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Step-by-Step Solution
Step 1: Understand the Wheatstone Bridge Configuration
In a Wheatstone bridge, we have four resistors forming a quadrilateral, and a galvanometer connected between the two opposite corners. Here, each of the four resistors (P, Q, R, and S) is given as 10 Ω.
Step 2: Identify the Balance Condition
The Wheatstone bridge is said to be balanced if
$ \frac{P}{Q} = \frac{S}{R} $.
In our case, each of P, Q, R, and S is 10 Ω, so
$ \frac{10}{10} = \frac{10}{10} = 1 $.
Thus, the bridge is perfectly balanced and no current flows through the galvanometer.
Step 3: Simplify the Balanced Bridge
Because no current flows through the galvanometer in a balanced Wheatstone bridge, we can effectively ignore the galvanometer branch for calculating the equivalent resistance across the battery. Hence, the circuit reduces to two parallel combinations of resistors, each branch having two 10 Ω resistors in series (making 20 Ω on each branch). These two 20 Ω branches are then connected in parallel.
Step 4: Calculate the Equivalent Resistance
The combined resistance of two parallel 20 Ω resistors is given by:
$$
R_{eq}
= \frac{(20)\times(20)}{20 + 20}
= \frac{400}{40}
= 10 \,\Omega .
$$
Step 5: State the Final Answer
The equivalent resistance of the entire arrangement across the battery is therefore 10 Ω.
