NCERT Exemplar · Class 12 Physics Current Electricity

Only charge is conserved at the junction: \[I_1 = I_2 + I_3 \] Surface charges at the junction set up fields that turn the carriers: \[\mathbf{F}_{\text{net}} = \frac{d\mathbf{p}}{dt} \neq 0 \] Answer: No; charge is conserved, momentum is not.

Since \(\displaystyle v_d\) is far below the thermal speed, \(\displaystyle E\) hardly changes \(\displaystyle \tau\): \[\mathbf{v}_d = -\frac{e\tau}{m}\mathbf{E} \] \[\mathbf{J} = -ne\,\mathbf{v}_d = \frac{ne^2\tau}{m}\,\mathbf{E} = \sigma\mathbf{E},\qquad \tau \text{ constant} \Rightarrow \sigma \text{ constant} \Rightarrow J \propto E \] \[\rho = \frac{1}{\sigma} = \frac{m}{ne^2\tau} \] \[T\uparrow \Rightarrow \text{ion vibrations}\uparrow \Rightarrow \tau\downarrow \Rightarrow \rho\uparrow \] Answer: constant \(\displaystyle \tau\) gives Ohm's law; temperature changes \(\displaystyle \tau\), hence \(\displaystyle \rho\).

Null point, \(\displaystyle I_g = 0\): \[\frac{P}{Q} = \frac{R}{S} \;\Rightarrow\; R = \frac{P}{Q}\,S \] Independent of \(\displaystyle R_G\), emf and internal resistance. Off balance, loop rule on A-B-D: \[I_1 P + I_g R_G = I_2 R \;\Rightarrow\; R = \frac{I_1 P + I_g R_G}{I_2} \] Answer: the null method needs no current readings; otherwise \(\displaystyle I_1\), \(\displaystyle I_2\), \(\displaystyle I_g\) and \(\displaystyle R_G\) must be measured accurately.

\[R_{\text{strip}} = \rho\,\frac{\ell_s}{A_s},\qquad A_s\uparrow \Rightarrow R_{\text{strip}}\downarrow \] \[R_{\text{strip}} \ll R_{\text{wire}} \;\Rightarrow\; V_{\text{strip}} = I R_{\text{strip}} \approx 0 \] \[k = \frac{V_{AB}}{L} = \frac{I R_{\text{wire}}}{L},\qquad \varepsilon = k\,l \] Answer: negligible strip resistance adds no potential drop, so the gradient stays uniform; strip lengths are not counted in the balance length \(\displaystyle l\).

Cu and Al: low \(\displaystyle \rho\), ductile. Al is cheaper and lighter but needs a larger area: \[\rho_{\mathrm{Cu}} = 1.7\times10^{-8}\ \Omega\,\mathrm{m},\qquad \rho_{\mathrm{Al}} = 2.7\times10^{-8}\ \Omega\,\mathrm{m} \] \[R = \rho\,\frac{l}{A},\qquad P = I^2R \] \[\text{same } l,\ R:\quad \frac{A_{\mathrm{Al}}}{A_{\mathrm{Cu}}} = \frac{\rho_{\mathrm{Al}}}{\rho_{\mathrm{Cu}}} = \frac{2.7}{1.7} \approx 1.6 \] Answer: low resistivity (small \(\displaystyle I^2R\) loss), ductility, cost, weight.

Manganin, constantan: high resistivity, tiny \(\displaystyle \alpha\).\[R_T = R_0\,[\,1+\alpha\,(T-T_0)\,],\qquad \alpha\ \text{in}\ \mathrm{K^{-1}} \]\[\alpha_{\text{alloy}} \ll \alpha_{\text{Cu}} \;\Rightarrow\; R_T \approx R_0 \]Answer: Alloys have a very small temperature coefficient of resistance, so the coil's resistance stays almost constant as it heats.

The same current \(\displaystyle I\) flows through the cables.\[P = VI \;\Rightarrow\; I = \frac{P}{V} \]\[P_{\text{wasted}} = I^{2}R_C = \frac{P^{2}R_C}{V^{2}} \]\[V\uparrow \;\Rightarrow\; I\downarrow \;\Rightarrow\; P_{\text{wasted}}\downarrow \]Step \(\displaystyle V\) up with a transformer before transmission; thicker cables (smaller \(\displaystyle R_C\)) also help.Answer: Power wasted \(\displaystyle =I^{2}R_C=P^{2}R_C/V^{2}\); reduce it by transmitting at higher \(\displaystyle V\) (smaller \(\displaystyle I\)), or by lowering \(\displaystyle R_C\).

Wire \(\displaystyle AB\): resistance \(\displaystyle R_{AB}\), length \(\displaystyle L\); internal resistance \(\displaystyle r\) of \(\displaystyle E\); cell \(\displaystyle E_1\) in the galvanometer branch; balance length \(\displaystyle \ell = AJ\).\[I = \frac{E}{R + R_{AB} + r} \]\[k = \frac{I\,R_{AB}}{L} \]\[E_1 = k\,\ell \;\Rightarrow\; \ell = \frac{E_1}{k} \]\[R\uparrow \;\Rightarrow\; I\downarrow \;\Rightarrow\; k\downarrow \;\Rightarrow\; \ell\uparrow \]Balance exists only while \(\displaystyle E_1 < I R_{AB}\).Answer: J shifts towards B.

Jockey at \(\displaystyle J\), \(\displaystyle \ell = AJ\); \(\displaystyle A\) is the positive end of \(\displaystyle E\), so potential falls along \(\displaystyle AB\) at gradient \(\displaystyle k\); \(\displaystyle \Delta V\) is the net pd on \(\displaystyle G\).\[V_J = V_A - k\,\ell \]\[\text{(i)}\;\; X=+:\quad \Delta V = E_1 - k\,\ell \;\downarrow\ \text{as}\ \ell\uparrow,\ \text{never zero} \;\Rightarrow\; E_1 > V_{AB} \approx E \]\[\text{(ii)}\;\; X=-:\quad |\Delta V| = E_1 + k\,\ell \;\uparrow\ \text{as}\ \ell\uparrow \]Answer: (i) Positive terminal of \(\displaystyle E_1\) at \(\displaystyle X\); \(\displaystyle E_1 > E\). (ii) Negative terminal of \(\displaystyle E_1\) at \(\displaystyle X\).

\[I = \frac{E}{R + r} \]\[V = IR = \frac{E\,R}{R + r} \]\[R=0:\; V=0 \qquad R=r:\; V=\frac{E}{2} \qquad R\to\infty:\; V\to E \]Answer: \(\displaystyle V\) rises from zero, equals \(\displaystyle E/2\) at \(\displaystyle R=r\), and approaches \(\displaystyle E\) asymptotically.