Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Current Electricity question

2006 · Shift 0 · Q119
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Current Electricity
  5. /2006 · Shift 0 · Q119

Current Electricity question

2006 · Shift 0 · Q119

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
In a Wheatstone's bridge, three resistance P,QP, QP,Q and RRR connected in the three arms and the fourth arm is formed by two resistances S1{S_1}S1​ and S2{S_2}S2​ connected in parallel. The condition for the bridge to be balanced will be
  1. A
    PQ=2RS1+S2{P \over Q} = {{2R} \over {{S_1} + {S_2}}}QP​=S1​+S2​2R​
  2. B
    PQ=R(S1+S2)S1S2{P \over Q} = {{R\left( {{S_1} + {S_2}} \right)} \over {{S_1}{S_2}}}QP​=S1​S2​R(S1​+S2​)​
  3. C
    PQ=R(S1+S2)2S1S2{P \over Q} = {{R\left( {{S_1} + {S_2}} \right)} \over {2{S_1}{S_2}}}QP​=2S1​S2​R(S1​+S2​)​
  4. D
    PQ=RS1+S2{P \over Q} = {R \over {{S_1} + {S_2}}}QP​=S1​+S2​R​
View written solutionFree

Correct answer: B

  1. Balance condition of Wheatstone bridge

For a Wheatstone bridge to be balanced,

PQ=RS\frac{P}{Q} = \frac{R}{S}QP​=SR​

where SSS is the resistance in the fourth arm.

  1. Find the equivalent resistance of the fourth arm

The fourth arm consists of two resistances S1S_1S1​ and S2S_2S2​ connected in parallel.

So their equivalent resistance is

S=S1S2S1+S2S = \frac{S_1 S_2}{S_1 + S_2}S=S1​+S2​S1​S2​​
  1. Substitute into the balance condition

Using

PQ=RS\frac{P}{Q} = \frac{R}{S}QP​=SR​

we get

PQ=RS1S2S1+S2\frac{P}{Q} = \frac{R}{\dfrac{S_1 S_2}{S_1 + S_2}}QP​=S1​+S2​S1​S2​​R​

Simplifying,

PQ=R⋅S1+S2S1S2\frac{P}{Q} = R \cdot \frac{S_1 + S_2}{S_1 S_2}QP​=R⋅S1​S2​S1​+S2​​

Thus,

PQ=R(S1+S2)S1S2\frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1 S_2}QP​=S1​S2​R(S1​+S2​)​
  1. Compare with options

This matches Option B.

  1. Final answer

The correct condition for balance is

PQ=R(S1+S2)S1S2\boxed{\frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1 S_2}}QP​=S1​S2​R(S1​+S2​)​​
PreviousNext

More from Current Electricity

  • The current I drawn from the 5 volt source will be Includes diagram2006 · MCQ
  • Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are R1​ and R2​(R1​>R1​). If the potential difference across the source having internal…2005 · MCQ
  • A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10- divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms…2005 · MCQ
  • The resistance of hot tungsten filament is about 10 times the cold resistance. What will be resistance of 100W and 200V lamp when not in use ?2005 · MCQ
  • An energy source will supply a constant current into the load if its internal resistance is2005 · MCQ
  • In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be - Includes diagram2005 · MCQ
  • A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be2005 · MCQ
  • Two voltmeters, one of copper and another of silver, are joined in parallel. When a total charge q flows through the voltmeters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are Z1​…2005 · MCQ