A student studies the Wheatstone bridge in the figure…

A student studies the Wheatstone bridge in the figure. The arms are $AB = 10\ \Omega$ , $BC = 20\ \Omega$ and $AD = 15\ \Omega$ . The arm $DC$ is a resistance box S. A galvanometer of resistance $20\ \Omega$ joins B to D. The cell has emf $6.0\ \text{V}$ and internal resistance $1.0\ \Omega$ , with its positive terminal towards C. First she sets S so that the galvanometer shows no deflection. Then she changes S to $24\ \Omega$ .

A Wheatstone bridge drawn as a diamond with corners A on the left, B at the top, C on the right and D at the bottom. Resistor AB is 10 ohm, BC is 20 ohm, AD is 15 ohm, and DC is a resistance box marked S. A galvanometer of resistance 20 ohm joins B to D. A cell of emf 6.0 V and internal resistance 1.0 ohm joins A to C below the diamond, its positive terminal towards C.

Take the connecting wires as having no resistance. For the current in the galvanometer, call it positive when it flows from B to D. Use Kirchhoff's rules or the potentials of the junctions, and keep the internal resistance of the cell in your working, because it changes the potential difference at C a little. The galvanometer is part of the network once the bridge is unbalanced.

What value of S gives no deflection?
With S set to $24\ \Omega$ , what current does the cell supply?
With S set to $24\ \Omega$ , what is the current in the galvanometer? Give its sign.
With S set to $24\ \Omega$ , what is the potential difference $V_B - V_D$ across the galvanometer?
With S set to $24\ \Omega$ , what is the potential difference between the terminals of the cell?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

(a) Balance. No current in the galvanometer means $V_B = V_D$ . Then $\dfrac{AB}{BC} = \dfrac{AD}{DC}$ , so $\dfrac{10}{20} = \dfrac{15}{S}$ and $S = 30\ \Omega$ .

Method for (b) to (e). Take A as the zero of potential. Let the potentials of B, C and D be $V_B$ , $V_C$ and $V_D$ . The cell gives $V_C = E - Ir$ . Write the junction rule at B, at D and at C, and solve the three equations. The galvanometer sits between B and D, so it is part of the network.

(b) The cell. Solving gives $V_C = 5.665\ \text{V}$ , so $I = \dfrac{6.0 - 5.665}{1.0}\ \text{A} = 0.335\ \text{A}$ .

(c) The galvanometer. The potentials are $V_B = 1.942\ \text{V}$ and $V_D = 2.104\ \text{V}$ . So the current is $\dfrac{V_B - V_D}{G} = \dfrac{-0.162\ \text{V}}{20\ \Omega} = -8.1\ \text{mA}$ , which is from D to B.

(d) Across the galvanometer. $V_B - V_D = -0.162\ \text{V}$ .

(e) Cell terminals. The terminal potential difference is $V_C - V_A = 5.665\ \text{V}$ . It is less than the emf $6.0\ \text{V}$ because the cell supplies current through its internal resistance.

Answer: (a) $30\ \Omega$ . (b) $0.335\ \text{A}$ . (c) $-8.1\ \text{mA}$ . (d) $-0.162\ \text{V}$ . (e) $5.665\ \text{V}$ .

Check. Use a Thevenin equivalent for the galvanometer. With the galvanometer removed, the open potential difference across B and D is $-0.291\ \text{V}$ , and the resistance seen from B and D is $15.900\ \Omega$ . Then $I_G = \dfrac{-0.291\ \text{V}}{(15.900 + 20)\ \Omega} = -8.1\ \text{mA}$ , the same. With $S = 30\ \Omega$ the same method gives exactly zero.

Watch out. A bridge is balanced by the ratio of the resistors, not by their sizes, and the cell and its internal resistance do not change the balance. Once it is unbalanced, the galvanometer's own resistance matters.

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