Answer the questions.[Phy-5M]

Question 15 Marks

i. What is meant by resistance of a conductor? Define its SI unit.
ii. List two factors on which the resistance of a rectangular conductor depends.
iii. How will the resistance of a wire be affected if its
I. length is doubled, and
II. radius is also doubled?
Give justification for your answer.

Answer

i. Resistance is the quality of a conductor that causes it to resist the flow of an electric current through it. It is the proportion of the potential difference between ends to the current flowing. The SI unit is ohm ( $\Omega$ ).
ii. Two factors on which the resistance of a rectangular conductor depends are:
I. Length of conductor
II. Area if cross-section
iii. $\quad$ I. $R=\frac{\rho l}{A}$
Where, $\rho=$ electrical resistivity
$1=$ length of the conductor
$A =$ cross-sectional area of the conductor
Hence, if the length is double then
$\begin{array}{l}\Rightarrow R_1=\rho \frac{(2 \mid)}{A} \\\therefore R_1=2(R)\end{array}$
So, if the length of the resistance gets doubled then resistance also gets doubled.
II. Now, when the radius is double then
$\begin{array}{l}\Rightarrow R_2=\frac{\rho l}{A} \\\Rightarrow R_2=\frac{\rho l}{\pi(2 r)^2} \\\therefore R_2=\frac{1}{4}(R)\end{array}$
So, if the radius gets doubled then resistance will be $\left(\frac{1}{4}\right)^{\text {th }}$ of initial resistance.

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Question 25 Marks

i. For the combination of resistors shown in the following figure, find the equivalent resistance between M & N.

ii. State Joule’s law of heating.
iii. Why we need a 5 A fuse for an electric iron which consumes 1 kW power at 220 V?
iv. Why is it impracticable to connect an electric bulb and an electric heater in series?

Answer

i. Equivalent resistance between M and N
$=\left[\frac{\left(R_3 \times R_4\right)}{\left(R_3+R_4\right)}\right]$
ii. Joule's law of heating states that when a current $i$ passes through a conductor of resistance $r$ for time $t$ then the heat developed in the conductor is equal to the product of the square of the current, the resistance, and time. This can be expressed as : $H=l^2 Rt$
iii. We need a fuse of 5 A for an electric iron which consumes 1 kW power at 220 V . It is because:
Given $P =1000 W, V =220 V$
As we know, $P = V \times t$
Or $I =\frac{P}{V}$
Or $I =\frac{1000}{220 V}=4.5 A$
Hence, 4.54 ampere current flows in the circuit, the fuse should be of 5 A .
iv. As you know, in a series circuit the current is constant throughout the electric circuit. Thus it is obviously impracticable to comnect an electric bulb and an electric heater in series because they need currents of widely different values to operate properly.

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Science Answer the questions.[Phy-5M]

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