A) 3 : 30 : 1 : 10
B) 30 : 3 : 10 : 1
C) 30 : 10 : 1 : 3
D) 30 : 1 : 3 : 10
Correct Answer: B
Solution :
Let i be the total current passing through balanced Wheatstone bridge. Current through arms of resistances P and Q in series is \[{{i}_{1}}=\frac{i\times 330}{330+110}\] \[=\frac{3}{4}i\] and current through arms of resistances R and S in series is \[{{i}_{1}}=\frac{i\times 110}{330+110}=\frac{1}{4}i\] \[\therefore \] Ratio of heat developed per sec \[{{H}_{P}}:{{H}_{Q}}:{{H}_{R}}:{{H}_{S}}\] \[={{\left( \frac{3}{4}i \right)}^{2}}\times 100:{{\left( \frac{3}{4}i \right)}^{2}}\times 10:{{\left( \frac{1}{4}i \right)}^{2}}\times 300\] \[:{{\left( \frac{1}{4}i \right)}^{2}}\times 30\] \[=30:3:10:1\]You need to login to perform this action.
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