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એક વ્હીસ્ટન બ્રિજની ચાર ભુજાઓ $P,Q,R$ અને $S$ ના અવરોધો અનુક્રમે $10\,Ω,\,30\,Ω,\,30\,Ω $ અને $90\,Ω$ છે.કોષનો $emf $ અને આંતરિક અવરોધ અનુક્રમે $7\,V$ અને $5\,Ω $ છે.જો ગેલ્વેનોમિટરનો અવરોધ $50\,Ω $ હોય,તો કોષમાંથી નીકળતો પ્રવાહ ................ $A$ હશે.
$0.2$
$0.1$
$2$
$1$
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
$I=\frac{7 \mathrm{V}}{35\, \Omega}=\frac{1}{5} \mathrm{A}=0.2 \,\mathrm{A}$
