Answer the questions.[Phy-3M]

Question 513 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 523 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 533 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 543 Marks

A resistor has a resistance of 176 ohms. How many of these resistors should be connected in parallel so that their combination draws a current of 5 amperes from a 220 volt supply line?

Answer

I = 5A
V = 220V
$\text{R}=\frac{\text{V}}{\text{I}}=\frac{220}{5}=44\Omega$
Required resistance is less that $176\Omega,$ so the resistor should be connected in parallel.
Let the required no. be n.
$\text{R}_\text{eq}=\frac{176}{\text{n}}=44$
$\text{n}=\frac{176}{44}=4$

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Question 553 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 = 220V, t = 30sec
we know that
Total resistance, R = 40 + 60 = 100 ohms
By ohm's law
V = IR
$\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=\frac{220}{100}=2.2\text{amp}$
Putting the value of I, R and t in eq. H = I²RT
H = 2.2² × 100 × 30
H = 14520J.

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Question 563 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 573 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,
R₁ = 3 ohm, R₂ = 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 583 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 593 Marks

An electric iron is connected to the mains power supply of 220V. When the electric iron is adjusted at ‘minimum heating’ it consumes a power of 360W but at ‘maximum heating’ it takes a power of 840W. Calculate the current and resistance in each case.

Answer

Given: V = 220V, P_min = 360W, P_max = 840W
For minimum heating case:
We know that
P_min = VI
360 = 220XI
I = 1.63 amp
$\text{R}=\frac{\text{V}}{\text{I}}$
$\text{R}=\frac{220}{1.63}$
R = 134.96 ohms
For maximum heating case:
We know that
P_max = VI
840 = 220XI
I = 3.81 amp
$\text{R}=\frac{\text{V}}{\text{I}}$
$\text{R}=\frac{220}{3.81}$
R = 57.74 ohms

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Question 603 Marks

An electric heater is connected to the 230V mains supply. A current of 8A flows through the heater.

How much charge flows around the circuit each second?
How much energy is transferred to the heater each second?

Answer

P.d. = 230V, I = 8A
$\text{I}=\frac{\text{Q}}{\text{t}}$
$8=\frac{\text{Q}}{1}$
Q = 8 × 1 = 8C
So, 8C of charge flows around the circuit each secound.
Energy transfred = Work done
$\text{P.d}.=\frac{\text{Work done}}{\text{Charge moved}}$
$230=\frac{\text{Work done}}{8}$
Work done = 230 × 8 = 1840J
Energy trandferred = 1840J.

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Question 613 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 623 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 × Q
= 10 × 10⁶ × 10 = 100 × 10⁶ = 100MJ
Energy - Work done = 100MJ

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Science Answer the questions.[Phy-3M]