NCERT Solutions Class 12 Physics Chapter 3 Current Electricity with Video Explanation - Param Himalaya
Question 3.1 : The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery?
Solution :
$$ I_{max} = \frac{E}{r}=\frac{12}{0.4}$$
$$I_{max} = 30 A$$
Question 3.2 : A battery of e.m.f 10V and internal resistance 3Ω is connected to a resistor. If The current in the circuit is 0.5A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ?
Solution :
$$ I = \frac{E}{R+r}$$
$$R+r = \frac{E}{I}$$
$$R+r= \frac{10}{.5}=20$$
$$Or\;R = 20 - r = 20 - 3$$
$$R = 17 Ω$$
$$Terminal \ Voltage = IR$$
$$V= IR$$
$$V=0.5×17=8.5Volt$$
Question 3.3 : At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 x 10-4 °C-1.
Solution:
$$R_{2} = R_{1}(1+\alpha \Delta T)$$
$$\Delta T=\frac{R_{2}-R_{1}}{\alpha R_{1}}$$
$$\Delta T=\frac{117-100}{(1.7\times10^{-4})\times100}$$
$$\Delta T=\frac{17\times10^{2}}{1.7}=1000^{0}C$$
$$T_{2} -T_{1}=1000^{0}C$$
$$T_{2}=1000+T_{1}=1000+27$$
$$T_{2}=1027^{0}C$$
Question 3.4 : A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10 -7m2 , and its resistance is measured to be 5.0Ω . What is the resistivity of the material at the temperature of the experiment?
Solution:
Using
$$R= \rho \frac{l}{a}$$
we get
$$\rho =\frac{R\times a}{l}$$
$$i.e. \rho = \frac{5 \times 6 \times 10^{-7}}{15}$$
$$\rho = 2 \times 10^{-7} \Omega m.$$
Question 3.5 : A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
Solution :
$$R_{2}=R_{1}(1+\alpha \Delta T)$$
$$\alpha = \frac{R_{2}-R_{1}}{R_{1}\Delta T}$$
$$\alpha = \frac{2.7-2.1}{2.1\times (100-27.5)}$$
$$\alpha = \frac{0.6}{2.1(72.5)}$$
$$\alpha = 0.0039^{0}C^{-1}$$
Question 3.6 : A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 x 10-4 °C-1.
Solution :
$$R_{1}=\frac{230}{3.2} = 71.88 \Omega$$
$$R_{2}=\frac{230}{2.8}=82.14 \Omega$$
Using
$$\alpha =\frac{R_{2}-R_{1}}{R_{1}\Delta T}$$ ,
We get
$$\Delta T = \frac{R_{2}-R_{1}}{R_{1}\alpha }$$
$$T_{2}-T_{1}= \frac{R_{2}-R_{1}}{R_{1}\alpha }$$
$$T_{2}=T_{1}+\frac{R_{2}-R_{1}}{R_{1}\alpha }$$
$$T_{2}= \frac{82.14-71.88}{71.88 \times1.7 \times10^{-4}} + 27$$
$$T_{2}= 27 + 840 =867^{0}C$$
3.7 Determine the current in each branch of the network shown in Fig. 3.20.
Solution :
$I_{1}$= Current flowing through the outer circuit
$I_{2}$ = Current flowing through branch AB
$I_{3}$ = Current flowing through branch AD
$I_{2} - I_{4}$ = Current flowing through branch BC
$I_{3} + I_{4}$ = Current flowing through branch CD
$I_{4}$ = Current flowing through branch BD
For the closed circuit ABDA , potential is zero i.e.,
Loop ABDA : Applying Kirchhoff's voltage law to this loop , we get ,
$$10I_{2}+5I_{4}-5I_{3}=0$$
$$2I_{2}+I_{4}-I_{3}=0$$
$$I_{3} = 2 I_{2}+ I_{4} .....(1)$$
Loop BCDB : Applying Kirchhoff's voltage law to this loop , we get ,
$$5(I_{2}-I_{4}) - 10 (I_{3}+I_{4})-5I_{4} = 0$$
$$5I_{2}-5I_{4})-10I_{3}-10I_{4}-5I_{4} = 0$$
$$5I_{2}-10I_{3}-20I_{4} =0$$
$$I_{2} = 2 I_{3} + 4 I_{4} ....... (2)$$
Loop ADCFEA : Applying Kirchhoff's voltage law to this loop , we get ,
$$-10+10(I_{1}+I_{2})+5(I_{2}-I_{4})= 0$$
$$10I_{1}+15I_{2}-5I_{4} = 10$$
$$2I_{1}+3I_{2}-I_{4} = 2 ........ (3)$$
From equation (1) and (2) , we obtain
$$I_{3} = 2 (2I_{3}+4I_{4}) + I_{4}$$
$$I_{3} = 4I_{3} + 8I_{4} + I_{4}$$
$$-3I_{3} = 9I_{4}$$
$$I_{3} = -3 I_{4} ......... (4)$$
Putting equation (4) in equation (1) , we obtain
$$I_{3} = 2 I_{2} + I_{4}$$
$$-3 I_{4} = 2 I_{2}.+ I_{4}$$
$$-4 I_{4} = 2 I_{2}$$
$$2 I_{2} = -4 I_{4}$$
$$I_{2} = -2 I_{4} ...........(5)$$
It is evident from the given fiqure that
$$I_{1} = I_{3}+ I_{2} .........(6)$$
Putting equation (6) in equation (3) , we obtain
$$3I_{2} + 2 (I_{3}+I_{2}) - I_{4} =2$$
$$5I_{2}+ 2I_{3}- I_{4} =2 .......(7)$$
Putting equation (4) and (5) in equation (7) , we obtain
$$5(-2I_{4}+2(-3I_{4})- I_{4} =2$$
$$-10I_{4}-6I_{4}-I_{4} = 2$$
$$17 I_{4} = -2$$
$$I_{4} = \frac{-2}{17}$$
Equation (4) reduces to
$$I_{3} = -3I_{4}$$
$$= -3(\frac{-2}{17}) = \frac{6}{17}A$$
$$I_{2} = -2 (I_{4})$$
$$I_{2} = -2 (\frac{-2}{17})$$
$$I_{2} = \frac{4}{17}A$$
$$I_{2} - I_{4} = \frac{4}{17}-(\frac{-2}{17})A = \frac{6}{17}A$$
$$I_{3} + I_{4} = \frac{6}{17}+(\frac{-2}{17})A = \frac{4}{17}A$$
$$I_{1} = I_{3} + I_{2}$$
$$= \frac{6}{17} + \frac{4}{17}A$$
$$= \frac{10}{17}A$$
current in branch
$$AB= \frac{4}{17}A$$
$$BC = \frac{6}{17}A$$
$$CD= \frac{4}{17}A$$
$$AD = \frac{6}{17}A$$
$$BD = \frac{-2}{17}A$$
Total Current =
$$I_{1} = I_{2}+I_{3}$$
$$I_{1} = \frac{6}{17} + \frac{4}{17}$$
$$I_{1} = \frac{10}{17}$$
Question 3.8 : A storage battery of emf 8.0 V and internal resistance 0.5 $\Omega$ is being charged by a 120 V dc supply using a series resistor of 15.5 $\Omega$. What is the terminal voltage of These battery during charging? What is the purpose of having a series resistor in the charging circuit?
Solution :
During Charging
$$V=E+I(r+R) =\frac{E-V}{r+R}$$
$$I= \frac{120-8}{.5+15.5} =\frac{112}{16}$$
$$I=7A$$
Terminal Voltage ,
$$V=E+Ir = 8+7\times 0.5$$
$$V=11.5 Vz$$
Question 3.9 : The number density of free electrons in a copper conductor estimated in Example 3.1 is $8.5\times 10^{28} m^{-3}$. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is $2.0\times 10^{-6} m^{2}$ and it is carrying a current of 3.0 A.
Solution :
Using ,
$$v_{d} = \frac{I}{neA}$$
$$v_{d} = \frac{3}{(8.5\times 10^{28})(1.6 \times10^{-19})(2 \times 10^{-6})}$$
$$v_{d} = 1.1\times 10^{-4} m s^{-1}$$
Time taken,
$$t = \frac{l}{v_{d}} = \frac{3}{1.1 \times10^{-4}}$$
$$t= 2.72 \times10^{4} s.$$