Question

In Wheatstone bridge, resistance P is formed by a parallel combination of \(p_{1}\) and \(p_{2}\). The Q resistance is formed by a parallel combination of \(q_{1}\) and \(q_{2}\). The remaining two resistances are R and S respectively, and in the balanced condition, \(\frac{R}{S}\) is

• A. \(\frac{(p_{1}p_{2})}{(q_{1}q_{2})(p_{1}+p_{2})}\)
• B. \(\frac{(p_{1}p_{2})(q_{1}+q_{2})}{(q_{1}q_{2})(p_{1}+p_{2})}\)
• C. \(\frac{(q_{1}q_{2})(q_{1}+q_{2})}{(p_{1}p_{2})(p_{1}+p_{2})}\)
• D. \(\frac{(p_{1}p_{2})(q_{1}+q_{2})}{(p_{1}+p_{2})}\)

Hint:

Find the equivalent resistances of the parallel arms P and Q, then apply the balance condition.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In a balanced Wheatstone bridge, \(\dfrac{P}{Q}=\dfrac{R}{S}\) as per the arrangement in the figure.

Step 2: Equivalent resistances
\(P=\dfrac{p_1p_2}{p_1+p_2}\) and \(Q=\dfrac{q_1q_2}{q_1+q_2}\).

Step 3: Calculation
\(\dfrac{R}{S}=\dfrac{P}{Q}=\dfrac{p_1p_2}{p_1+p_2}\times\dfrac{q_1+q_2}{q_1q_2}\).

Step 4: Result
\(\dfrac{R}{S}=\dfrac{(p_1p_2)(q_1+q_2)}{(q_1q_2)(p_1+p_2)}\).

Final Answer:
R divided by S equals p1 p2 (q1 + q2) divided by q1 q2 (p1 + p2). \[ \boxed{\text{(B) }\dfrac{(p_{1}p_{2})(q_{1}+q_{2})}{(q_{1}q_{2})(p_{1}+p_{2})}} \]