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Physics · 2025 · 3 marks
CBSE 2025 · Region 1 · Set 1 · Q22
(a) Two batteries of emf's 3 V & 6 V and internal resistances 0.2 Ω & 0.4 Ω are connected in parallel. This combination is connected to a 4 Ω…
Two batteries of emf's $\displaystyle 3$ V & $\displaystyle 6$ V and internal resistances $\displaystyle 0.2 \Omega \& 0.4 \Omega$ are connected in parallel. This combination is connected to a $\displaystyle 4 \Omega$ resistor. Find:
(i)
the equivalent emf of the combination
(ii)
the equivalent internal resistance of the combination
(iii)
the current drawn from the combination
(i)
A conductor of length $\displaystyle l$ is connected across an ideal cell of emf E. Keeping the cell connected, the length of the conductor is increased to $\displaystyle 2 l$ by gradually stretching it. If $\displaystyle R$ and $\displaystyle R^{\prime}$ are initial and final values of resistance and $\displaystyle v_{\mathrm{d}}$ and $\displaystyle v_{\mathrm{d}}^{\prime}$ are initial and final values of drift velocity, find the relation between (i) $\displaystyle R^{\prime}$ and $\displaystyle R$ and (ii) $\displaystyle v_{\mathrm{d}}^{\prime}$ and $\displaystyle v_{\mathrm{d}}$.
(ii)
When electrons drift in a conductor from lower to higher potential, does it mean that all the'free electrons'of the conductor are moving in the same direction?
Marking-scheme solution
(i)
Because $\displaystyle E_{eq} = \dfrac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$
$\displaystyle E_{eq} = \dfrac{3 \times 0.4 + 6 \times 0.2}{0.6} = 4\ \text{V}$
(ii)
$\displaystyle r_{eq} = \dfrac{r_1 r_2}{r_1 + r_2}$
$\displaystyle r_{eq} = \dfrac{0.2 \times 0.4}{0.2 + 0.4} = 0.133\,\Omega$
(iii)
$\displaystyle I = \dfrac{E}{R + r_{eq}}$
$\displaystyle I = \dfrac{4}{4 + 0.13} = \dfrac{4}{4.13}\ A$
$\displaystyle I = 0.9\,A$
(i)
$\displaystyle l' = 2l$
$\displaystyle Al = A'l'$ = volume of the wire
$\displaystyle Al = A'(2l)$
$\displaystyle \dfrac{A}{2} = A'$
$\displaystyle R = \dfrac{\rho l}{A}$
$\displaystyle R' = \dfrac{\rho l'}{A'}$
$\displaystyle R' = \dfrac{\rho (2l)}{A/2}$
$\displaystyle \dfrac{R'}{R} = 4$
Alternatively
$\displaystyle R' = n^2 R$
$\displaystyle n = 2$
$\displaystyle R' = 4R$
(ii)
$\displaystyle \mathrm{v_d} = \dfrac{eE}{m}\tau$
$\displaystyle \mathrm{v_d} = \dfrac{eV}{ml}\tau$
$\displaystyle \mathrm{v_d}' = \dfrac{eV}{ml'}\tau$
$\displaystyle \dfrac{\mathrm{v_d}'}{\mathrm{v_d}} = \dfrac{l}{l'} = \dfrac{1}{2}$
(ii)
No
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