Series and parallel circuits and energy in circuits
By the end of this chapter you will be able to
- Add resistances in series, and use the total resistance with V = IR
- [H] Find the total resistance of two resistors in parallel, and say why it is less than the smaller one
- Explain what happens to the current and pd when a component is added in series or in parallel
- Calculate the power and the energy transferred in a circuit
- [H] Calculate the output pd of a potential divider
You should already know
- Ohm’s law: R = V / I (Chapter 6)
- Rules for current and pd in series and parallel (Chapter 5)
Where this leads
- The mains supply and the cost of energy (Chapter 8)
- Electromagnetism and the transformer (Chapter 20)
GCSE Physics. Content follows the GCSE (9-1) subject content for England (Edexcel and AQA). Content references: Calculate the total resistance of resistors in series (R = R1 + R2), [H] Calculate the total resistance of two resistors in parallel (1/R = 1/R1 + 1/R2), Predict how adding components changes the current, the pd and the brightness of lamps, [H] Use a potential divider to find an output pd, and describe LDR and thermistor sensor circuits, Calculate power using P = VI and P = I squared R, and energy using E = Pt.
Exercise P Do you remember?
Method. Use I = V / R.
- Divide. I = 5 / 10 = 0.5 A.5 / 10 = 0.5
- Write the answer. 0.5 A
Check. 0.5 x 10 = 5 V. ✓
Method. Recall the series rule.
- Recall. In series the current is the same at every point.current
- Write the answer. current
Check. ✓
Method. Multiply the numbers.
- Multiply. 12 x 2 = 24, then 24 x 3 = 72.12 x 2 x 3 = 72
- Write the answer. 72
Check. 72 / 3 = 24. ✓
Method. Add the fractions with the same denominator.
- Add. 1/4 + 1/4 = 2/4 = 1/2.2/4 = 1/2
- Write the answer. 1/2
Check. 1/2 = 0.5. ✓
Method. Recall the unit.
- Recall. Power is measured in watts, W, which is joules per second.watt (W)
- Write the answer. watt
Check. ✓
Series circuits and total resistance
In a series circuit, the components are joined one after another in a single loop. The current is the same at every point. The total resistance is the sum of the resistances:
Rtotal = R1 + R2 + ...
Adding another component in series increases the total resistance, so the current falls for the same cell pd.
Key ideas
- Series: R total = R1 + R2 + ...
- Series: the current is the same at every point.
- Series: the pd of the cell is shared between the components.
- Adding a component in series increases the total resistance and reduces the current.
Method. Add the resistances.
- Add. R total = 4 + 6 = 10 ohm.4 + 6 = 10
- Check. The total resistance is bigger than either resistor alone.10 > 6
- Write the answer. 10 ohm
Check. ✓
Method. Use I = V / R with the total resistance.
- Take the numbers. V = 12 V and R total = 10 ohm.V = 12, R = 10
- Divide. I = 12 / 10 = 1.2 A.12 / 10 = 1.2
- Write the answer. 1.2 A
Check. 1.2 x 10 = 12 V. ✓
Method. Add all three resistances.
- Read. The resistances are 5 ohm, 12 ohm and 3 ohm.5, 12, 3
- Add. R total = 5 + 12 + 3 = 20 ohm.5 + 12 + 3 = 20
- Write the answer. 20 ohm
Check. ✓
Watch out
Watch out: in series, add the resistances. Do not add the reciprocals (that is for parallel, in Section 7.2).
Exam tip
Exam tip. In a series circuit, say “the current is the same” and “the total resistance is the sum”.
Exercise 7A Series resistance and current
Method. Add the resistances.
- Add. 2 + 3 = 5 ohm.2 + 3 = 5
- Write the answer. 5 ohm
Check. ✓
Method. Add the resistances.
- Add. 10 + 15 = 25 ohm.10 + 15 = 25
- Write the answer. 25 ohm
Check. ✓
Method. Add the three resistances.
- Add. 4 + 4 + 4 = 12 ohm.4 + 4 + 4 = 12
- Write the answer. 12 ohm
Check. ✓
Method. Find the total resistance, then use I = V / R.
- Total. R = 6 + 4 = 10 ohm.10 ohm
- Divide. I = 10 / 10 = 1 A.10 / 10 = 1
- Write the answer. 1 A
Check. ✓
Method. Total resistance is 8 ohm; use I = V / R.
- Total. R = 2 + 6 = 8 ohm.8 ohm
- Divide. I = 4 / 8 = 0.5 A.4 / 8 = 0.5
- Write the answer. 0.5 A
Check. 0.5 x 8 = 4 V. ✓
Method. The current is the same at every point in series.
- Recall. The current is the same everywhere, so 0.25 A.0.25 A
- Write the answer. 0.25 A
Check. ✓
Method. The pd is shared equally between identical components.
- Divide. 6 / 2 = 3 V each.6 / 2 = 3
- Write the answer. 3 V
Check. 3 + 3 = 6 V. ✓
Method. Use V = I x R for the resistor.
- Multiply. V = 0.5 x 9 = 4.5 V.0.5 x 9 = 4.5
- Write the answer. 4.5 V
Check. 4.5 / 9 = 0.5 A. ✓
Method. Adding resistance in series increases the total.
- Recall. The total resistance rises, so the current falls for the same cell pd.decreases
- Write the answer. decreases
Check. ✓
Method. Think about the loop.
- Loop. In series there is one loop, and removing a part breaks it.one loop
- Current. With no complete loop, there is no current.no current
- Write the answer. Accept any sentence that says the loop is broken, so there is no current.
Check. ✓
Method. Add the resistances.
- Check. For 4 ohm and 6 ohm in series, the total is 10 ohm, which is more than 6 ohm.10 > 6
- Write the answer. no
Check. ✓
Method. Add all four resistances, then use I = V / R.
- Total. R = 1 + 2 + 3 + 4 = 10 ohm.10 ohm
- Divide. I = 20 / 10 = 2 A.20 / 10 = 2
- Write the answer. 2 A
Check. 2 x 10 = 20 V. ✓
Method. Think about the total resistance and the current.
- Resistance. Adding a lamp increases the total resistance.rises
- Current. So the current falls and the lamp is dimmer.dimmer
- Write the answer. Accept any sentence that says the total resistance rises, so the current falls, and the lamp is dimmer.
Check. ✓
Parallel circuits and total resistance
In a parallel circuit, each branch has the same pd across it as the cell. The current splits at each junction, so the total current is the sum of the branch currents.
Adding a branch in parallel decreases the total resistance, because there is more than one path for the current.
[H] Higher tier only. For two resistors in parallel the total resistance is found from 1 / Rtotal = 1 / R1 + 1 / R2. The total is always less than the smallest resistor. Take the reciprocal at the end: R = 1 / (sum of reciprocals).
Key ideas
- Parallel: each branch has the full pd of the cell.
- Parallel: the total current is the sum of the branch currents.
- [H] Parallel: 1 / R total = 1 / R1 + 1 / R2.
- Adding a branch in parallel reduces the total resistance and increases the total current.
Method. Use 1/R = 1/R1 + 1/R2, then take the reciprocal.
- Reciprocals. 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2.1/R = 1/2
- Invert. R = 2 ohm.R = 2
- Write the answer. 2 ohm
Check. The total (2 ohm) is less than the smallest resistor (3 ohm). ✓
Method. Use 1/R = 1/R1 + 1/R2.
- Reciprocals. 1/R = 1/4 + 1/4 = 2/4 = 1/2.1/R = 1/2
- Invert. R = 2 ohm.R = 2
- Write the answer. 2 ohm
Check. Two equal resistors in parallel give half the resistance. ✓
Method. Use 1/R = 1/12 + 1/4.
- Reciprocals. 1/R = 1/12 + 3/12 = 4/12 = 1/3.1/R = 1/3
- Invert. R = 3 ohm.R = 3
- Write the answer. 3 ohm
Check. Check: 12 V / 3 ohm = 4 A, which is 1 A + 3 A. ✓
Watch out
Watch out: [H] when you use 1/R, you find the reciprocal first. The answer is R, so invert the fraction at the end. Writing R = 1/6 + 1/3 is wrong.
Exam tip
Exam tip. [H] Check that your parallel answer is smaller than the smallest resistor. If it is not, you have made a mistake.
Exercise 7B Parallel resistance and branch currents
Method. Each branch has the same pd as the cell.
- Recall. Each branch in parallel has the full cell pd.6 V
- Write the answer. 6 V
Check. ✓
Method. Use I = V / R for each branch and add.
- 6 ohm. I = 6 / 6 = 1 A.1 A
- 3 ohm. I = 6 / 3 = 2 A.2 A
- Total. 1 + 2 = 3 A.1 + 2 = 3
- Write the answer. 3 A
Check. 3 A from the cell, 6 V / 2 ohm = 3 A. ✓
Method. Add the branch currents.
- Add. 0.2 + 0.7 = 0.9 A.0.2 + 0.7 = 0.9
- Write the answer. 0.9 A
Check. ✓
Method. Use 1/R = 1/10 + 1/10.
- Reciprocals. 1/R = 2/10 = 1/5.1/R = 1/5
- Invert. R = 5 ohm.R = 5
- Write the answer. 5 ohm
Check. ✓
Method. Use 1/R = 1/2 + 1/2.
- Reciprocals. 1/R = 1/2 + 1/2 = 1.1/R = 1
- Invert. R = 1 ohm.R = 1
- Write the answer. 1 ohm
Check. ✓
Method. Use 1/R = 1/12 + 1/4.
- Reciprocals. 1/12 + 3/12 = 4/12 = 1/3.1/3
- Invert. R = 3 ohm.R = 3
- Write the answer. 3 ohm
Check. ✓
Method. Use 1/R = 1/15 + 1/5.
- Reciprocals. 1/15 + 3/15 = 4/15.4/15
- Invert. R = 15 / 4 = 3.75 ohm.15 / 4 = 3.75
- Write the answer. 3.75 ohm
Check. Smaller than 5 ohm. ✓
Method. Adding a branch in parallel adds another path.
- Recall. More paths reduce the total resistance.decrease
- Write the answer. decrease
Check. ✓
Method. Think about the paths for the current.
- Paths. The current can use either branch.two paths
- Result. So the combined resistance is lower than either.lower
- Write the answer. Accept any sentence that says the current has two paths, so the combined resistance is lower.
Check. ✓
Method. Parallel branches each have the full pd.
- Recall. Each lamp has 12 V across it.12 V
- Write the answer. 12 V
Check. ✓
Method. Use I = V / R for each lamp.
- 3 ohm. I = 6 / 3 = 2 A.2 A
- 6 ohm. I = 6 / 6 = 1 A.1 A
- Compare. 2 A is more than 1 A.2 > 1
- Write the answer. 3
Check. ✓
Method. Find the total resistance first, then the current.
- Total R. 1/R = 1/3 + 1/6 = 3/6 = 1/2, so R = 2 ohm.2 ohm
- Current. I = 6 / 2 = 3 A.6 / 2 = 3
- Write the answer. 3 A
Check. The current from the branches is 2 A + 1 A = 3 A. ✓
Power and energy in circuits
The power of a component is the energy it transfers each second. It can be found from the pd and the current:
P = V x I
and, when the resistance is known, from the current and the resistance:
P = I2 x R
The energy transferred in time t is E = P x t.
| Equation | Use it when you know | Unit check |
|---|---|---|
| P = V x I | the pd and the current | V x A = W |
| P = I x I x R (I squared R) | the current and the resistance | A squared x ohm = W |
| P = V squared / R | the pd and the resistance (higher tier) | V squared / ohm = W |
| E = P x t | the power and the time | W x s = J |
Key ideas
- P = V x I, in watts when V is in volts and I in amperes.
- P = I squared x R, with the current squared.
- E = P x t, with t in seconds, gives energy in joules.
- Higher tier: P = V squared / R can also be used [H].
Method. Use P = V x I.
- Take the numbers. V = 230 V and I = 0.5 A.V = 230, I = 0.5
- Multiply. P = 230 x 0.5 = 115 W.230 x 0.5 = 115
- Write the answer. 115 W
Check. 115 / 230 = 0.5 A. ✓
Method. Use P = I x I x R.
- Square. I squared = 2 x 2 = 4.2 squared = 4
- Multiply. P = 4 x 5 = 20 W.4 x 5 = 20
- Write the answer. 20 W
Check. 20 / 2 = 10 V, and 2 x 10 = 20 W. ✓
Method. Use E = P x t, with time in seconds.
- Convert. 10 minutes = 600 s.600 s
- Multiply. E = 60 x 600 = 36000 J.60 x 600 = 36000
- Write the answer. 36000 J
Check. 36000 / 600 = 60 W. ✓
Watch out
Watch out: use the current squared in P = I squared R. 2 A gives I squared = 4, not 2 x 2 = 2 times 2 the power.
Exam tip
Exam tip. Write the equation, then the numbers with units, then calculate. Check the unit of your answer.
Exercise 7C Power and energy in circuits
Method. Use P = V x I.
- Multiply. P = 6 x 2 = 12 W.6 x 2 = 12
- Write the answer. 12 W
Check. ✓
Method. Use P = V x I.
- Multiply. P = 12 x 0.5 = 6 W.12 x 0.5 = 6
- Write the answer. 6 W
Check. ✓
Method. Use P = V x I.
- Multiply. P = 230 x 10 = 2300 W.230 x 10 = 2300
- Write the answer. 2300 W
Check. ✓
Method. Use P = I x I x R.
- Square. 4 squared = 16.16
- Multiply. P = 16 x 3 = 48 W.16 x 3 = 48
- Write the answer. 48 W
Check. 48 / 4 = 12 V, and 4 x 12 = 48 W. ✓
Method. Use P = I squared R.
- Square. 0.5 squared = 0.25.0.25
- Multiply. P = 0.25 x 10 = 2.5 W.0.25 x 10 = 2.5
- Write the answer. 2.5 W
Check. ✓
Method. Use E = P x t.
- Multiply. E = 40 x 30 = 1200 J.40 x 30 = 1200
- Write the answer. 1200 J
Check. 1200 / 30 = 40 W. ✓
Method. Convert the time, then use E = P x t.
- Convert. 2 minutes = 120 s.120 s
- Multiply. E = 2000 x 120 = 240000 J.2000 x 120 = 240000
- Write the answer. 240000 J
Check. 240000 / 120 = 2000 W. ✓
Method. Find the power, then the energy.
- Power. P = 12 x 2 = 24 W.24 W
- Energy. E = 24 x 120 = 2880 J.24 x 120 = 2880
- Write the answer. 2880 J
Check. 2880 / 120 = 24 W. ✓
Method. Compare the energy in 60 s.
- Heater. E = 1000 x 60 = 60000 J.60000 J
- Lamp. E = 60 x 60 = 3600 J.3600 J
- Write the answer. heater
Check. ✓
Method. Use P = I squared R.
- Square. 3 squared = 9.9
- Multiply. P = 9 x 8 = 72 W.9 x 8 = 72
- Write the answer. 72 W
Check. 72 / 8 = 9 = 3 squared. ✓
Method. Square the pd, then divide by R.
- Square. 12 squared = 144.144
- Divide. P = 144 / 12 = 12 W.144 / 12 = 12
- Write the answer. 12 W
Check. 12 V / 12 ohm = 1 A, and 12 x 1 = 12 W. ✓
Method. Use the definition of power.
- Define. Power is energy transferred each second.each second
- Link. A brighter lamp transfers more energy to light each second.more energy
- Write the answer. Accept any sentence that says the lamp transfers more energy each second.
Check. ✓
Method. Find the power, then E = P t.
- Power. P = 6 x 0.25 = 1.5 W.1.5 W
- Energy. E = 1.5 x 60 = 90 J.1.5 x 60 = 90
- Write the answer. 90 J
Check. 90 / 60 = 1.5 W. ✓
Potential dividers and sensors
[H] Higher tier only. In a potential divider, two resistors share the pd of the supply. The output pd across R2 is found from Vout = Vin x R2 / (R1 + R2). The larger resistor takes the larger share of the pd.
A potential divider is a series circuit used to give an output pd. If one resistor is a thermistor or an LDR, the output pd changes with temperature or light. The changing pd can then switch a circuit on or off.
| Sensor | Input quantity | Resistance change | Output |
|---|---|---|---|
| thermistor | temperature | falls as it gets hotter | pd across it rises or falls |
| LDR | light level | falls in brighter light | pd across it rises or falls |
Key ideas
- [H] Vout = Vin x R2 / (R1 + R2).
- [H] The bigger the resistance, the bigger the share of the pd.
- A thermistor or an LDR is used as one resistor, so the output pd changes with the input.
- The output pd can be used to switch a circuit on at a set level.
Method. Use Vout = Vin x R2 / (R1 + R2).
- Take the numbers. Vin = 9 V, R1 = 20 ohm, R2 = 10 ohm.Vin = 9, R1 = 20, R2 = 10
- Calculate. Vout = 9 x 10 / (20 + 10) = 90 / 30 = 3 V.90 / 30 = 3
- Write the answer. 3 V
Check. R1 takes the other 6 V, which is twice as big: 6 = 2 x 3. ✓
Method. Use the potential divider formula.
- Take the numbers. Vin = 6 V, R1 = 100 ohm, R2 = 200 ohm.Vin = 6
- Calculate. Vout = 6 x 200 / (100 + 200) = 1200 / 300 = 4 V.1200 / 300 = 4
- Write the answer. 4 V
Check. 4 V + 2 V = 6 V. ✓
Method. Think about the share of the pd.
- Look. The thermistor is R2, and its resistance falls when hot.R2 falls
- Decide. A smaller R2 takes a smaller share of the pd, so the output falls.falls
- Write the answer. falls
Check. Use the formula: with R2 smaller, Vout is smaller. ✓
Watch out
Watch out: [H] the output pd is taken across R2, not across R1. Check which resistor is across the voltmeter before you use the formula.
Exam tip
Exam tip. [H] Write the formula, then substitute, then check that the output is less than the input.
Exercise 7D Potential dividers and sensor circuits
Method. Equal resistors share the pd equally.
- Calculate. Vout = 10 x 5 / (5 + 5) = 50 / 10 = 5 V.50 / 10 = 5
- Write the answer. 5 V
Check. ✓
Method. Use the formula.
- Calculate. Vout = 12 x 10 / (30 + 10) = 120 / 40 = 3 V.120 / 40 = 3
- Write the answer. 3 V
Check. ✓
Method. Use the formula.
- Calculate. Vout = 10 x 40 / (10 + 40) = 400 / 50 = 8 V.400 / 50 = 8
- Write the answer. 8 V
Check. The other 2 V is across R1. ✓
Method. The smaller resistor takes a smaller share of the pd.
- Recall. The thermistor has less resistance, so less pd across it.falls
- Write the answer. falls
Check. ✓
Method. Smaller LDR resistance takes a smaller share.
- Recall. The LDR now has less resistance, so the pd across it falls.falls
- Write the answer. falls
Check. ✓
Method. Recall the light sensor.
- Recall. A light-dependent resistor (LDR) changes resistance with light.LDR
- Write the answer. LDR
Check. ✓
Method. Use R2 / (R1 + R2) with R1 = R2.
- Fraction. R2 / (R1 + R2) = R / (R + R) = 1/2.1/2
- Write the answer. 1/2
Check. ✓
Method. Half the pd across R2 means the resistors are equal.
- Look. Vout / Vin = 6 / 12 = 1/2, so R2 = R1.1/2
- Ratio. R2 / R1 = 1.R2 / R1 = 1
- Write the answer. 1
Check. ✓
Method. Think about the LDR in the dark.
- Dark. In the dark the LDR resistance rises.rises
- Switch. Its share of the pd rises, which switches on the light.switches on
- Write the answer. Accept any sentence that says the LDR resistance rises in the dark, so its pd rises and switches on the light.
Check. ✓
Method. Think about what changes.
- Resistance. The thermistor resistance changes with temperature.changes
- Share. So its share of the pd changes too.share changes
- Write the answer. Accept any answer that says the thermistor resistance changes with temperature, so its share of the pd changes.
Check. ✓
Chapter summary
| Skill | Method |
|---|---|
| Series resistance | R = R1 + R2 + ...; current is the same; pd is shared |
| [H] Parallel resistance | 1/R = 1/R1 + 1/R2; the total is smaller than the smallest resistor |
| Power | P = V x I; P = I squared x R; E = P x t |
| [H] Potential divider | Vout = Vin x R2 / (R1 + R2); thermistors and LDRs change the output |
Can you…?
- I can add resistances in series
- [H] I can find the total resistance of two resistors in parallel
- I can calculate power using P = VI and P = I squared R
- I can calculate energy using E = Pt
- [H] I can calculate the output pd of a potential divider
- I can explain how an LDR or thermistor changes a circuit
Key terms
- series circuit
- a circuit with one loop; the current is the same at every point
- parallel circuit
- a circuit with branches; each branch has the full pd of the cell
- power
- the rate of energy transfer, in watts
- potential divider
- two resistors in series that share the pd of a supply to give an output pd
- [H] reciprocal
- one divided by a number; used in 1/R for parallel resistors
Exercise 7Z Check-up
Method. Add the resistances.
- Add. 7 + 5 = 12 ohm.7 + 5 = 12
- Write the answer. 12 ohm
Check. ✓
Method. Total R = 15 ohm, then I = V / R.
- Total. R = 10 + 5 = 15 ohm.15 ohm
- Divide. I = 15 / 15 = 1 A.15 / 15 = 1
- Write the answer. 1 A
Check. ✓
Method. Use 1/R = 1/6 + 1/6.
- Reciprocals. 1/R = 2/6 = 1/3.1/3
- Invert. R = 3 ohm.R = 3
- Write the answer. 3 ohm
Check. ✓
Method. Use P = V x I.
- Multiply. P = 4 x 3 = 12 W.4 x 3 = 12
- Write the answer. 12 W
Check. ✓
Method. Square the current, then multiply by R.
- Square. 3 squared = 9.9
- Multiply. P = 9 x 2 = 18 W.9 x 2 = 18
- Write the answer. 18 W
Check. ✓
Method. Convert 5 minutes to 300 s, then E = P t.
- Convert. 5 minutes = 300 s.300 s
- Multiply. E = 40 x 300 = 12000 J.40 x 300 = 12000
- Write the answer. 12000 J
Check. ✓
Method. More resistance in series.
- Recall. The total resistance rises, so the current falls.decrease
- Write the answer. decrease
Check. ✓
Method. Recall the parallel rule.
- Recall. Each branch has the full pd of the cell.pd
- Write the answer. pd
Check. ✓
Method. Substitute the values.
- Calculate. Vout = 10 x 10 / 20 = 5 V.100 / 20 = 5
- Write the answer. 5 V
Check. ✓
Method. Think about the share.
- Share. The pd is shared in the ratio of the resistances.ratio
- Write the answer. Accept any sentence that says the share of the pd depends on the ratio of the resistances.
Check. ✓
Total 28 marks. Allow about 28 minutes.
Exercise 7X Exam-style questions
Method. Add all three.
- Add. 3 + 7 + 10 = 20 ohm.3 + 7 + 10 = 20
- Write the answer. 20 ohm
Check. ✓
Method. Convert minutes to seconds, then E = P t.
- Convert. 25 minutes = 1500 s.1500 s
- Multiply. E = 60 x 1500 = 90000 J.60 x 1500 = 90000
- Write the answer. 90000 J
Check. ✓
Method. Use 1/R = 1/6 + 1/12.
- Reciprocals. 1/R = 2/12 + 1/12 = 3/12 = 1/4.1/4
- Invert. R = 4 ohm.R = 4
- Write the answer. 4 ohm
Check. ✓
Method. Find the total resistance, the current, then the pd.
- Total. R = 5 + 10 = 15 ohm.15 ohm
- Current. I = 9 / 15 = 0.6 A.0.6 A
- pd. V across 10 ohm = 0.6 x 10 = 6 V.6 V
- Write the answer. 0.6, 6
Check. 5 x 0.6 + 10 x 0.6 = 9 V. ✓
Method. Find the power in watts, then convert.
- Power. P = 230 x 8 = 1840 W.1840 W
- Convert. 1840 W = 1.84 kW.1.84 kW
- Write the answer. 1.84 kW
Check. ✓
Method. Think about the LDR share of the pd.
- Resistance. In brighter light the LDR resistance falls.falls
- Share. So it takes a smaller share of the pd.smaller share
- Output. The pd across the LDR falls.pd falls
- Write the answer. Accept any answer that says the LDR resistance falls, so its share of the pd falls.
Check. ✓
Method. Compare the total resistance.
- Resistance. Adding a lamp in series increases the total resistance.increases
- Current. For the same cell pd, the current is smaller.smaller
- Brightness. A smaller current means less power, so the lamp is dimmer.dimmer
- Write the answer. Accept any answer that says the total resistance is higher, so the current is smaller.
Check. ✓
Method. Use Vout = Vin x R2 / (R1 + R2).
- Substitute. Vout = 12 x 8 / (4 + 8).12 x 8 / 12
- Calculate. Vout = 96 / 12 = 8 V.96 / 12 = 8
- Write the answer. 8 V
Check. The 4 ohm resistor takes 4 V. ✓
Method. Use P = I squared R.
- Square. 1.5 squared = 2.25.2.25
- Multiply. P = 2.25 x 5 = 11.25 W.2.25 x 5 = 11.25
- Write the answer. 11.25 W
Check. 11.25 / 5 = 2.25 = 1.5 squared. ✓
Method. Think about the pd across each lamp.
- Pd. Each parallel lamp keeps the full cell pd.full pd
- Brightness. So the brightness of each lamp is unchanged.unchanged
- Total. The total current from the cell increases.total increases
- Verdict. So the claim is not true for the lamps.not true
- Write the answer. Accept any answer that says each lamp keeps the full cell pd, so their brightness is unchanged, and the total current from the cell increases.
Check. ✓