Answer the questions.[Phy-3M]

Question 1013 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 1023 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 1033 Marks

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Answer

Here, voltage (V) = 220 V

R₁ = 100 Ω, R₂ = 50 Ω and R₃ = 500 Ω

$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}+\frac{1}{\text{R}_3}$

$\frac{1}{\text{R}}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}=\frac{16}{500}$

$\text{R}=\frac{500}{16}=31.25 \ \Omega$

The resistance of electric iron, which draws as much current as all three appliances take together $=\text{R} = 31.25 \Omega.$

Current passing through electric iron $ \text{(I)} = \frac{\text{V}}{\text{R}} = \frac{220}{31.25} = 7.04 \text{A.}$

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Question 1043 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 1053 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 1063 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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Question 1073 Marks

A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5Ω when connected to a 10V battery. Calculate the resistance of the electric lamp.
Now if a resistance of 10Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5Ω conductor and potential difference across the lamp will take place? Give reason.

Answer

Total resistance of circuit can be calculated as follows:

$\text{R}=\frac{\text{V}}{\text{I}}=\frac{10\text{V}}{1\text{A}}=10\Omega$

Since lamp and conductor are in series so resistance of lamp,

$=10\Omega-5\Omega=5\Omega$

The new resistance in parallel to earlier combination has same value, i.e. 10Ω as the resistance of series combination. This means that the amount of current would be equally divided into two branches. Hence, 0.5A current will flow through 5Ω conductor.

Now, resistance remains the same but current has become half. Using Ohm formula, potential difference across the lamp can be calculated as follows:

$\text{V}=\text{IR}=0.5\text{A}\times5\Omega=2.5\text{V}$

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Science Answer the questions.[Phy-3M]