current-electricity Question 27
Question: Q. 21. A cell of $\operatorname{emf} E$ and internal resistance $r$ is connected to two external resistances $R_{1}$ and $R_{2}$ and a perfect ammeter. The current in the circuit is measured in four different situations :
(i) without any external resistance in the circuit
(ii) with resistance $R_{1}$ only
(iii) with $R_{1}$ and $R_{2}$ in series combination
(iv) with $R_{1}$ and $R_{2}$ in parallel combination.
The currents measured in the four cases are $0.42 \mathrm{~A}$, 1.05 A, 1.4 A and 4.2 A, but not necessarily in that order. Identify the currents corresponding to the four cases mentioned above. A[Delhi I, II, III 2012]
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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]
