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The resistances of the four arms $P, Q, R$ and $S$ in a Wheatstone's bridge are $10\,ohm$, $30\,ohm$, $30\,ohm$ and $90\,ohm$, respectively. The e.m.f. and internal resistance of the cell are $7\,volt$ and $5\,ohm$ respectively. If the galvanometer resistance is $50\,ohm$, the current drawn from the cell will be ............... $A$
$0.2$
$0.1$
$2$
$1$
Solution
The a balanced Wheatstone,s birdge
For a balanced Wheatstone's bridge
$\frac{P}{Q}=\frac{R}{S}$
$\therefore \quad \frac{10 \,\Omega}{30\, \Omega}=\frac{30\, \Omega}{90\, \Omega}$ or $\frac{1}{3}=\frac{1}{3}$
It is a balanced Wheatstone's bridge. Hence no current flows in the galvanometer arm. Hence, resistance $50\, \Omega$ becomes ineffective.
$\therefore$ The equivalent resistance of the circuit is
$R_{\mathrm{eq}}=5\, \Omega+\frac{(10 \,\Omega+30\, \Omega)(30\, \Omega+90\, \Omega)}{(10\, \Omega+30\, \Omega)+(30 \,\Omega+90\, \Omega)} $
$=5\, \Omega+\frac{(40\, \Omega)(120\, \Omega)}{40 \,\Omega+120\, \Omega}=5 \,\Omega+30\, \Omega=35\, \Omega$
Current drawn from the cell is
$I=\frac{7 \mathrm{V}}{35\, \Omega}=\frac{1}{5} \mathrm{A}=0.2 \,\mathrm{A}$
