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?

Solution :

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 $117\Omega$.