Potentiometer Calculations | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

Potentiometer Calculations

The potential gradient is the p.d. change per unit length of the uniform wire.

At balance, the test e.m.f. equals the p.d. along its balance length.

From wire voltage to test e.m.f.

k=\frac{V_{\text{wire}}}{L_{\text{wire}}}

Symbols

$k$ = potential gradient (\text{V m}^{-1})
$V_{\text{wire}}$ = p.d. across the wire (V)
$L_{\text{wire}}$ = wire length (m)

E_x=kL_x

Symbols

$E_x$ = test-cell e.m.f. (V)
$k$ = potential gradient (\text{V m}^{-1})
$L_x$ = balance length (m)

Worked example

One balance length

A 1.20 V p.d. is maintained across a 1.50 m wire. A cell balances at 0.640 m. Then $k=1.20/1.50=0.800\ \text{V m}^{-1}$ and $E=0.800\times0.640=0.512\ \text{V}$ .

Compare two cells with the same gradient

Balance cell X and note the balance point. Then put cell Y in its place and balance again. The driver and the wire did not change, so the two e.m.f.s are in the same ratio as the two lengths.

\frac{E_x}{E_y}=\frac{L_x}{L_y}

Symbols

$E_x, E_y$ = cell e.m.f.s (V)
$L_x, L_y$ = balance lengths at the same gradient (m)

Extra internal resistance in the driver circuit lowers its current, the wire p.d. and the potential gradient.
The same test e.m.f. then needs a longer balance length. If the required length is beyond the wire, no null point is possible.

Your turn9702/23/M/J/25 Q7(b)(ii–iii)

Two cells balance at 48.0 cm and 64.0 cm with the driver circuit unchanged. The first e.m.f. is 1.20 V. Calculate the second e.m.f. Then state the effect of increasing the driver cell's internal resistance on its balance length. [3] (9702/23/M/J/25 Q7)

Show worked answer

$E_2=1.20\times64.0/48.0=1.60\ \text{V}$ . Increasing the driver internal resistance lowers the wire gradient, so the balance length becomes larger.

Potentiometer Calculations

The potential gradient is the p.d. change per unit length of the uniform wire.

At balance, the test e.m.f. equals the p.d. along its balance length.

From wire voltage to test e.m.f.

k=\frac{V_{\text{wire}}}{L_{\text{wire}}}

Symbols

$k$ = potential gradient (\text{V m}^{-1})
$V_{\text{wire}}$ = p.d. across the wire (V)
$L_{\text{wire}}$ = wire length (m)

E_x=kL_x

Symbols

$E_x$ = test-cell e.m.f. (V)
$k$ = potential gradient (\text{V m}^{-1})
$L_x$ = balance length (m)

Worked example

One balance length

A 1.20 V p.d. is maintained across a 1.50 m wire. A cell balances at 0.640 m. Then $k=1.20/1.50=0.800\ \text{V m}^{-1}$ and $E=0.800\times0.640=0.512\ \text{V}$ .

Compare two cells with the same gradient

Balance cell X and note the balance point. Then put cell Y in its place and balance again. The driver and the wire did not change, so the two e.m.f.s are in the same ratio as the two lengths.

\frac{E_x}{E_y}=\frac{L_x}{L_y}

Symbols

$E_x, E_y$ = cell e.m.f.s (V)
$L_x, L_y$ = balance lengths at the same gradient (m)

Extra internal resistance in the driver circuit lowers its current, the wire p.d. and the potential gradient.
The same test e.m.f. then needs a longer balance length. If the required length is beyond the wire, no null point is possible.

Your turn9702/23/M/J/25 Q7(b)(ii–iii)

Show worked answer

$E_2=1.20\times64.0/48.0=1.60\ \text{V}$ . Increasing the driver internal resistance lowers the wire gradient, so the balance length becomes larger.