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Questions on the potential divider and its output
A potential divider is two resistors in series, and the output across one of them is the supply pd shared in the ratio of the resistances. A sliding contact moves that output between zero and the full supply, and along a uniform wire the pd is proportional to length. A load lowers it.
27 original questions · 88 marks · the potential divider and its output notes · Electricity
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These questions cover both lessons on potential dividers: The potential divider and its output and Sensing circuits and the potentiometer.
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Explain why two resistors in series share the supply potential difference between them.
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The same current flows through both, and the pd across each is IR (1). The larger the resistance, the larger its share of the pd, so the supply voltage is divided between them in the ratio of their resistances (1).A 12 V supply is connected across a 4.0 kΩ resistor in series with an 8.0 kΩ resistor. Calculate the potential difference across the 8.0 kΩ resistor.
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Vout = Vin × R2/(R1 + R2) (1)
Vout = 12 × 8.0/(4.0 + 8.0) = 8.0 V (1)State the potential divider equation for the output taken across R2, and the assumption it relies on.
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Vout = Vin × R2/(R1 + R2), assuming the output draws negligible current, any load having a resistance far larger than the divider resistances (1).The output of a potential divider is taken across R2. State the value the output approaches (a) as R2 becomes very small and (b) as R2 becomes very large.
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(a) Zero (1). (b) The full supply pd, Vin (1).In a two-resistor potential divider the output across R2 is exactly one quarter of the supply pd. Determine the ratio R1 : R2.
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R2/(R1 + R2) = 1/4, so R1 + R2 = 4R2 (1); R1 = 3R2, so the ratio is 3 : 1 (1).State how the resistance of a light-dependent resistor changes as the light intensity falling on it increases.
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The resistance falls as the light intensity rises (1).A 10 V supply is connected across a 2.0 kΩ resistor (R1) in series with a 3.0 kΩ resistor (R2). Calculate the output potential difference taken across R2.
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Vout = 10 × 3.0/(2.0 + 3.0) (1)
Vout = 6.0 V (1)A potential divider has a 4.0 kΩ resistor as R1 and a 9.0 V supply. Calculate the value of R2 needed to give an output of 3.0 V across it.
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3.0 = 9.0 × R2/(4.0 + R2) (1)
3.0(4.0 + R2) = 9.0R2 (1)
R2 = 2.0 kΩ (1)A thermistor is connected in series with a fixed 4.0 kΩ resistor across a 6.0 V supply, with the output taken across the fixed resistor. When warm, the thermistor's resistance is 2.0 kΩ. Calculate the output pd, and state how it changes as the thermistor cools.
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Vout = 6.0 × 4.0/(2.0 + 4.0) (1)
Vout = 4.0 V (1)
As it cools the thermistor's resistance rises, taking a larger share of the pd (1)
The output across the fixed resistor therefore falls (1)Describe how a potentiometer (a potential divider with a sliding contact) can provide a continuously variable output voltage.
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A single resistive track is connected across the supply, and a sliding wiper taps off the output between one end and the wiper (1). Moving the wiper changes the proportion of the track, and so the resistance, in the output section (1), varying the output smoothly from zero up to the full supply voltage (1).A 12 V supply is connected across a uniform resistance track 20 cm long. A sliding contact touches the track 15 cm from the bottom end, and the output is taken between the bottom end and the contact. Calculate the output pd.
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For a uniform track the resistance, and therefore the pd, is proportional to length (1); Vout = 12 × 15/20 = 9.0 V (1).A thermistor is connected in series with a fixed 2.0 kΩ resistor across a 12 V supply. The output, taken across the fixed resistor, is measured as 8.0 V. Determine the resistance of the thermistor.
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pd across the thermistor = 12 − 8.0 = 4.0 V (1); current in the chain I = 8.0/2000 = 4.0 mA (1); R = 4.0/(4.0 × 10−3) = 1.0 kΩ (1).A potential divider consists of a 68 kΩ resistor and a 12 kΩ resistor in series across a 20 V supply. Calculate the current in the divider chain and hence the output pd across the 12 kΩ resistor.
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Total resistance = 68 + 12 = 80 kΩ (1); I = 20/80 kΩ = 0.25 mA (1); Vout = IR = 0.25 mA × 12 kΩ = 3.0 V (1).A potentiometer is built from a uniform resistance wire 100.0 cm long with a driving cell across its ends. A standard cell of emf 1.02 V balances at 51.0 cm. Without disturbing the driving circuit, the standard cell is replaced by a cell of unknown emf, which balances at 32.5 cm. Calculate the emf of the unknown cell.
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The wire is uniform, so the pd tapped off is proportional to the balance length and E1/E2 = l1/l2 (1). E2 = 1.02 × 32.5/51.0 (1) = 0.650 V (1).A fan is to switch on when a room gets too warm. An ntc thermistor is connected above a fixed resistor across a 9.0 V supply, with the output taken across the fixed resistor, and the switching unit turns the fan on when the output reaches half the supply pd. At the temperature at which the fan should switch on the thermistor's resistance is 8.0 kΩ. Determine the fixed resistance required, and explain whether the fan switches on above or below that temperature.
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The output is half the supply when the two resistances are equal, so the fixed resistor must be 8.0 kΩ, giving Vout = 9.0 × 8.0/(8.0 + 8.0) = 4.5 V (1)
Above that temperature the thermistor is warmer, so its resistance is lower and its share of the supply pd is smaller (1)
The share across the fixed resistor, which is the output, therefore rises past half the supply, so the fan switches on above that temperature (1)In a potentiometer experiment the galvanometer deflects in the same direction when the jockey is touched to the wire at each end in turn, so no balance point can be found. Suggest two reasons for this, and state one change that would allow a balance point to be found.
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The pd across the whole length of the wire is smaller than the emf being measured, so no length of it can balance that emf (1). Alternatively the cell being measured is connected the wrong way round, so its emf adds to the pd along the wire instead of opposing it (1). Either increase the pd across the wire, by using a driving cell of larger emf, or reverse the connections of the cell being measured (1).A street lamp is to light itself at dusk. A light-dependent resistor is connected in series with a fixed resistor across a 9.0 V supply, with the output taken across the LDR, and the switching unit turns the lamp on when the output reaches 6.0 V. At the light level at which the lamp should come on the LDR's resistance is 12 kΩ. Determine the fixed resistance required, and explain whether the lamp comes on as the light fades or as it brightens.
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Vout = Vin × RLDR/(RLDR + Rfixed), so 6.0 = 9.0 × 12/(12 + Rfixed) (1)
12 + Rfixed = 9.0 × 12/6.0 = 18 kΩ (1)
Rfixed = 6.0 kΩ, which gives Vout = 9.0 × 12/(12 + 6.0) = 6.0 V (1)
As the light fades the LDR's resistance rises, so its share of the supply pd, which is the output, rises past 6.0 V and the lamp comes on; in brighter light the LDR's resistance is lower, so with 3.0 kΩ the output is only 3.0 V and the lamp stays off (1)A light-dependent resistor (LDR) is placed in series with a fixed 2.0 kΩ resistor across a 6.0 V supply, with the output taken across the fixed resistor. The LDR's resistance is 10 kΩ in the dark and 1.0 kΩ in bright light. Calculate the output in each case.
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Dark: Vout = 6.0 × 2.0/(10 + 2.0) (1)
Vout = 1.0 V (1)
Light: Vout = 6.0 × 2.0/(1.0 + 2.0) = 4.0 V (1)
The output rises as the light gets brighter (1)Explain one advantage of using a potential divider rather than a variable resistor (rheostat) to control the voltage supplied to a component.
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A potential divider can deliver any output from zero up to the full supply voltage (1), whereas a series rheostat can only reduce the voltage and never bring it down to zero (1). The divider therefore gives full, controllable range (1).An automatic porch light must be off in daylight and on in darkness. An LDR (resistance 300 Ω in bright light, 90 kΩ in darkness) is connected above a fixed 5.0 kΩ resistor across a 5.0 V supply, and the output is taken across the LDR. The lamp switches on when the output exceeds 2.5 V. Deduce whether the circuit behaves as intended.
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Darkness: Vout = 5.0 × 90/(90 + 5.0) = 4.7 V (1); bright light: Vout = 5.0 × 0.30/(0.30 + 5.0) = 0.28 V (1); 4.7 V is above 2.5 V, so the lamp is on in darkness, and 0.28 V is below 2.5 V, so it is off in daylight (1); the circuit therefore behaves as intended (1).A potential divider is built from two equal 6.0 kΩ resistors across a 9.0 V supply, so the output across the lower resistor is 4.5 V. A third 6.0 kΩ resistor is now connected in parallel with the lower resistor. Determine the new output pd.
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The lower section becomes 6.0 kΩ in parallel with 6.0 kΩ = 3.0 kΩ (1); Vout = 9.0 × 3.0/(6.0 + 3.0) (1) = 3.0 V (1).A grower wants a circuit whose output voltage rises as a greenhouse cools, so that a switching circuit can turn a heater on during cold nights. Design a suitable potential divider using an ntc thermistor, explaining fully how it works. State which component the output is taken across and what could be changed to adjust the temperature at which the heater comes on.
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Connect the thermistor in series with a fixed resistor across the supply (1). Take the output across the thermistor (1). As the greenhouse cools, the ntc thermistor's resistance rises (1). Its share of the supply pd therefore rises, so the output voltage rises, which is the required direction (1). The output feeds the switching circuit, which turns the heater on when the output exceeds its threshold (1). Using a variable resistor in place of the fixed resistor (or changing its value) alters the ratio, setting the temperature at which the switch-on threshold is reached (1).In a potentiometer experiment the cell being measured is connected between one end of the uniform wire and a sliding contact, through a galvanometer, with its emf opposing the pd along the wire. The contact is moved until the galvanometer reads exactly zero. Explain why this null method gives the emf of the cell rather than its terminal potential difference, and state why the galvanometer needs no calibration.
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At the balance point the galvanometer reads zero, so no current flows in the branch containing the cell (1). With no current there is no lost pd across the cell's own internal resistance, so the pd it supplies is its full emf (1). The branch also draws no current from the wire, so the wire is not loaded and the pd tapped off is the plain length ratio of the driving pd (1). The galvanometer is only ever used to locate a zero and never to read a value, so any error in its calibration drops out of the result entirely (1).A thermistor whose resistance is 10 kΩ at 20 °C and 1.0 kΩ at 60 °C is connected in series with a fixed resistor across a 12 V supply, with the output taken across the fixed resistor. Two values are available for the fixed resistor, 1.0 kΩ and 3.2 kΩ. Calculate the change in the output pd between 20 °C and 60 °C for each value, and deduce which value gives the more sensitive circuit.
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With the 1.0 kΩ resistor: Vout = 12 × 1.0/(10 + 1.0) = 1.09 V at 20 °C and 12 × 1.0/(1.0 + 1.0) = 6.00 V at 60 °C (1)
Change = 4.91 V (1)
With the 3.2 kΩ resistor: Vout = 12 × 3.2/(10 + 3.2) = 2.91 V at 20 °C and 12 × 3.2/(1.0 + 3.2) = 9.14 V at 60 °C (1)
Change = 6.23 V (1)
The 3.2 kΩ resistor gives the larger change in output for the same change in temperature, so it makes the sensing circuit more sensitive and its output easier for a switching unit to distinguish (1)The sensor input of a frost controller is a logic input: it reads LOW when the potential difference applied to it is 0.8 V or less, reads HIGH when that pd is 2.0 V or more, and is undefined between those values. An ntc thermistor is connected in series with a fixed 3.0 kΩ resistor across a 5.0 V supply, and the output, taken across the fixed resistor, is connected to that input. The thermistor's resistance is 20 kΩ at 5 °C and 2.0 kΩ at 40 °C. Calculate the output pd at each of those two temperatures, and deduce whether the input reads a valid logic level at both.
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At 5 °C: Vout = 5.0 × 3.0/(20 + 3.0) = 0.65 V (1)
0.65 V is below 0.8 V, so the input reads LOW at 5 °C (1)
At 40 °C: the thermistor is warmer, so its resistance is lower and its share of the supply pd smaller; Vout = 5.0 × 3.0/(2.0 + 3.0) = 3.0 V (1)
3.0 V is above 2.0 V, so the input reads HIGH at 40 °C (1)
Both ends of the range give a valid level, so the controller can tell cold from warm. At temperatures in between the output passes through the band from 0.8 V to 2.0 V, where the input's reading is undefined, so the circuit sets no single switching temperature (1)Two potential dividers are connected side by side across the same 9.0 V supply. In the first, an ntc thermistor is connected above a fixed 4.7 kΩ resistor. In the second, a fixed 4.7 kΩ resistor is connected above a variable resistor set to 4.7 kΩ. The output of the circuit is the potential difference between the two junctions, the potential of the first junction minus that of the second. The thermistor's resistance is 4.7 kΩ at 25 °C and 1.5 kΩ at 60 °C. Calculate the output at each of those two temperatures, and explain two advantages of this arrangement over taking the output from the thermistor divider alone.
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At 25 °C the thermistor and the fixed resistor are equal, so the first junction sits at V = 9.0 × 4.7/(4.7 + 4.7) = 4.5 V (1)
The second divider also has two equal resistances, so its junction sits at 4.5 V as well, and the output is 4.5 − 4.5 = 0 V (1)
At 60 °C the thermistor is warmer, so its resistance and its share of the supply pd are smaller and the first junction rises: V = 9.0 × 4.7/(1.5 + 4.7) = 6.8 V (1)
The second junction is unchanged, so the output is 6.8 − 4.5 = 2.3 V (1)
The output starts from zero at the temperature the variable resistor is set for, so the whole of it is the change the warming produced, instead of a small change riding on a standing 4.5 V, and moving the variable resistor sets the temperature at which the output is zero (1)
The zero also survives a change in the supply pd: both junctions are the same fraction of the supply at 25 °C, so a supply that drifts moves them together and the output stays zero at the same temperature, whereas the output of a single divider drifts with the supply (1)The circuit shown is a potential divider across a 9.0 V supply, with the output taken at the terminals marked. Determine the output potential difference with the switch S open, and again with S closed.
A potential divider whose output is taken across the upper resistor, with a load that can be switched across the output terminals. Mark scheme
From the diagram the output is taken across the upper 3.0 kΩ resistor, so that resistance is the numerator of the ratio (1)
With S open, Vout = 9.0 × 3.0/(3.0 + 6.0) (1)
Vout = 3.0 V (1)
With S closed the load is in parallel with the 3.0 kΩ resistor, (6.0 × 3.0)/(6.0 + 3.0) = 2.0 kΩ, and the ratio is re-applied (1)
Vout = 9.0 × 2.0/(2.0 + 6.0) = 2.3 V (2.25 V), below the unloaded value (1)
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