Class 12th

3.5 Let T be the room temperature and R be the resistance at room temperature and $T_1$ be the required temperature and $R_1$ be the resistance at that temperature and α be the coefficient of resistor.

We have:

T = 27 $?,$ R = 100 Ω, $T_1$ = ?, $R_1$ = 117 Ω, α = 1.70 $*{10}^{-4}$ ° ${\mathit{C}}^{-1}$

We know the relation of α can be given as

α = $\frac{R_1-R}{R(T_1-T)}$ = $\frac{117-100}{100(T_1-27)}$ = 1.70 $*{10}^{-4}$

or ( $T_1$ - 27) = $\frac{117-100}{100*1.70\mathrm{}*{10}^{-4}}$ = 1000

$T_1$ = 1000 +27 = 1027 $?$$$

3.4 (a) Let $R_1$ = 2 Ω, $R_2$ = 4 Ω, $R_3$ = 5 Ω

If the equivalent resistance is R, then $\frac{1}{R}$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$ + $\frac{1}{R_3}$ = $\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{5}$ = $\frac{10+5+4}{20}$ = $\frac{19}{20}$

R = $\frac{20}{19}$ = 1.05 Ω

(b) The EMF of the battery = 20 V

Current through $R_{1,}$ $I_1$ = $\frac{V}{R_{1,}}$ = $\frac{20}{2}$ = 10 A

Current through $R_{2,}$ $I_2$ = $\frac{V}{R_{2,}}$ = $\frac{20}{4}$ = 5A

Current through $R_{3,}$ $I_3$ = $\frac{V}{R_{3,}}$ = $\frac{20}{5}$ = 4 A

Total current I = $I_1$ + $I_2$ + $I_3$ = 10 + 5 + 4 = 19 A

3.3 (a) The equivalent resistance of the resistor in series is given by

R = 1 + 2 + 3 = 6 Ω

(b) From Ohm's law, I = $\frac{V}{R}$ we get I = $\frac{12}{6}$ = 2 A.

Potential drop across 1 Ω resistor = I $*R$ = 2 $*1$ = 2 V

Potential drop across 2 Ω resistor = I $*R$ = 2 $*2$ = 4 V

Potential drop across 3 Ω resistor = I $*R$ = 2 $*3$ = 6 V

3.2 EMF of the battery = 10 V

Internal resistance of the battery, r = 3 Ω

Current in the circuit, I = 0.5 A

Let the resistance of the resistor be R

According to Ohm's law

I = $\frac{E}{(R+r)}$

R + r = $\frac{E}{I}$ or R = $\frac{E}{I}$ - r = $\frac{10}{0.5}$ - 3 = 17 Ω

Terminal voltage of the battery when the circuit is closed is given by

V = IR = 0.5 $*17=8.5V$

Therefore the resistance is 17 Ω and the terminal voltage is 8.5 V

3.1 EMF of the battery, E = 12 V

Internal resistance of the battery, r = 0.4 Ω

Let the maximum current drawn = I

According to OHM's law, E = Ir

So I = $\frac{E}{r}$ = $\frac{12}{0.4}$ amp = 30 amp

Therefore, the maximum current can be drawn is 30 ampere.

students can check the table for the principal values for all ITFs below;

To understand this, Assume you have a bucket that has infinite number of apples and if your mother asks "give me the apple". How will you figure out which one is "The Apple", she is asking for.

Similarly any inversre trigonometric functions behaves like a Many-one Function; which means,

For Example $\sin^ {-1}\left (\frac {2} {3}\right)$ can have many solutions, we need to fix one solutions which can be used as standerd value for the function.

• A standerd value (Angles) of any inverse trigonometric value lies between a fiexed range is known as principal value. 

For Ex; The value of $\sin^ {-1}\left (\frac {2} {3}\right)$ will always lie between –? /2 to? /2.

$y={\sin }^{?1}\left(\frac{2}{3}\right)?\sin \left(y\right)=\frac{2}{3}?y?[?\frac{?}{2},\frac{?}{2}]$