Wheatstone bridge
A Wheatstone bridge is balanced when P/Q = R/S, and then no current flows through the galvanometer. With P = 10 Ω, Q = 20 Ω and R = 15 Ω it balances at S = QR/P = 30 Ω.
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| R | I | V | Power | |
|---|---|---|---|---|
| P | 10 Ω | 0.2 A | 2 V | 0.4 W |
| Q | 20 Ω | 0.2 A | 4 V | 0.8 W |
| R | 15 Ω | 0.13 A | 2 V | 0.27 W |
| S | 30 Ω | 0.13 A | 4 V | 0.53 W |
| G | 5 Ω | 0 A | 0 V | 0 W |
Step by step
- P/Q = 10 / 20 = 0.5 and R/S = 15 / 30 = 0.5.
- The ratios are equal, so B and D sit at the same potential and no current flows through the galvanometer.
- With G carrying nothing, P and Q are in series (10 + 20 = 30 Ω) and so are R and S (15 + 30 = 45 Ω). The two paths in parallel give 18 Ω between A and C.
- Cell current: I = E / R = 6 / 18 = 1/3 ≈ 0.33 A.
Balance it
P from A to B, Q from B to C, R from A to D, S from D to C, the galvanometer G from B to D. It balances when P/Q = R/S; then no current flows through G.
- P/Q = 10 / 20 = 0.5 and R/S = 15 / 30 = 0.5.
- The ratios are equal, so B and D sit at the same potential and no current flows through the galvanometer.
- With G carrying nothing, P and Q are in series (10 + 20 = 30 Ω) and so are R and S (15 + 30 = 45 Ω). The two paths in parallel give 18 Ω between A and C.
- Cell current: I = E / R = 6 / 18 = 1/3 ≈ 0.33 A.
Why it works
- Four resistors P, Q, R and S form a loop ABCD, with a cell across A and C and a galvanometer across B and D.
- At balance the galvanometer reads zero, so B and D are at the same potential.
- Then the drop across P equals the drop across R, and across Q equals across S: I₁P = I₂R and I₁Q = I₂S.
- Dividing the two gives P/Q = R/S.
- Knowing three of the resistances gives the fourth: S = QR/P. A meter bridge uses exactly this.
Wheatstone bridge: quick answers
What is the balance condition of a Wheatstone bridge?
P/Q = R/S. Then the points B and D are at the same potential and the galvanometer shows no deflection.
Can the cell and the galvanometer be swapped?
Yes. The balance condition stays P/Q = R/S, so a balanced bridge stays balanced.
When is the bridge most sensitive?
When the four resistances are of the same order. Then a small change in S gives the largest galvanometer current.