 # Current Electricity

## Physics

### NCERT

1   The storage battery of a car has an emf of $12 V$. If the internal resistance of the battery is $0.4 \Omega$, what is the maximum current that can be drawn from the battery?

##### Solution :

Emf of the battery, $E = 12 V$$\\ Internal resistance of the battery, r = 0.4 \Omega$$\\$ Maximum current drawn from the battery =$I$$\\ According to Ohm’s law, E = Ir\\ I=\dfrac{E}{r}\\ =\dfrac{12}{0.4}=30A$$\\$ The maximum current drawn from the given battery is $30 A.$

2   A battery of emf $10 V$ and internal resistance $3\Omega$ is connected to a resistor. If the current in the circuit is $0.5 A$, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Emf of the battery, $E = 10 V$$\\ Internal resistance of the battery, r = 3 \Omega$$\\$ Current in the circuit, $I = 0.5 A$$\\ Resistance of the resistor =R$$\\$ The relation for current using Ohm’s law is,$\\$ $I=\dfrac{E}{R+1}\\ R+r=\dfrac{E}{I}\\ =\dfrac{10}{0.5}=20\Omega\\ \therefore R=20-3=17\Omega$$\\ Terminal voltage of the resistor = V$$\\$ According to Ohm’s law,$\\$ $V=IR\\ =0.5*17\\ 8.5V$$\\ Therefore, the resistance of the resistor is 17 \Omega and the terminal voltage is 8.5 V. 3 (a) Three resistors 1\Omega , 2\Omega and 3\Omega are combined in series. What is the total resistance of the combination?\\ (b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor. ##### Solution : (a) Three resistors of resistances 1 \Omega , 2 \Omega and 3\Omega are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances.\\ Total resistance =1 + 2 + 3 = 6 \Omega$$\\$ (b) Current flowing through the circuit $= I$$\\ Emf of the battery, E =12V$$\\$ Total resistance of the circuit, $R = 6\Omega$$\\ The relation for current using Ohm’s law is,\\ I=\dfrac{E}{R}\\ =\dfrac{12}{6}=2A$$\\$ Potential drop across $1\Omega$ resistor $=V_1$$\\ From Ohm’s law, the value of V_1 can be obtained as V_1 = 2 * 1 = 2V ....(i)$$\\$ Potential drop across $2\Omega$ resistor $=V_2$$\\ Again, from Ohm’s law, the value of V_2 can be obtained as\\ V_2=2*2=4V....(ii)$$\\$ Potential drop across $3\Omega$ resistor $=V_3$$\\ Again, from Ohm’s law, the value of V_3 can be obtained as\\ V_3 = 2 * 3 = 6V ....(iii)$$\\$ Therefore, the potential drop across $1 \Omega , 2 \Omega$ and $3\Omega$ resistors are $2 V, 4 V,$ and $6 V$ respectively.

4   a) Three resistors $2 \Omega , 4 \Omega$ and $5\Omega$ are combined in parallel. What is the total resistance of the combination?$\\$ (b) If the combination is connected to a battery of emf $20 V$ and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

##### Solution :

(a) There are three resistors of resistances,$\\$ $R_1 = 2 \Omega , R_2 = 4 \Omega$ , and $R_3 = 5 \Omega$$\\ They are connected in parallel. Hence, total resistance (R) of the combination is given by,\\ \dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\\ =\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{5}\\ =\dfrac{10+5+4}{20}=\dfrac{19}{20}\\ \therefore R=\dfrac{20}{19}\Omega$$\\$ Therefore, total resistance of the combination is $\dfrac{20}{19}\Omega.$$\\ (b) Emf of the battery, V = 20 V$$\\$ Current $( I_1 )$ flowing through resistor $R_1$ is given by,$\\$ $I_1=\dfrac{V}{R_1}\\ =\dfrac{20}{2}=10A$$\\ Current ( I_2 ) flowing through resistor R_2 is given by,\\ I_2=\dfrac{V}{R_2}\\ =\dfrac{20}{4}=5A$$\\$ Current $( I_3)$ flowing through resistor $R_3$ is given by,$\\$ $I_3=\dfrac{V}{R_3}\\ =\dfrac{20}{5}=4A$$\\ Total current,I = I_1 + I_2 + I_3 + 10 + 5 + 4 = 19 A$$\\$ Therefore, the current through each resistor is $10 A, 5 A,$ and $4 A$ respectively and the total current is $19 A.$

5   At room temperature $(27.0 ^o C )$ the resistance of a heating element is $100\Omega$ . What is the temperature of the element if the resistance is found to be $117\Omega$ , given that the temperature coefficient of the material of the resistor is $1.70 * 10 ^{-4}$$^ o$$ C ^ {- 1 }.$

##### Solution :

Room temperature,$T = 27^oC$$\\ Resistance of the heating element at T , R =100 \Omega$$\\$ Let $T_1$ is the increased temperature of the filament.$\\$ Resistance of the heating element at $T_1 , R_1 = 117 \Omega$$\\ Temperature co-efficient of the material of the filament,\\ \alpha=1.70*10^{-4}$$^oC^{-1}$$\\ \alpha is given by the relation,\\ \alpha=\dfrac{R_1-R}{R(T_1-T)}\\ T_1-T=\dfrac{R_1-R}{R\alpha}\\ T_1-27=\dfrac{117-100}{100(1.7*10^{-4})}\\ T_1-27=1000\\ T_1=1027^oC$$\\$ Therefore, at $1027^o C$ , the resistance of the element is $117\Omega$.

6   A negligibly small current is passed through a wire of length $15 m$ and uniform cross-section $6.0 * 10^{- 7 }m ^2$ , and its resistance is measured to be $5.0\Omega$. What is the resistivity of the material at the temperature of the experiment?

Length of the wire,$l = 15m$$\\ Area of cross-section of the wire, a =6.0 * 10 ^{- 7} m ^2$$\\$ Resistance of the material of the wire, $R = 5.0 \Omega $$\\ Resistivity of the material of the wire =\rho$$\\$ Resistance is related with the resistivity as $\\$ $R=\rho\dfrac{l}{A}\\ \rho =\dfrac{RA}{l}\\ =\dfrac{5*6*10^{-7}}{15}=2*10^{-7}\Omega m$$\\ Therefore, the resistivity of the material is 2*10^{-7}\Omega m 7 A silver wire has a resistance of 2.1\Omega at 27.5 ^o C , and a resistance of 2.7 \Omega at 100 ^o C . Determine the temperature coefficient of resistivity of silver. ##### Solution : Temperature, T_ 1 = 27.5 ^o C$$\\$ Resistance of the silver wire at $T_ 1 , R_ 1 = 2.1 \Omega $$\\ Temperature, T_ 2 = 100 ^o C$$\\$ Resistance of the silver wire at $T_ 2 , R _2 = 2.7 \Omega $$\\ Temperature coefficient of silver =\alpha$$\\$ It is related with temperature and resistance as $\\$ $\alpha =\dfrac{R_2-R_1}{R_1(T_2-T_1)}\\ =\dfrac{2.7-2.1}{2.1(100 -27.5)} =0.0039^oC^{-1}$$\\ Therefore, the temperature coefficient of silver is 0.0039^o C^{-1} 8 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 ^o C ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 *{10 ^{-4} }^{o}C^{- 1} ##### Solution : Supply voltage, V = 230 V$$\\$ Initial current drawn, $I _1 = 3.2A$$\\ Initial resistance =R_ 1 , which is given by the relation,\\ R_1=\dfrac{V}{I}\\ =\dfrac{230}{3.2}=71.87 \Omega$$\\$ Steady state value of the current,$I_ 2 = 2.8A$$\\ Resistance at the steady state =R_ 2 , which is given as\\ R_2=\dfrac{230}{2.8}=82.14 \Omega$$\\$ Temperature co-efficient of nichrome, $a = 1.70*{10^{-4 }}^oC ^{-1}$$\\ Initial temperature of nichrome, T_ 1 = 27.0 ^oC$$\\$ Study state temperature reached by nichrome $= T _2$$\\ T_ 2 can be obtained by the relation for \alpha ,\\ 1 \alpha =\dfrac{R_2-R_1}{R_1(T_2-T_1)}\\ T_2-27^oC=\dfrac{82.14-71.87}{71.87*1.7810^{-4}}=840.5\\ T_2=840.5+27=867.5^oC$$\\$ Therefore, the steady temperature of the heating element is $867.5^oC$

9   Determine the current in each branch of the network shown in figure:

##### Solution :

Current flowing through various branches of the circuit is represented in the given figure.

$I _1 =$ 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.,$\\$ $10 I_ 2 + 5 I _4- 5 I_ 3 = 0\\ 2 I_ 2 + I _4 - I_ 3 = 0\\ I _3 = 2 I_ 2 + I_ 4 \quad ............(1)$$\\ For the closed circuit BCDB, potential is zero i.e.,\\ 5 ( I_ 2 - I_ 4 )- 10 ( I_ 3 +I _4 ) - 5 I_ 4 = 0\\ 5 I _2 + 5 I_ 4 - 10 I_ 3 - 10 I_ 4 - 5 I_ 4 = 0\\ 5 I_ 2 - 10 I_ 3 - 20 I_ 4 = 0\\ I_ 2 = 2 I _3 + 4 I_ 4 \quad ...... (2 )$$\\$ For the closed circuit $ABCFEA$, potential is zero i.e.,$\\$ $-10 + 10 (I _1 ) +10 ( I_ 2 )+ 5 ( I_ 2 - I_ 4 )= 0\\ 10 = 15 I_ 2 + 10 I_ 1 - 5 I_ 4\quad ...... (3 ) 3 I_ 2 + 2 I _1 - I_ 4 = 2$$\\ From equations (1) and (2), we obtain\\ I_ 3 = 2 ( 2 I_ 3 + 4 I_ 4)+ I_ 4\\ I _3 = 4 I _3 + 8 I_ 4 + I _4\\ -3 I_ 3 = 9 I_ 4\\ - 3 I_ 4 =+ I _3 \quad ...... (4 )$$\\$ Putting equation (4) in equation (1), we obtain$\\$ $I _3 = 2 I_ 2 + I_ 4\\ - 4 I_ 4 = 2 I_ 2\\ I_ 2 =-2 I_ 4\quad ...... (5 )$$\\ It is evident from the given figure that,\\ I_ 1 = I_ 3 + I_ 2 \quad ....... (6 )$$\\$ Putting equation (6) in equation (1), we obtain$\\$ $3 I_ 2 + 2 ( I_ 3 + I_ 2 ) -I_ 4 = 2\\ 5 I _2 + 2 I_ 3 - I_ 4 = 2\quad ...... (7 )$$\\ Putting equations (4) and (5) in equation (7), we obtain\\ 5 (-2 I_ 4 ) + 2 (-3 I_ 4 )- I _4 = 2\\ - 10 I_ 4 - 6 I_ 4 - I_ 4 = 2\\ 17 I _4 =- 2$$\\$

$I_4=\dfrac{-2}{17}A$$\\ Equation(4) reduces to\\ I_3=-3(I_4)\\ =-3(\dfrac{-2}{17})=\dfrac{6}{17}A\\ I_2 =-2(I_4)\\ =-2(\dfrac{-2}{17})=\dfrac{4}{17}A\\ I_2-I_4=\dfrac{4}{17}-(\dfrac{-2}{17})=\dfrac{6}{17}A\\ I_3+I_4=\dfrac{6}{17}+(\dfrac{-2}{17})=\dfrac{6}{17}A\\ I_1=I_3+I_2\\ =\dfrac{6}{17}+\dfrac{4}{17}=\dfrac{10}{17}A$$\\$ Therefore, current in branch $AB=\dfrac{4}{17}A$$\\ In brance BC=\dfrac{6}{17}A$$\\$ In brance $CD=\dfrac{-4}{17}A$$\\ In brance AD=\dfrac{6}{17}A$$\\$ In brance $BD=(\dfrac{-2}{17})A$$\\ Total current=\dfrac{4}{17}+\dfrac{6}{17}+\dfrac{-4}{17}+\dfrac{6}{17}+\dfrac{-2}{17}=\dfrac{10}{17}A$$\\$

10   (a) In a metre bridge [Fig. $3.27]$, the balance point is found to be at $39.5 cm$ from the end $A,$ when the resistor $Y$ is of $12.5\Omega$ . Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?$\\$ (b) Determine the balance point of the bridge above if X and Y are interchanged.$\\$ (c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?$\\$

##### Solution :

A metre bridge with resistors X and Y is represented in the given figure.

(a) Balance point from end $A, l_ 1 = 39.5 cm$$\\ Resistance of the resistor Y = 12.5 \Omega$$\\$ Condition for the balance is given as,$\\$ $\dfrac{X}{Y}=\dfrac{100-l_1}{l_1}\\ X=\dfrac{100-39.5}{39.5}*12.5=8.2 \Omega$$\\ Therefore, the resistance of resistor X is 8.2\Omega .\\ The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula.\\ (b) If X and Y are interchanged, then l_ 1 and 100 - l_ 1 get interchanged.\\ The balance point of the bridge will be 100 - l_ 1 from A.\\ 100 - l_ 1 = 100 - 39.5 = 60.5 cm$$\\$ Therefore, the balance point is $60.5 cm$ from A.$\\$ (c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection. Hence, no current would flow through the galvanometer.