Solution:

Ans. Use the formulae and values come as follows :

(i) $i=\frac{E}{r}$

(ii) $i=\frac{E}{r+R_{1}}$

(iii) $i=\frac{E}{r+R_{1}+R_{2}}$

(iv) $i=\frac{E}{r+\frac{R_{1} R_{2}}{R_{1}+R_{2}}}$

(i) $i=4.2 \mathrm{~A}$

(ii) $i=1.05 \mathrm{~A}$

(iii) $i=0.42 \mathrm{~A}$

(iv) $i=1.4 \mathrm{~A}$ AI Q. 22. In the circuit shown in the figure, find the current through each resistor.

A [Delhi I, II, III 2015]

Ans. Total emf in the circuit $=8 \mathrm{~V}-4 \mathrm{~V}=4 \mathrm{~V} \quad 1 / 2$

Total resistance of the circuit $=8 \Omega \quad 1 / 2$

Hence current flowing in the circuit

$$ =\frac{V}{R}=\frac{4}{8} \mathrm{~A}=0.5 \mathrm{~A} $$

Current flowing through the resistors :

Current through $0.5 \Omega, 1.0 \Omega$ and $4.5 \Omega$ is $0.5 \mathrm{~A}$

Current through $3.0 \Omega$ is $\frac{2}{3} \mathrm{I}=\frac{1}{3} \mathrm{~A}$

Current through $6.0 \Omega$ is $\frac{1}{3} \mathrm{I}=\frac{1}{6} \mathrm{~A}$ [AI Q. 1. Derive the expression for the current density of a conductor in terms of the conductivity and applied electric field. Explain, with reason how the mobility of electrons in a conductor changes when the potential difference applied is doubled, keeping the temperature of the conductor constant. A [Delhi Comptt. I, II, III 2017]

Ans. Derivation of current density

Explanation with reason the change in mobility of electrons

Using Ohm’s law,

$$ V=I R=\frac{I \rho l}{A} $$

Potential difference $(V)$, across the ends of a conductor of length ’ $l$ ‘, where field ’ $E$ ’ is applied is given by

$$ \begin{align*} V & =E l \ \therefore \quad E l & =\frac{I \rho l}{A} \end{align*} $$

But current density,

$$ J=\frac{I}{A} $$

$$ \begin{aligned} & E l=J \rho l=\frac{J l}{\sigma} \ & {\left[\because \frac{1}{\rho}=\sigma\right] 1 / 2} \ & \mu=\frac{v_{d}}{E} \ & v_{d}=\frac{e V \tau}{m l} \end{aligned} $$

As potential is doubled, drift velocity also gets doubled, therefore, no change in mobility.

[CBSE Marking Scheme 2017]