Why Parallel Resistors Have Equal Voltage but Different Power

Connect a 100 Ω resistor and a 200 Ω resistor in parallel across a 6.0 V supply. A voltmeter across either resistor reads exactly 6.0 V. Does that mean both resistors dissipate the same power?

No. The 100 Ω resistor dissipates 0.36 W, while the 200 Ω resistor dissipates 0.18 W. The lower-resistance resistor draws twice the current, so it transfers electrical energy twice as quickly.

This is one of the most important distinctions in parallel circuits: equal voltage does not mean equal current or equal power. To understand why, you need to look at what conservation of charge and energy actually tells you about a circuit.

01First, understand what the circuit guarantees

Imagine two resistors connected between the same two junctions, A and B.

Both resistors start at junction A and finish at junction B. This means they have the same potential difference across them.

However, the current can split when it reaches junction A and recombine at junction B. The amount of current in each branch depends on its resistance.

Two conservation laws explain why this happens.

1. Conservation of charge: Kirchhoff’s current law

Charge cannot simply disappear at a junction.

If current enters a junction, the total current leaving must equal the total current entering, provided there is no charge accumulating there.

For two parallel branches:

\[ I_{\text{total}}=I_1+I_2 \]

Here:

• \(I_{\text{total}}\) is the current supplied to the junction.
• \(I_1\) is the current through the first branch.
• \(I_2\) is the current through the second branch.

The current splits between the branches according to their resistances.

It does not necessarily split equally.

For example, a 100 Ω resistor will draw more current than a 200 Ω resistor if both have the same voltage across them.

2. Conservation of energy: Kirchhoff’s voltage law

Now consider the voltage.

Voltage represents the change in electrical potential energy per unit charge. A charge travelling around a complete circuit cannot gain or lose net energy after returning to its starting point.

This is the basis of Kirchhoff’s voltage law.

The sum of the potential differences around a complete loop is zero.

Using the convention that a rise in potential is positive and a drop is negative:

\[ +\mathcal{E}-V_1=0 \]

where:

• \(\mathcal{E}\) is the supply’s electromotive force (emf), in volts.
• \(V_1\) is the voltage drop across the first branch resistor.

For an ideal supply with negligible wire resistance:

\[ V_1=\mathcal{E} \]

The same reasoning applies to the second branch:

\[ V_2=\mathcal{E} \]

Therefore:

\[ \boxed{V_1=V_2} \]

Both resistors have the same voltage because they connect to the same two junctions.

Importantly, this does not mean every resistor within a multi-resistor branch receives the full supply voltage. It is the voltage across the complete branch that equals the supply’s terminal voltage.

What about equivalent resistance?

Once you understand the junction and loop relationships, you can simplify the parallel network.

For two resistors in parallel:

\[ \frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2} \]

Here, \(R_p\) is the equivalent resistance of the parallel combination.

This equation helps calculate total current. It does not remove the need to calculate individual branch currents when finding power.

For a refresher on the broader method, see Parallel Circuits: Voltage, Current, Resistance and Power.

02Why equal voltage does not mean equal power

The key is understanding what power actually measures.

Voltage tells you how much energy is transferred per coulomb of charge.

Current tells you how much charge passes a point each second.

Power tells you how much energy is transferred each second.

The relationship is:

\[ P=VI \]

where:

• \(P\) is electrical power, measured in watts (W).
• \(V\) is potential difference, measured in volts (V).
• \(I\) is current, measured in amperes (A).

You can see why voltage alone is insufficient.

Two resistors can transfer the same amount of energy per coulomb while allowing different amounts of charge to pass each second.

Imagine two conveyor belts carrying boxes that each contain the same amount of energy. The belt moving more boxes each second transfers more energy per second.

In a circuit, the charge is not used up as it passes through a resistor. The analogy simply illustrates the difference between energy per unit of charge and the rate of charge flow.

Deriving the power relationship

For an ohmic resistor, Ohm’s law gives:

\[ V=IR \]

Rearranging:

\[ I=\frac{V}{R} \]

Substitute this into the power equation:

\[ P=VI \]

\[ \boxed{P=\frac{V^2}{R}} \]

This is the crucial equation for comparing resistors in parallel.

If two resistors have the same voltage across them, their powers depend on their resistances.

\[ P\propto\frac{1}{R} \]

This means:

• A smaller resistance dissipates more power at the same voltage.
• A larger resistance dissipates less power at the same voltage.
• If both voltage and resistance are equal, their powers are equal.

There is another useful form:

\[ P=I^2R \]

This form is particularly useful when the current through a resistor is known.

Be careful: the statement that a lower resistance dissipates more power is only valid when the voltage is held constant. At equal current, the opposite relationship applies: a larger resistance dissipates more power.

Comparing the three power equations

For parallel resistors, \(P=V^2/R\) is often the quickest option because the branch voltages are equal.

03Test the prediction: a parallel resistor investigation

Consider an investigation designed to test whether equal voltage produces equal power.

Use the following apparatus:

• A regulated 6.0 V DC power supply.
• A 100 Ω resistor rated at least 1 W.
• A 200 Ω resistor rated at least 1 W.
• Two ammeters.
• One voltmeter.
• Connecting wires.

Connect the positive terminal of the power supply to junction A.

From A, construct two parallel branches. Place one ammeter in series with the 100 Ω resistor in the first branch, and the second ammeter in series with the 200 Ω resistor in the other branch.

Join the branches at junction B and connect B back to the negative terminal.

Finally, connect the voltmeter across A and B.

The ammeters measure the current in each individual branch. The voltmeter measures the common potential difference across both resistors.

What should the investigation show?

Before calculating, predict the readings.

Both resistors are connected across the same two junctions, so the voltmeter should read 6.0 V across each.

But should the ammeters show equal currents?

No. Using Ohm’s law:

For the 100 Ω resistor:

\[ I_1=\frac{6.0}{100} =0.060\text{ A} \]

For the 200 Ω resistor:

\[ I_2=\frac{6.0}{200} =0.030\text{ A} \]

The first branch carries twice the current.

We can now calculate the power in each resistor.

\[ P_1=VI_1 \]

\[ P_1=6.0\times0.060 \]

\[ \boxed{P_1=0.36\text{ W}} \]

For the second resistor:

\[ P_2=VI_2 \]

\[ P_2=6.0\times0.030 \]

\[ \boxed{P_2=0.18\text{ W}} \]

The predicted readings are summarised below.

Illustrative data: calculated ideal readings, not recorded experimental measurements.

The total current is:

\[ I_{\text{total}}=0.060+0.030 \]

\[ I_{\text{total}}=0.090\text{ A} \]

This agrees with conservation of charge at the junction.

The total power supplied is:

\[ P_{\text{total}}=VI_{\text{total}} \]

\[ P_{\text{total}}=6.0\times0.090 =0.54\text{ W} \]

The sum of the individual resistor powers is also:

\[ P_1+P_2=0.36+0.18 =0.54\text{ W} \]

Energy is accounted for.

Why measuring total current alone is insufficient

Imagine a first experimental design that places just one ammeter in the main supply wire.

It would measure the total current of 0.090 A, but it would not reveal how that current divides between the two branches.

A student might incorrectly assume that each branch receives half the total current.

To improve the design, place an ammeter in each branch, or move a single ammeter between the branches while keeping the supply voltage stable. Measuring both branch currents simultaneously is preferable because it avoids errors caused by changes in the supply between readings.

In a real experiment, small differences from the predicted values may occur because of meter resistance, wire resistance, supply internal resistance, and changes in resistor temperature.

The calculations above assume an ideal supply, negligible meter and wire resistance, and constant ohmic resistance.

06Questions and solutions

Question 1

An ideal 9.0 V supply is connected directly across two ohmic resistors, 18 Ω and 36 Ω, in parallel.

Calculate:

a) The current through each resistor.

b) The power dissipated by each resistor.

c) Which resistor dissipates more power, and by what factor?

Solution 1

The 18 Ω resistor dissipates twice the power of the 36 Ω resistor. Both resistors have 9.0 V across them, but the 18 Ω resistor draws twice the current.

a) Calculate the branch currents.

Because the resistors are in parallel:

\[ V_{18}=V_{36}=9.0\text{ V} \]

Using Ohm’s law:

\[ I_{18}=\frac{9.0}{18} =0.50\text{ A} \]

\[ I_{36}=\frac{9.0}{36} =0.25\text{ A} \]

b) Calculate the power in each resistor.

Using \(P=VI\):

\[ P_{18}=9.0\times0.50 \]

\[ \boxed{P_{18}=4.5\text{ W}} \]

For the 36 Ω resistor:

\[ P_{36}=9.0\times0.25 \]

\[ \boxed{P_{36}=2.25\text{ W}} \]

c) Compare the powers.

\[ \frac{P_{18}}{P_{36}} = \frac{4.5}{2.25} =2 \]

Therefore, the 18 Ω resistor dissipates twice the power.

The important point is that equal voltage does not force equal power. The lower-resistance branch carries more current, resulting in a greater rate of energy transfer.

Question 2

An ideal 18.0 V supply is connected to a 3.0 Ω resistor in series with a parallel combination of 9.0 Ω and 18.0 Ω resistors.

Calculate:

a) The equivalent resistance of the parallel combination.

b) The total current supplied.

c) The voltage across each parallel resistor.

d) The current and power in each parallel resistor.

A student claims that each parallel resistor receives the full 18.0 V supply voltage. Explain the error.

Solution 2

The 9.0 Ω resistor dissipates 16 W, and the 18.0 Ω resistor dissipates 8 W. Neither receives the full 18.0 V supply voltage because the series resistor also has a voltage drop.

a) Calculate the equivalent parallel resistance.

\[ \frac{1}{R_p} = \frac{1}{9}+\frac{1}{18} \]

\[ \frac{1}{R_p}=\frac{3}{18} \]

\[ \boxed{R_p=6.0\ \Omega} \]

b) Calculate the total circuit resistance and current.

The 3.0 Ω resistor is in series with the parallel combination.

\[ R_{\text{total}}=3.0+6.0 \]

\[ R_{\text{total}}=9.0\ \Omega \]

Using Ohm’s law:

\[ I_{\text{total}}=\frac{18.0}{9.0} \]

\[ \boxed{I_{\text{total}}=2.0\text{ A}} \]

c) Calculate the voltage across the parallel network.

The voltage drop across the series resistor is:

\[ V_3=IR \]

\[ V_3=2.0\times3.0=6.0\text{ V} \]

Using Kirchhoff’s voltage law:

\[ V_p=18.0-6.0 \]

\[ \boxed{V_p=12.0\text{ V}} \]

Both parallel resistors therefore have 12.0 V across them.

The student’s error is assuming that the supply voltage must appear across every resistor. Only the complete parallel network receives the remaining 12.0 V.

d) Calculate the branch currents and powers.

For the 9.0 Ω resistor:

\[ I_9=\frac{12.0}{9.0} =1.33\text{ A} \]

\[ P_9=\frac{12.0^2}{9.0} \]

\[ \boxed{P_9=16\text{ W}} \]

For the 18.0 Ω resistor:

\[ I_{18}=\frac{12.0}{18.0} =0.667\text{ A} \]

\[ P_{18}=\frac{12.0^2}{18.0} \]

\[ \boxed{P_{18}=8.0\text{ W}} \]

The junction law provides a useful check:

\[ 1.33+0.667\approx2.0\text{ A} \]

The branch currents add to the total current, as required.

Question 3

An ideal 20.0 V supply is connected across two parallel branches.

• Branch A contains a 2.0 Ω resistor in series with an 8.0 Ω resistor.
• Branch B contains a single 10.0 Ω resistor.

Calculate:

a) The current through each branch.

b) The voltage across each of the three resistors.

c) The power dissipated by each resistor.

A student argues that all three resistors must have 20.0 V across them because they are in a parallel circuit. Explain why this reasoning is incorrect.

Solution 3

The two complete branches each carry 2.0 A, but the three individual resistors do not have equal voltages. Their powers are 8.0 W, 32 W, and 40 W respectively.

The key distinction is between a parallel branch and the individual components inside that branch.

a) Calculate the current through each branch.

For Branch A, the two series resistors combine:

\[ R_A=2.0+8.0 \]

\[ R_A=10.0\ \Omega \]

For Branch B:

\[ R_B=10.0\ \Omega \]

Both complete branches have 20.0 V across them.

Therefore:

\[ I_A=\frac{20.0}{10.0} =2.0\text{ A} \]

\[ I_B=\frac{20.0}{10.0} =2.0\text{ A} \]

b) Calculate the voltage across each resistor.

In Branch A, the two resistors are in series, so they carry the same current of 2.0 A.

For the 2.0 Ω resistor:

\[ V_2=IR \]

\[ V_2=2.0\times2.0 \]

\[ \boxed{V_2=4.0\text{ V}} \]

For the 8.0 Ω resistor:

\[ V_8=2.0\times8.0 \]

\[ \boxed{V_8=16.0\text{ V}} \]

For the 10.0 Ω resistor in Branch B:

\[ \boxed{V_{10}=20.0\text{ V}} \]

Check the voltage drops around Branch A:

\[ 4.0+16.0=20.0\text{ V} \]

This satisfies Kirchhoff’s voltage law.

c) Calculate the power dissipated by each resistor.

For the 2.0 Ω resistor:

\[ P_2=I^2R \]

\[ P_2=2.0^2\times2.0 \]

\[ \boxed{P_2=8.0\text{ W}} \]

For the 8.0 Ω resistor:

\[ P_8=2.0^2\times8.0 \]

\[ \boxed{P_8=32\text{ W}} \]

For the 10.0 Ω resistor in Branch B:

\[ P_{10}=\frac{V^2}{R} \]

\[ P_{10}=\frac{20.0^2}{10.0} \]

\[ \boxed{P_{10}=40\text{ W}} \]

The student’s mistake is treating every resistor inside a parallel branch as though it connects directly across the supply.

Only the complete branches share the same voltage. The individual series resistors divide their branch’s voltage according to their resistances.

Question 4

An ideal 12.0 V supply is connected directly across two resistors, 6.0 Ω and 12.0 Ω, in parallel.

A third resistor, 4.0 Ω, is then added in parallel with the original two resistors.

Calculate:

a) The initial total current and total power supplied.

b) The new total current and total power supplied.

c) The power dissipated by the original 6.0 Ω resistor before and after the additional resistor is connected.

d) Explain why the total power increases while the power dissipated by the original 6.0 Ω resistor remains unchanged.

Solution 4

The total current doubles from 3.0 A to 6.0 A, and the total power doubles from 36 W to 72 W. However, the power in the original 6.0 Ω resistor remains at 24 W.

The reason is that an ideal voltage supply maintains the same potential difference across every parallel branch, even when another branch is added.

a) Calculate the initial total current and power.

Initially, the 6.0 Ω and 12.0 Ω resistors are in parallel.

Their currents are:

\[ I_6=\frac{12.0}{6.0}=2.0\text{ A} \]

\[ I_{12}=\frac{12.0}{12.0}=1.0\text{ A} \]

Using Kirchhoff’s current law:

\[ I_{\text{total}}=2.0+1.0 \]

\[ \boxed{I_{\text{total}}=3.0\text{ A}} \]

The total power supplied is:

\[ P_{\text{total}}=VI \]

\[ P_{\text{total}}=12.0\times3.0 \]

\[ \boxed{P_{\text{total}}=36\text{ W}} \]

b) Calculate the new total current and power.

The additional 4.0 Ω resistor is connected across the same two junctions.

Its current is:

\[ I_4=\frac{12.0}{4.0} =3.0\text{ A} \]

The new total current is:

\[ I_{\text{total}}=2.0+1.0+3.0 \]

\[ \boxed{I_{\text{total}}=6.0\text{ A}} \]

The new total power supplied is:

\[ P_{\text{total}}=12.0\times6.0 \]

\[ \boxed{P_{\text{total}}=72\text{ W}} \]

c) Compare the power in the original 6.0 Ω resistor.

Before adding the new branch:

\[ P_6=\frac{12.0^2}{6.0} \]

\[ P_6=24\text{ W} \]

After adding the new branch:

\[ P_6=\frac{12.0^2}{6.0} \]

\[ P_6=24\text{ W} \]

Therefore:

\[ \boxed{\Delta P_6=0\text{ W}} \]

d) Explain the result.

The additional branch increases the total current drawn from the supply, but it does not change the voltage across the existing 6.0 Ω resistor.

Its resistance remains 6.0 Ω, and its voltage remains 12.0 V.

Consequently, its power remains unchanged.

The extra power supplied by the source is used by the additional resistor:

\[ P_4=\frac{12.0^2}{4.0} =36\text{ W} \]

The total power therefore increases by 36 W, even though the original branches continue operating at exactly the same power.

This result depends on the ideal voltage supply assumption. A real source with internal resistance behaves differently, as the next question demonstrates.

Question 5

A battery has an emf of 12.0 V and an internal resistance of 2.0 Ω.

It supplies two external parallel branches:

• Branch A contains a selectable resistor, \(R_A\).
• Branch B contains a fixed 6.0 Ω resistor.

Initially, \(R_A=3.0\ \Omega\). A switch then changes the resistance in Branch A to \(6.0\ \Omega\), without disconnecting Branch B.

Assume the battery’s emf and internal resistance remain constant, the external resistors are ohmic, and the circuit has reached steady state.

Calculate and compare the two operating states.

a) Find the total current, terminal voltage, and power dissipated by each external resistor when \(R_A=3.0\ \Omega\).

b) Repeat the calculations when \(R_A=6.0\ \Omega\).

c) A student makes two predictions:

• Doubling \(R_A\) must halve its power.
• The power in Branch B must remain constant because its resistance has not changed.

Evaluate both predictions.

d) Calculate the total power delivered by the battery and the power dissipated by its internal resistance in both states. Explain how the results satisfy conservation of energy.

Solution 5

Both student predictions are incorrect. When \(R_A\) increases from 3.0 Ω to 6.0 Ω, its power decreases from 12 W to 8.64 W, rather than halving. Meanwhile, the power in the unchanged 6.0 Ω resistor increases from 6.0 W to 8.64 W.

The reason is that the battery has internal resistance. Changing one external branch changes the total current, which changes the voltage lost inside the battery and therefore changes the voltage across both external branches.

a) Calculate the initial operating state when \(R_A=3.0\ \Omega\).

First, calculate the equivalent resistance of the external parallel network:

\[ \frac{1}{R_p}= \frac{1}{3.0}+\frac{1}{6.0} \]

\[ \boxed{R_p=2.0\ \Omega} \]

The internal resistance is in series with the external network.

Therefore:

\[ R_{\text{total}}=2.0+2.0 \]

\[ R_{\text{total}}=4.0\ \Omega \]

The total current supplied by the battery is:

\[ I_{\text{total}}=\frac{\mathcal E}{R_{\text{total}}} \]

\[ I_{\text{total}}=\frac{12.0}{4.0} \]

\[ \boxed{I_{\text{total}}=3.0\text{ A}} \]

Now calculate the terminal voltage.

The terminal voltage is the emf minus the voltage drop across the internal resistance:

\[ V_t=\mathcal E-I_{\text{total}}r \]

where \(r\) is the internal resistance.

\[ V_t=12.0-(3.0\times2.0) \]

\[ \boxed{V_t=6.0\text{ V}} \]

Both external branches therefore have 6.0 V across them.

For Branch A:

\[ I_A=\frac{6.0}{3.0} =2.0\text{ A} \]

\[ P_A=VI=6.0\times2.0 \]

\[ \boxed{P_A=12\text{ W}} \]

For Branch B:

\[ I_B=\frac{6.0}{6.0} =1.0\text{ A} \]

\[ P_B=6.0\times1.0 \]

\[ \boxed{P_B=6.0\text{ W}} \]

The branch currents add to 3.0 A, consistent with Kirchhoff’s current law.

b) Calculate the new operating state when \(R_A=6.0\ \Omega\).

The two external resistors are now both 6.0 Ω.

Their equivalent resistance is:

\[ \frac{1}{R_p}= \frac{1}{6.0}+\frac{1}{6.0} \]

\[ \boxed{R_p=3.0\ \Omega} \]

Including the internal resistance:

\[ R_{\text{total}}=3.0+2.0 \]

\[ R_{\text{total}}=5.0\ \Omega \]

The new total current is:

\[ I_{\text{total}}=\frac{12.0}{5.0} \]

\[ \boxed{I_{\text{total}}=2.4\text{ A}} \]

The terminal voltage is now:

\[ V_t=\mathcal E-I_{\text{total}}r \]

\[ V_t=12.0-(2.4\times2.0) \]

\[ \boxed{V_t=7.2\text{ V}} \]

Both parallel resistors now have 7.2 V across them.

Their currents are:

\[ I_A=I_B=\frac{7.2}{6.0} \]

\[ I_A=I_B=1.2\text{ A} \]

Their powers are therefore:

\[ P_A=7.2\times1.2 \]

\[ \boxed{P_A=8.64\text{ W}} \]

And:

\[ P_B=7.2\times1.2 \]

\[ \boxed{P_B=8.64\text{ W}} \]

c) Evaluate the student’s predictions.

The first prediction was that doubling \(R_A\) must halve its power.

This would be correct if the voltage across \(R_A\) remained constant.

However, the voltage increased from 6.0 V to 7.2 V.

Using the power relationship:

\[ P=\frac{V^2}{R} \]

The initial power was:

\[ P_{A,i}=\frac{6.0^2}{3.0} =12\text{ W} \]

The final power was:

\[ P_{A,f}=\frac{7.2^2}{6.0} =8.64\text{ W} \]

Therefore, the power does not halve.

In fact:

\[ \frac{P_{A,f}}{P_{A,i}} = \frac{8.64}{12} =0.72 \]

The final power is 72% of its initial value.

The second prediction was that the power in Branch B must remain constant because its resistance is unchanged.

This is also incorrect.

Branch B shares the same two external junctions as Branch A. When the terminal voltage increases, the voltage across Branch B increases as well.

Its power therefore increases:

\[ P_{B,i}=6.0\text{ W} \]

\[ P_{B,f}=8.64\text{ W} \]

The increase occurs because the voltage across the unchanged resistor has increased.

The interesting result is that Branch B dissipates more power even though the total current supplied by the battery has decreased.

d) Check conservation of energy.

The battery’s total electrical power output is determined by its emf and total current:

\[ P_{\text{source}}=\mathcal E I_{\text{total}} \]

The internal resistance also dissipates power as thermal energy:

\[ P_{\text{internal}}=I_{\text{total}}^2r \]

Initially:

\[ P_{\text{source}}=12.0\times3.0 =36\text{ W} \]

The internal resistance dissipates:

\[ P_{\text{internal}}=3.0^2\times2.0 =18\text{ W} \]

The external resistors dissipate:

\[ P_{\text{external}}=12+6 =18\text{ W} \]

Therefore:

\[ 36=18+18 \]

After changing \(R_A\):

\[ P_{\text{source}}=12.0\times2.4 =28.8\text{ W} \]

The internal resistance now dissipates:

\[ P_{\text{internal}}=2.4^2\times2.0 =11.52\text{ W} \]

The external resistors dissipate:

\[ P_{\text{external}}=8.64+8.64 =17.28\text{ W} \]

Therefore:

\[ 28.8=11.52+17.28 \]

Energy is conserved in both states.

The particularly interesting result is that the power in Branch B increases by 2.64 W, while the power in Branch A decreases by 3.36 W. The total external power therefore decreases by 0.72 W.

This is why you cannot predict how power changes in a real parallel circuit simply by looking at one resistor’s resistance. You must account for how the entire circuit affects the terminal voltage.

07The key idea to take into mixed circuits

When comparing power in parallel resistors, remember three things:

1. Voltage: Components connected across the same two junctions have the same potential difference.
2. Current: Current divides at junctions according to the resistance of each branch.
3. Power: Equal voltage does not guarantee equal power. For ohmic resistors at the same voltage, \(P=V^2/R\).

The next challenge is recognising which components actually share the same two junctions in a mixed series-parallel circuit. That distinction is essential when comparing the power and brightness of different bulbs.

For further practice, see How to Compare Bulb Brightness in Series-Parallel Circuits.