Cells, the Meter Bridge and the Potentiometer: Quick Revision

A brief note on Cells, the Meter Bridge and the Potentiometer. Read it before a test. Try each question at the end before you open its answer.

Cells in combination

• Series, aiding: $E = E_1 + E_2$. Opposing: $E = E_1 - E_2$. Always $r = r_1 + r_2$.
• Charging cell: $V = E + Ir$. Discharging cell: $V = E - Ir$.
• Two cells in parallel: $E = \dfrac{E_1r_2 + E_2r_1}{r_1 + r_2}$ and $r = \dfrac{r_1r_2}{r_1 + r_2}$.
• $m$ rows of $k$ cells: $E = kE_0$ and $r = \dfrac{kr_0}{m}$.
• Greatest current: $R = \dfrac{kr_0}{m}$, so $m = \sqrt{\dfrac{Nr_0}{R}}$.

The meter bridge

• Balance: $\dfrac{X}{R} = \dfrac{l}{100 - l}$, with $l$ measured from the side of $X$.
• Does not depend on: the emf of the cell or the galvanometer.
• Best accuracy: a balance point near $50\ \text{cm}$. The fractional error is $\dfrac{100\,\delta l}{l(100 - l)}$.
• Interchange: the balance point becomes about $100 - l$. Average the two values of $X$ to reduce the end corrections.
• Resistivity: $\rho = \dfrac{\pi d^2 X}{4L}$.

The potentiometer

• Gradient: $k = \dfrac{IR_w}{L}$. Potential: $V = kl$.
• Balance: $E = kl$, so $\dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}$.
• Internal resistance: $r = R\,\dfrac{l_1 - l_2}{l_2}$, with $l_1$ for the open cell.
• True emf: at balance no current is drawn, so nothing is lost inside the cell.
• More sensitive: a smaller gradient, with the balance still on the wire. The driver emf must be larger than the emf measured.

A worked example

A cell balances at $300\ \text{cm}$ when open, and at $250\ \text{cm}$ with a $5.0\ \Omega$ resistor across it. Find its internal resistance.

1. The rule. $r = R\,\dfrac{l_1 - l_2}{l_2}$.
2. The numbers. $r = 5.0\ \Omega \times \dfrac{300 - 250}{250} = 1.0\ \Omega$.

Practice with new numbers

A program to try

The program finds the emf, resistance and current of a group of $20$ cells in $5$ rows.

Mistakes to avoid

• Subtracting resistances for opposing cells. Only the emfs subtract.
• Using $kr$ for a group. The rows are in parallel, so it is $kr/m$.
• Turning the bridge ratio upside down. The length on the side of $X$ goes on top.
• Putting $l_1$ on the bottom in the internal resistance formula. The closed length $l_2$ goes there.

Check yourself

1. What is the emf of two cells of $5.0\ \text{V}$ and $3.0\ \text{V}$ joined in series, opposing?

2. A meter bridge has a $20\ \Omega$ box in the right gap and balances at $25\ \text{cm}$. What is $X$?

3. Two cells balance at $200\ \text{cm}$ and $300\ \text{cm}$. What is the ratio of their emfs?

4. Why can a potentiometer not balance a cell of $3\ \text{V}$ if its driver cell is $2\ \text{V}$?