Questions on current-voltage characteristics

Question 1

State Ohm's law.

Accepted Answer

For a metallic conductor at constant temperature (1), the current through it is directly proportional to the potential difference across it, I ∝ V (1).

Question 2

A point on a component's I–V graph reads V = 4.0 V and I = 0.20 A. Calculate the resistance at that point.

Accepted Answer

R = V/I = 4.0/0.20 (1) R = 20 Ω (1)

Question 3

Sketch the current–voltage characteristic of an ohmic conductor at constant temperature.

Accepted Answer

A straight line passing through the origin, current proportional to voltage (1), with the same gradient for positive and negative pd (1).

Question 4

When the I–V characteristic of a component is measured, the meters are assumed to be ideal. State the resistance of an ideal ammeter and of an ideal voltmeter.

Accepted Answer

Ideal ammeter: zero resistance (1). Ideal voltmeter: infinite resistance (1).

Question 5

State how the resistance of a component is found from a point on its I–V characteristic.

Accepted Answer

R is the ratio V/I of the coordinates at that point, not the gradient of the curve (1).

Question 6

A student plots the characteristic of an ohmic conductor with V on the vertical axis and I on the horizontal axis. Describe the graph obtained and state what its gradient represents.

Accepted Answer

A straight line through the origin (1). The gradient is V/I, which for an ohmic conductor is its (constant) resistance (1).

Question 7

Describe the shape of the I–V characteristic of a filament lamp and explain its shape.

Accepted Answer

The curve is S-shaped through the origin (1) and gets shallower as the voltage rises: for a given increase in V the current increases less (1). As current rises the filament heats up, so its resistance increases, curving the graph away from a straight line (1).

Question 8

Describe the I–V characteristic of a semiconductor diode.

Accepted Answer

In forward bias almost no current flows until about 0.6 V (1), after which the current rises steeply (1). In reverse bias only a negligible current flows, so the diode conducts in only one direction (1).

Question 9

A filament lamp gives the readings V = 2.0 V, I = 0.40 A and V = 8.0 V, I = 0.80 A. Calculate the resistance at each point and comment on the change.

Accepted Answer

At 2.0 V: R = 2.0/0.40 (1) R = 5.0 Ω (1) At 8.0 V: R = 8.0/0.80 = 10 Ω (1) The resistance has increased, because the filament is hotter at the higher voltage (1)

Question 10

A component gives currents of 0.10 A, 0.20 A and 0.30 A at 2.0 V, 4.0 V and 6.0 V. Show that it is ohmic.

Accepted Answer

R = V/I for each pair (1) 2.0/0.10 = 4.0/0.20 = 6.0/0.30 = 20 Ω in every case (1) The resistance is constant, so I ∝ V and the component is ohmic (1)

Question 11

In forward bias a diode passes 35 mA when the pd across it is 0.62 V. In reverse bias a pd of 5.0 V drives a current of only 1.0 μA through it. Calculate the resistance of the diode at each point, and comment on the values.

Accepted Answer

Use of R = V/I with the currents converted to amperes (1); forward: R = 0.62/0.035 = 18 Ω (1); reverse: R = 5.0/(1.0 × 10⁻⁶) = 5.0 × 10⁶ Ω (1); the resistance is hundreds of thousands of times larger in reverse, so the diode conducts in one direction only (1).

Question 12

A student measures the I–V characteristic of a metal wire at 20 °C, then repeats the measurement with the wire held at a steady 80 °C in an oven. Describe and explain how the two graphs differ.

Accepted Answer

Both are straight lines through the origin, since each is measured at a constant temperature (1). The 80 °C line is shallower on an I–V graph: less current at every pd, so a higher resistance (1). At the higher temperature the metal ions vibrate with greater amplitude, so the electrons collide with them more frequently (1).

Question 13

The characteristic of a filament lamp is unchanged in shape when the pd is reversed, but the characteristic of a diode is not. Explain both observations.

Accepted Answer

The filament's heating, and therefore its resistance, depends only on the size of the current, not its direction (1), so reversing the pd reverses the current and traces the same shaped curve (1). A diode conducts only in forward bias, beyond a threshold of about 0.6 V; in reverse it passes almost no current, so the two directions give completely different curves (1).

Question 14

Describe how the resistance of a negative temperature coefficient (NTC) thermistor changes with temperature, and explain this in terms of charge carriers.

Accepted Answer

Its resistance decreases as temperature rises (1). Heating releases many more free charge carriers, electrons, in the semiconductor (1), so more current flows for a given voltage (1); this effect outweighs the increased vibration of the lattice, so the resistance falls (1).

Question 15

A filament lamp operates at 12 V with a current of 2.0 A. Calculate (a) its operating resistance and (b) its power. When barely lit it reads 1.0 V and 0.50 A; calculate its resistance then.

Accepted Answer

(a) R = 12/2.0 = 6.0 Ω (1) (b) P = VI = 12 × 2.0 (1) P = 24 W (1) When barely lit: R = 1.0/0.50 = 2.0 Ω, much lower, as the filament is cool (1)

Question 16

Explain why a filament lamp does not obey Ohm's law.

Accepted Answer

Ohm's law requires constant temperature (1). As the pd increases, the filament current heats it, raising its temperature and hence its resistance (1), so the current is not proportional to the voltage and the component is non-ohmic (1).

Question 17

A control circuit needs a component whose resistance changes by less than 5% between pds of 2.0 V and 6.0 V. Component P passes 0.40 A at 2.0 V and 1.00 A at 6.0 V. Component Q passes 0.100 A at 2.0 V and 0.294 A at 6.0 V. Deduce which component is suitable.

Accepted Answer

P: R = 2.0/0.40 = 5.0 Ω and R = 6.0/1.00 = 6.0 Ω (1); Q: R = 2.0/0.100 = 20.0 Ω and R = 6.0/0.294 = 20.4 Ω (1); P's resistance rises by 20% but Q's by only about 2% (1); Q meets the 5% requirement and P does not, so Q should be chosen (1).

Question 18

Describe an experiment to obtain the full I–V characteristic of a filament lamp, including negative values of pd. Include the circuit used, the measurements taken and how the results are processed.

Accepted Answer

Connect the lamp in series with an ammeter and a variable supply (1). Use a potential divider arrangement so the pd across the lamp can be varied from zero up to the maximum, with a voltmeter connected across the lamp (1). Record pairs of V and I readings over the whole range (1). Reverse the connections to the supply to obtain negative values of V and I (1). Plot I against V for both polarities (1). The graph is a curve through the origin that flattens at larger pds as the filament heats and its resistance rises (1).

Question 19

A student says: 'Below its threshold a diode has no resistance, because there is no current.' Explain the student's error and state what actually happens to the diode's resistance below the threshold.

Accepted Answer

Resistance is the ratio R = V/I (1). Below the threshold the pd is non-zero while the current is almost zero, so the ratio V/I is extremely large, not zero (1). The diode's resistance is therefore very high (it behaves as an open switch) (1).

Question 20

The graph shows the characteristic of a filament lamp. Determine the resistance of the lamp at the point where the current in it is 1.5 A. A student measures the gradient of the graph at that point and gives the result as the resistance. State, with a reason, whether the student is right.

Accepted Answer

Reading the coordinates of the point off the graph, V = 7.5 V when I = 1.5 A (allow 7.4 V to 7.6 V) (1)
R = V/I = 7.5/1.5 (1)
R = 5.0 Ω (allow 4.9 Ω to 5.1 Ω) (1)
The student is wrong, because the characteristic is a curve, so the gradient of the tangent (about 8 Ω here) and the ratio V/I are different quantities (1)