Equivalent resistance

Definition

Equivalent resistance

The resistance of a single resistor that could replace a combination of resistors and draw the same current from the same supply.

Several resistors in series can be replaced, in a calculation, by one resistor with the same overall effect on the circuit.
For resistors in series, the equivalent resistance is the sum of the individual resistances, $Req=R1+R2+R3R_{\text{eq}} = R_1 + R_2 + R_3$ .
This works because the same current flows through every resistor, and the potential differences across them add up to the supply potential difference.
The current drawn from the supply is found by dividing the supply potential difference by the equivalent resistance, $\dfrac{V_{\text{supply}}}{R_{\text{eq}}}$ .
Adding another resistor in series increases the equivalent resistance, so for the same supply the current falls.

The equations you need

All of these equations must be recalled, so learn each one with its units.
Charge: $Q = I t$ , with charge in coulombs, $C\text{C}$ , current in amperes, $A\text{A}$ , and time in seconds, $s\text{s}$ .
Potential difference: $V = I R$ , with potential difference in volts, $V\text{V}$ , and resistance in ohms, $Ω\Omega$ .
Energy from charge: $E = Q V$ , with energy in joules, $J\text{J}$ .
Power from potential difference: $P = V I$ , with power in watts, $W\text{W}$ .
Power from resistance: $P = I^2R$ .
Energy from power: $E = P t$ , with energy in joules and time in seconds, or energy in $h\text{kW h}$ with power in $kW\text{kW}$ and time in hours.

Solving series circuit problems

Add the resistances to find the equivalent resistance of the circuit.
Use $\dfrac{V}{R_{\text{eq}}}$ with the supply potential difference to find the current, which is the same through every resistor.
Use $V = I R$ with each resistor's own resistance to find the potential difference across it.
Use $P = V I$ or $P = I^2R$ to find the power of each resistor or of the whole circuit.
Use $E = P t$ or $E = Q V$ to find the energy transferred in a given time, and $Q = I t$ to find the charge that flows.
Check the answers using conservation of energy.
1. The potential differences across the resistors add up to the supply potential difference.
2. The powers of the resistors add up to the power supplied by the source.

Comparing resistors in series

Every resistor in a series circuit carries the same current, so by $P = I^2R$ the resistor with the larger resistance transfers energy at the greater rate.
The larger resistor also has the larger potential difference across it, so each coulomb transfers more energy in that resistor.
When an extra resistor is added in series, the current falls, so the power of each of the original resistors decreases.

Example

Energy and power in two series resistors

A $V6.0\ \text{V}$ supply is connected in series with a $Ω2.0\ \Omega$ resistor and a $Ω4.0\ \Omega$ resistor for $5.0$ minutes.

R_{\text{eq}} = 2.0\ \Omega + 4.0\ \Omega = 6.0\ \Omega

\dfrac{6.0\ \text{V}}{6.0\ \Omega} = 1.0\ \text{A}

V_1 = 1.0\ \text{A} \times 2.0\ \Omega = 2.0\ \text{V}, \quad V_2 = 1.0\ \text{A} \times 4.0\ \Omega = 4.0\ \text{V}

P_2 = I^2R_2 = (1.0\ \text{A})^2 \times 4.0\ \Omega = 4.0\ \text{W}

P_{\text{total}} = VI = 6.0\ \text{V} \times 1.0\ \text{A} = 6.0\ \text{W}

\times 60 = 300\ \text{s}

6.0\ \text{W} \times 300\ \text{s} = 1800\ \text{J}

1.0\ \text{A} \times 300\ \text{s} = 300\ \text{C}

The energy can be checked with $300\ \text{C} \times 6.0\ \text{V} = 1800\ \text{J}$ , which agrees.

Example

Finding an unknown resistance

A lamp is connected in series with an $Ω8.0\ \Omega$ resistor and a $V12\ \text{V}$ battery, and the potential difference across the lamp is $V4.0\ \text{V}$ .

V_{\text{resistor}} = 12\ \text{V} - 4.0\ \text{V} = 8.0\ \text{V}

\dfrac{8.0\ \text{V}}{8.0\ \Omega} = 1.0\ \text{A}

R_{\text{lamp}} = \dfrac{4.0\ \text{V}}{1.0\ \text{A}} = 4.0\ \Omega

P_{\text{lamp}} = VI = 4.0\ \text{V} \times 1.0\ \text{A} = 4.0\ \text{W}

In $2.0$ minutes, the charge through the lamp and the energy it transfers are found next.

1.0\ \text{A} \times 120\ \text{s} = 120\ \text{C}

120\ \text{C} \times 4.0\ \text{V} = 480\ \text{J}

Common Mistake

Do not use the supply potential difference with the resistance of one resistor, because each value in $V = I R$ must belong to the same part of the circuit.
Do not confuse $E = P t$ with $\dfrac{E}{t}$ when rearranging, and always convert minutes or hours into seconds when working in joules.
Do not round values part way through a calculation, because rounding early can change the final answer.

Exam technique

Multi-step circuit problems

Before you start a long calculation, list the values you know and the quantity you need, then pick the equation that links them.
Show every step with its substitution, because method marks are given even if an earlier slip carries an error forward.
Use a second equation to check your final energy or power where you can, such as $E = P t$ checked against $E = Q V$ .

Self review

State how to find the equivalent resistance of resistors in series.
Write down the equation linking energy, charge and potential difference.
Calculate the current from a $V9.0\ \text{V}$ battery connected to $Ω1.0\ \Omega$ and $Ω2.0\ \Omega$ resistors in series.
Explain which of two resistors in series transfers energy at the greater rate.
Calculate the charge that flows when a current of $A0.50\ \text{A}$ passes for $4.0$ minutes.

Equivalent resistance

Definition

Equivalent resistance

The resistance of a single resistor that could replace a combination of resistors and draw the same current from the same supply.

Several resistors in series can be replaced, in a calculation, by one resistor with the same overall effect on the circuit.
For resistors in series, the equivalent resistance is the sum of the individual resistances, $Req=R1+R2+R3R_{\text{eq}} = R_1 + R_2 + R_3$ .
This works because the same current flows through every resistor, and the potential differences across them add up to the supply potential difference.
The current drawn from the supply is found by dividing the supply potential difference by the equivalent resistance, $\dfrac{V_{\text{supply}}}{R_{\text{eq}}}$ .
Adding another resistor in series increases the equivalent resistance, so for the same supply the current falls.

The equations you need

All of these equations must be recalled, so learn each one with its units.
Charge: $Q = I t$ , with charge in coulombs, $C\text{C}$ , current in amperes, $A\text{A}$ , and time in seconds, $s\text{s}$ .
Potential difference: $V = I R$ , with potential difference in volts, $V\text{V}$ , and resistance in ohms, $Ω\Omega$ .
Energy from charge: $E = Q V$ , with energy in joules, $J\text{J}$ .
Power from potential difference: $P = V I$ , with power in watts, $W\text{W}$ .
Power from resistance: $P = I^2R$ .
Energy from power: $E = P t$ , with energy in joules and time in seconds, or energy in $h\text{kW h}$ with power in $kW\text{kW}$ and time in hours.

Solving series circuit problems

Add the resistances to find the equivalent resistance of the circuit.
Use $\dfrac{V}{R_{\text{eq}}}$ with the supply potential difference to find the current, which is the same through every resistor.
Use $V = I R$ with each resistor's own resistance to find the potential difference across it.
Use $P = V I$ or $P = I^2R$ to find the power of each resistor or of the whole circuit.
Use $E = P t$ or $E = Q V$ to find the energy transferred in a given time, and $Q = I t$ to find the charge that flows.
Check the answers using conservation of energy.
1. The potential differences across the resistors add up to the supply potential difference.
2. The powers of the resistors add up to the power supplied by the source.

Comparing resistors in series

Every resistor in a series circuit carries the same current, so by $P = I^2R$ the resistor with the larger resistance transfers energy at the greater rate.
The larger resistor also has the larger potential difference across it, so each coulomb transfers more energy in that resistor.
When an extra resistor is added in series, the current falls, so the power of each of the original resistors decreases.

Example

Energy and power in two series resistors

A $V6.0\ \text{V}$ supply is connected in series with a $Ω2.0\ \Omega$ resistor and a $Ω4.0\ \Omega$ resistor for $5.0$ minutes.

R_{\text{eq}} = 2.0\ \Omega + 4.0\ \Omega = 6.0\ \Omega

\dfrac{6.0\ \text{V}}{6.0\ \Omega} = 1.0\ \text{A}

V_1 = 1.0\ \text{A} \times 2.0\ \Omega = 2.0\ \text{V}, \quad V_2 = 1.0\ \text{A} \times 4.0\ \Omega = 4.0\ \text{V}

P_2 = I^2R_2 = (1.0\ \text{A})^2 \times 4.0\ \Omega = 4.0\ \text{W}

P_{\text{total}} = VI = 6.0\ \text{V} \times 1.0\ \text{A} = 6.0\ \text{W}

\times 60 = 300\ \text{s}

6.0\ \text{W} \times 300\ \text{s} = 1800\ \text{J}

1.0\ \text{A} \times 300\ \text{s} = 300\ \text{C}

The energy can be checked with $300\ \text{C} \times 6.0\ \text{V} = 1800\ \text{J}$ , which agrees.

Example

Finding an unknown resistance

A lamp is connected in series with an $Ω8.0\ \Omega$ resistor and a $V12\ \text{V}$ battery, and the potential difference across the lamp is $V4.0\ \text{V}$ .

V_{\text{resistor}} = 12\ \text{V} - 4.0\ \text{V} = 8.0\ \text{V}

\dfrac{8.0\ \text{V}}{8.0\ \Omega} = 1.0\ \text{A}

R_{\text{lamp}} = \dfrac{4.0\ \text{V}}{1.0\ \text{A}} = 4.0\ \Omega

P_{\text{lamp}} = VI = 4.0\ \text{V} \times 1.0\ \text{A} = 4.0\ \text{W}

In $2.0$ minutes, the charge through the lamp and the energy it transfers are found next.

1.0\ \text{A} \times 120\ \text{s} = 120\ \text{C}

120\ \text{C} \times 4.0\ \text{V} = 480\ \text{J}

Common Mistake

Do not use the supply potential difference with the resistance of one resistor, because each value in $V = I R$ must belong to the same part of the circuit.
Do not confuse $E = P t$ with $\dfrac{E}{t}$ when rearranging, and always convert minutes or hours into seconds when working in joules.
Do not round values part way through a calculation, because rounding early can change the final answer.

Exam technique

Multi-step circuit problems

Before you start a long calculation, list the values you know and the quantity you need, then pick the equation that links them.
Show every step with its substitution, because method marks are given even if an earlier slip carries an error forward.
Use a second equation to check your final energy or power where you can, such as $E = P t$ checked against $E = Q V$ .

Self review

State how to find the equivalent resistance of resistors in series.
Write down the equation linking energy, charge and potential difference.
Calculate the current from a $V9.0\ \text{V}$ battery connected to $Ω1.0\ \Omega$ and $Ω2.0\ \Omega$ resistors in series.
Explain which of two resistors in series transfers energy at the greater rate.
Calculate the charge that flows when a current of $A0.50\ \text{A}$ passes for $4.0$ minutes.

3.2.13 Energy and power in resistor circuits

Equivalent resistance

The equations you need

Solving series circuit problems

Comparing resistors in series

Energy and power in two series resistors

Finding an unknown resistance

Multi-step circuit problems

Review 3.2.13 Energy and power in resistor circuits by teaching Genie

3.2.13 Energy and power in resistor circuits

Equivalent resistance

The equations you need

Solving series circuit problems

Comparing resistors in series

Energy and power in two series resistors

Finding an unknown resistance

Multi-step circuit problems

Review 3.2.13 Energy and power in resistor circuits by teaching Genie

3.2.13 Energy and power in resistor circuits

Equivalent resistance

The equations you need

Solving series circuit problems

Comparing resistors in series

Energy and power in two series resistors

Finding an unknown resistance

Multi-step circuit problems

Review 3.2.13 Energy and power in resistor circuits by teaching Genie

3.2.13 Energy and power in resistor circuits Revision Guide

3.2.13 Energy and power in resistor circuits

Equivalent resistance

The equations you need

Solving series circuit problems

Comparing resistors in series

Energy and power in two series resistors

Finding an unknown resistance

Multi-step circuit problems

Review 3.2.13 Energy and power in resistor circuits by teaching Genie

3.2.13 Energy and power in resistor circuits Revision Guide