WHEATSTONE BRIDGE
Wheatstone bridge is an arrangement of four resistances which can be used to measure one of them in terms of rest. Here arms AB and BC are called ratio arm and arms AC and BD are called conjugate arms.
Balanced Wheatstone bridge: The bridge is said to be balanced when deflection in galvanometer is zero i.e. no current flows through the galvanometer or in other words VB = VD. In the balanced condition P/Q = R/S, on mutually changing the position of cell and galvanometer this condition will not change.
Unbalanced Wheatstone bridge: If the bridge is not balanced current will flow from D to B if VD > VB i.e. (VA – VD) < (VA – VB) which gives PS > RQ.
Applications of Wheatstone bridge: Meter bridge, post office box and Carey Foster bridge are instruments based on the principle of Wheatstone bridge and are used to measure unknown resistance.
METER BRIDGE: In case of Meter Bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, bridge is balanced. If in balanced position of bridge AB = l,
BC = (100 – l)
So that .Q/P = (100–l)/l
Also P/Q = R/S ⇒ S = (100–l)/l R
Solved example 1: In Wheatstone bridge P = 9 ohm, Q = 11 ohm, R = 4 ohm and S = 6 ohm. How much resistance must be put in parallel to the resistance S to balance the bridge
(A) 24 ohm (B) 44/9 ohm (C) 26.4 ohm (D) 18.7 ohm
Solution: (C) (For balancing bridge)
⇒ S' = 4×11/9 = 44/9 ⇒ 1/S' = 1/r + 1/6
⇒ 9/44 – 1/6 = 1/r ⇒ r = 132/5 = 26.4 Ω
Solved example 2: A voltmeter having a resistance of 998 ohms is connected to a cell of emf 2 volt and internal resistance 2 ohm. The error in the measurement of emf will be
(A) 4 ×10–1 volt (B) 2 ×10–3 volt
(C) 4 ×10–3 volt (D) 2 ×10–1 volt
Solution: (C) Error in measurement = Actual value – Measured value
Actual value = 2A
i = 2/998+2 = 1/500 A
Since E = V + ir = ⇒ V = E – ir = 2 – 1/500 × 2 = 998/500 V
Measured value = 998/500 V ⇒ Error = 2 – 998/500 = 4 × 10–3 volt.
