Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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1Exercises8 questions
Q.3.1The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?v
SolutionMaximum current is drawn when the external resistance is negligible, so the current is limited only by internal resistance. Thus $I_{\max}=\epsilon/r=12/0.4=30\,\text{A}$.
Q.3.2A battery of emf 10 V and internal resistance 3 Ω 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?v
SolutionFor a cell with internal resistance, $I=\epsilon/(R+r)$. Hence $R=\epsilon/I-r=10/0.5-3=17\,\Omega$. The terminal voltage is $V=IR=0.5\times17=8.5\,\text{V}$, also equal to $\epsilon-Ir=10-0.5\times3$.
Answer:The external resistance is $17\,\Omega$ and the terminal voltage is $8.5\,\text{V}$.
Q.3.3At 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 × 10–4 °C–1.v
SolutionUse $R=R_0[1+\alpha(T-T_0)]$. With $R_0=100\,\Omega$ at $T_0=27.0^\circ\text{C}$ and $R=117\,\Omega$, $117/100=1+(1.70\times10^{-4})(T-27)$. Thus $T-27=0.17/(1.70\times10^{-4})=1000^\circ\text{C}$, so $T=1027^\circ\text{C}$.
Answer:$1027\,^{\circ}\text{C}$.
Q.3.4A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10–7 m2, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?v
SolutionFrom $R=\rho l/A$, $\rho=RA/l=(5.0)(6.0\times10^{-7})/15=2.0\times10^{-7}\,\Omega\,\text{m}$.
Answer:$2.0\times10^{-7}\,\Omega\,\text{m}$.
Q.3.5A 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.v
SolutionUsing $R=R_0[1+\alpha(T-T_0)]$, $\alpha=(R-R_0)/[R_0(T-T_0)]$. Thus $\alpha=(2.7-2.1)/[2.1(100-27.5)]=0.6/152.25=3.94\times10^{-3}\,^{\circ}\text{C}^{-1}$.
Answer:$3.9\times10^{-3}\,^{\circ}\text{C}^{-1}$.
Q.3.6A 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 × 10–4 °C–1.v
SolutionThe initial resistance at room temperature is $R_0=230/3.2=71.875\,\Omega$. The steady resistance is $R=230/2.8=82.143\,\Omega$. Hence $R/R_0=1+\alpha(T-27)$ gives $82.143/71.875=1+(1.70\times10^{-4})(T-27)$. Therefore $T-27=0.142857/(1.70\times10^{-4})=840^\circ\text{C}$, so $T\approx867^\circ\text{C}$.
Answer:$8.7\times10^2\,^{\circ}\text{C}$ approximately.
Q.3.8A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?v
SolutionDuring charging, the supply emf opposes the battery emf, so $I=(120-8.0)/(15.5+0.5)=112/16=7.0\,\text{A}$. The terminal voltage of a battery being charged is $V=\epsilon+Ir=8.0+7.0(0.5)=11.5\,\text{V}$. The series resistor prevents an excessively large current and dissipates the remaining voltage safely.
Answer:The terminal voltage is $11.5\,\text{V}$. The series resistor limits the charging current and drops the excess supply voltage.
Q.3.9The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 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 × 10–6 m2 and it is carrying a current of 3.0 A.v
SolutionDrift speed is $v_d=I/(neA)$. With $I=3.0\,\text{A}$, $n=8.5\times10^{28}\,\text{m}^{-3}$, $e=1.6\times10^{-19}\,\text{C}$ and $A=2.0\times10^{-6}\,\text{m}^2$, $v_d=3.0/[(8.5\times10^{28})(1.6\times10^{-19})(2.0\times10^{-6})]=1.1\times10^{-4}\,\text{m s}^{-1}$. Time to drift $3.0\,\text{m}$ is $t=L/v_d=3.0/(1.1\times10^{-4})=2.7\times10^4\,\text{s}$.
Answer:$2.7\times10^4\,\text{s}$, or about $7.6\,\text{h}$.