
$\text{D}-\ 2.3\times10^{3}\Omega\text{ m}$
Which material is:

Good conductor.
Resistor.
Insulator, and
Semiconductor

Answer

a. Good conductor $= C \left(10 \times 10^{-8}\right.$ ohm $)$.
b. Resistor $= A \left(110 \times 10^{-8} ohm \right)$.
c. Insulator = B $\left(1 \times 10^{10} ohm \right)$.
d. Semiconductor $= D \left(2.3 \times 10^3 ohm \right)$.

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Question 353 Marks

Distinguish between good conductors, resistors and insulators. Name two good conductors, two resistors and two insulators.

Answer

Those substances which have very low electrical resistance are called as good conductors. E.g., copper and aluminium.
Those substances which have comparatively high resistance than conductors are known as resistors. E.g., nichrome and manganin.Those substances which have infinitely high electrical resistance are called insulators. E.g., rubber and wood.

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Question 363 Marks

What is a voltmeter? How is a voltmeter connected in the circuit to measure the potential difference between two points. Explain with the help of a diagram.

Answer

A voltmeter is a device which is used to measure the potential difference between two points in an electric circuit. Voltmeter is always connected in parallel across the two points where the potential difference is to be measured.

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Question 373 Marks

Calculate the power used in the $2Ω$ resistor in each of the following circuits:

A 6V battery in series with $1Ω$ and $2Ω$ resistors.
A 4V battery in parallel with $12Ω$ and $2Ω$ resistors.

Answer

V = 6 Volt, R1 = $\text{R}_1=1\Omega,$ $\text{R}_2=2\Omega$

Equivalent resistance $=\text{R}_1+\text{R}_2=1+2=3\Omega$

Total current, $\text{I}=\frac{\text{V}}{\text{R}}=\frac{6}{3}=2\text{A}$

Current through $R _2= I _2= I =2 A$
Voltage across $R _2= V _2= I _2 R _2=2 \times 2=4$
Power used in $R _2= I _2 V_2=2 \times 4=8 W$
i. $V =4$ Volt, $R _1=12 \Omega R _2=2 \Omega$
Voltage across $R _2= V _2= V =4 V$
Current across $R _2= I _2=\frac{ V _2}{ R _2}=\frac{4}{2}=2 A$
Poower used in $R _2= I _2 V_2=2 \times 4=8 W$

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Question 383 Marks

How does the resistance of a conductor depend on:

Length of the conductor?
Area of cross-section of the conductor?
Temperature of the conductor?

Answer

Resistance of a conductor increases (or decreases) with increase (or decrease) in the length of the conductor.
Resistance of a conductor decreases (increases) with increase (decrease) in the area of cross-section of the conductor.
Resistance of a conductor increases on raising the temperature and decreases on lowering the temperature.

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Question 393 Marks

Two resistances when connected in parallel give resultant value of 2 ohm; when connected in series the value becomes 9 ohm. Calculate the value of each resistance.

Answer

Two resistance when connected in series, resultant value is 9 ohms. Two resistance when connected in parallel, resultant values is 2 ohms.
Let the two resistance be $R_1$ and $R_2$
If connected in series, then
$9=R_1+R_2$
$R_1=9-R_2$
If connected in parallel, then
$\frac{1}{2}=\frac{1}{R_1}+\frac{1}{R_2}$
From aboves equations we get that
$\frac{1}{2}=\frac{\left(R_1+R_2\right)}{R_1 R_2}$
$\frac{1}{2}=\frac{9}{\left(9-R_2\right) R_2}$
$9 R_2-R_2^2=18$
$R_2^2-9 R_2+18=0$
$\left(R_2-6\right)\left(R_2-3\right)=0$
$R_2=6,3$
So if $R_2 6$ ohms, then $R_1=9-6=3$ ohms.
If $R_2=3$ ohms, then $R_1=9-3=6$ ohms.

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Question 403 Marks

A flash of lightning carries 10C of charge which flows for 0.01s. What is the current? If the voltage is 10MV, what is the energy?

Answer

Q = 10C
t = 0.01s
$\text{I}=\frac{\text{Q}}{\text{t}}=\frac{10}{0.01}=1000\text{A}$
$\text{P}.\text{d}=\frac{\text{W}}{\text{Q}}$
$W=P . d \times Q$
$=10 \times 10^6 \times 10=100 \times 10^6=100 MJ$
Energy - Work done $=100 MJ$

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Question 413 Marks

How will you connect three resistors of resistances $2Ω, 3Ω$ and $6Ω $ obtain a total n si stance of:

$4 Ω,$
$1Ω$

Answer

Connects 2 ohms resistor in series with a parallel combinations of 3 ohms and 6 ohms.
Connects 2 ohms, 3 ohms, and 6 ohms in parallel.

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Question 423 Marks

If the charge on an electron is 1.6 × 10 coulombs, how many electrons should pass through a conductor in 1 second to constitute 1 ampere current?

Answer

We know that
$\text{I}=\frac{\text{Q}}{\text{T}}$
$\Rightarrow1\text{A}=\frac{\text{Q}}{1\text{s}}$
$\Rightarrow\text{Q}=1\text{C}$
Now, when charge is $1.6 \times 10^{-19}$ Coulamb, number of electrone $=1$
When charge is 1 Coulamb, number of electrone $=\frac{1}{1.6\times10^{-19}}=0.625\times10^{19}=625\times10^{18}$

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Question 433 Marks

What is an ammeter? How is it connected in a circuit? Draw a diagram to illustrate your answer.

Answer

Ammeter is a device used for the measurement of electric current. It is always connected in series with the circuit in which the current is to be measured.

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Question 443 Marks

Derive the expression for the heat produced due to a current ‘I’ flowing for a time interval ‘t’ through a resistor ‘R’ having a potential difference ‘V’ across its ends. With which name is this relation known?

Answer

when an electric charge Q moves against a.p.d.V, the amount of work done is given by
W = Q × V .......(1)
We know, current, $\text{I}=\frac{\text{I}}{\text{T}}$
Q = I × t .....(2)
By ohm's law, $\frac{\text{V}}{\text{I}}=\text{R}$
V = I × R ........(3)
Putting eqs. (2) and (3) in eq (1),
W = I × t × I × R
$W=I^2 R T$
Assuming that all the electrical work done is converted into heat energy, we get Heat produced, $H = I ^2 Rt$ joules This relation is known as Joule's law of heating,

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Question 453 Marks

What possible values of resultant resistance one can get by combining two resistances, one of value 2 ohm and the other 6 ohm?

Answer

$R _1=2 ohm , R _2=6 ohm$
Case I: (parallel combination)
$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$
$\frac{1}{R}=\frac{1}{2}+\frac{1}{6}=\frac{4}{6}$
$R=\frac{6}{4}=1.5 ohm$
CaseII: (Series combination)
$R=R_1+R_2=2+6=80 h m$

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Question 463 Marks

Which uses more energy: a 250W TV set in 1 hour or a 1200W to aster in 10 minutes?

Answer

Case1: TV set
P = 250W = 0.25 KWh
t = 1h
Energy consumed = P × t = 0.25 × 1 = 0.25KWh
Case2: Toaster
P = 1200W = 1.2KW, t = 10min $=\frac{10}{60}=\frac{1}{6}\text{h}$
Energy consumed $=\text{P}\times\text{t}=1.2\times\big(\frac{1}{6}\big)=0.2\text{kWh}$
Thus, TV uses energy.

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Question 473 Marks

The resistors $R_1, R_2, R_3$ and $R_4$ in the figure given below are all equal in value.

What would you expect the voltmeters A, B and C to read assuming that the connecting wires in the circuit have negligible resistance?

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Question 483 Marks

What would be the effect on the resistance of a metal wire of:

Increasing its length?
Increasing its diameter?
Increasing its temperature?

Answer

Resistance will increase.
Resistance will decrease.
Resistance will increase.

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Question 493 Marks

A p.d. of 10V is needed to make a current of 0.02A flow through a wire. What p.d. is needed to make a current of 250mA flow through the same wire?

Answer

In first case,
I = 0.02 amp, V = 10 volt
V = IR
10 = 0.02 × R
$\text{R}=\frac{10}{0.02}=500 \ \text{ohm}$
In second case,
I = 250 × 10-3 amp, R = 500 ohm
V = IR
V = 250 × 10 - 3 × 500
V = 125 volt.

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Question 503 Marks

In the circuit diagram given below, the current flowing across 5 ohm resistor is 1 amp. Find the current flowing through the other two resistors.

Answer

Given:
1 amp current is flowing through 5ohm resistor.
We know that in case of parallel connection, the p.d. across each resistor is same and is equal to the voltage applied.
Therefore, applied voltage, V = IR = 1 × 5 = 5V
So,
Current through 4 ohm resistor = $\frac{\text{V}}{\text{R}}=\frac{5}{4}=1.25\text{A}$
Current through 10 ohm resistor = $\frac{\text{V}}{\text{R}}=\frac{5}{10}=0.5\text{A}$

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