Why Equal Current Does Not Mean Equal Power in Series Resistors

A 10 Ω resistor and a 30 Ω resistor are connected in series. The same current flows through both. Does that mean they dissipate the same amount of power?

No. The 30 Ω resistor dissipates three times as much power as the 10 Ω resistor, even though the current through each is identical. The reason is that power depends on both current and voltage. In a series circuit, current is conserved along the path, but the voltage drop across each resistor can be different.

To understand why, start with charge conservation and energy conservation. Then use those principles to work out where the electrical energy goes.

01Same current, different power

Imagine charge flowing around a single closed circuit containing two resistors. There are no branches, so every coulomb that passes through the first resistor must also pass through the second.

The current is the rate at which charge flows. If 0.20 C of charge passes a point every second, the current at that point is 0.20 A. The same amount of charge must pass every other point in the unbranched circuit each second.

But consider what happens to the energy carried by that charge. Each resistor transfers electrical energy into thermal energy. The amount transferred depends on the energy lost by each coulomb as it passes through the resistor.

A useful distinction is:

• Current tells you how much charge passes a point each second.
• Voltage tells you how much energy is transferred per coulomb of charge.
• Power tells you how much energy is transferred each second.

The current can be identical while the energy transferred per coulomb is different. That is why equal current does not mean equal power.

For a useful comparison, imagine that each coulomb of charge transfers 1.5 J of energy in the first resistor and 4.5 J in the second. The same number of coulombs passes through both resistors each second, but the second resistor transfers three times as much energy per second.

This is the key idea: equal charge flow does not mean equal energy transfer per second.

02Follow charge first: why the current is equal

Before calculating power, establish what happens at the connections between components.

Kirchhoff’s current law states that the total current entering a junction equals the total current leaving it. This follows from conservation of electric charge.

For example, if a junction receives 0.30 A and has two outgoing branches, those branches must carry currents that add to 0.30 A.

In a simple series circuit, however, there is only one path through the connection. There is nowhere for the current to split.

Therefore,

\[ I_1=I_2=I \]

where:

• \(I_1\) is the current through the first resistor, in amperes (A).
• \(I_2\) is the current through the second resistor, in amperes (A).
• \(I\) is their common current.

Notice what the current rule does not tell you. It does not say that the resistors have equal resistance, equal voltage drops, or equal power.

It only tells you that the same rate of charge flow passes through both.

03Follow energy around the loop: why the voltage is different

Now consider what happens to the electrical energy.

A battery transfers energy to charge as it moves around the circuit. Resistors transfer that electrical energy into thermal energy.

Voltage measures the energy transferred per coulomb of charge:

\[ V=\frac{\Delta E}{q} \]

Here, \(V\) is potential difference in volts (V), \(\Delta E\) is energy transferred in joules (J), and \(q\) is charge in coulombs (C).

A potential difference of 1 V means 1 J of energy is transferred per coulomb.

The energy gained from the battery must equal the energy transferred by the resistors around a complete loop.

This is Kirchhoff’s voltage law.

Choose a sign convention

Take conventional current to flow clockwise around the circuit.

• Travelling through the battery from its negative to positive terminal is a voltage rise, so write \(+V_s\).
• Travelling through a resistor in the direction of conventional current is a voltage drop, so write \(-V_1\) or \(-V_2\).

The signed sum of potential differences around the loop is zero:

\[ +V_s-V_1-V_2=0 \]

Rearranging:

\[ V_s=V_1+V_2 \]

The supply voltage is shared between the two resistors.

But there is no rule saying that each resistor must receive half.

Why the larger resistance gets the larger voltage drop

For an ohmic resistor, Ohm’s law gives:

\[ V=IR \]

The current is identical through both series resistors. Therefore, the voltage drop across each depends on its resistance.

For two resistors:

\[ V_1=IR_1 \]

\[ V_2=IR_2 \]

Dividing these equations gives:

\[ \frac{V_1}{V_2}=\frac{R_1}{R_2} \]

So the voltage drops are proportional to the resistances.

If one resistor has three times the resistance of the other, it receives three times the voltage drop.

This is the missing link between equal current and unequal power.

Only now: reducing the series resistance

We can now derive the familiar series resistance rule rather than simply memorising it.

Starting with conservation of energy:

\[ V_s=V_1+V_2 \]

Substitute Ohm’s law:

\[ V_s=IR_1+IR_2 \]

Factor out the common current:

\[ V_s=I(R_1+R_2) \]

Comparing this with \(V_s=IR_{\mathrm{eq}}\), where \(R_{\mathrm{eq}}\) is the equivalent resistance, gives:

\[ R_{\mathrm{eq}}=R_1+R_2 \]

The important order is:

1. Use charge conservation to establish the current.
2. Use energy conservation to establish the voltage relationship.
3. Apply Ohm’s law.
4. Reduce the resistance if you need the total circuit current.

For a fuller treatment of these foundations, see Series Circuits: Current, Voltage, Resistance and Power.

04Now connect voltage and current to power

Electrical power is the rate at which energy is transferred:

\[ P=\frac{\Delta E}{\Delta t} \]

where \(P\) is power in watts (W), and \(\Delta t\) is the time interval in seconds (s).

Since voltage is energy transferred per coulomb, and current is charge passing per second, we can combine the two:

\[ \boxed{P=IV} \]

This is the starting point for comparing the power dissipated by resistors.

For an ohmic resistor, substitute \(V=IR\):

\[ \boxed{P=I^2R} \]

Alternatively, substituting \(I=V/R\) gives:

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

All three equations describe electrical power:

Here’s the important decision.

In a series circuit, the current is common to both resistors. Therefore, \(P=I^2R\) is the most useful comparison.

For two resistors:

\[ P_1=I^2R_1 \]

\[ P_2=I^2R_2 \]

Dividing:

\[ \boxed{\frac{P_1}{P_2}=\frac{R_1}{R_2}} \]

For fixed ohmic resistors carrying the same current, the larger resistance dissipates more power.

If one resistance is three times the other, its power is also three times as large.

Worked numerical comparison: 10 Ω and 30 Ω

Consider an ideal 6.0 V DC supply connected to a 10 Ω resistor and a 30 Ω resistor in series.

Assume:

• Both resistors are ohmic and have constant resistance.
• Connecting-wire resistance is negligible.
• The supply maintains a constant 6.0 V across the circuit.
• The circuit has reached steady DC operation.

The calculations are:

Step 1: Find the common current.

The resistors are in series, so the same current passes through both.

\[ R_{\mathrm{eq}}=10+30=40\ \Omega \]

Using the loop equation:

\[ I=\frac{6.0}{40}=0.15\ \mathrm{A} \]

Step 2: Find the voltage across each resistor.

For the 10 Ω resistor:

\[ V_1=IR_1 \]

\[ V_1=0.15\times10=1.5\ \mathrm{V} \]

For the 30 Ω resistor:

\[ V_2=IR_2 \]

\[ V_2=0.15\times30=4.5\ \mathrm{V} \]

Check energy conservation:

\[ 1.5+4.5=6.0\ \mathrm{V} \]

Step 3: Calculate the power dissipated by each resistor.

For the 10 Ω resistor:

\[ P_1=I^2R_1 \]

\[ P_1=0.15^2\times10 \]

\[ P_1=0.225\ \mathrm{W} \]

For the 30 Ω resistor:

\[ P_2=I^2R_2 \]

\[ P_2=0.15^2\times30 \]

\[ P_2=0.675\ \mathrm{W} \]

The 30 Ω resistor dissipates three times as much power.

There is also an energy conservation check. The total power supplied by the battery is:

\[ P_s=V_sI=6.0\times0.15 \]

\[ P_s=0.90\ \mathrm{W} \]

The total power dissipated by the resistors is:

\[ P_1+P_2=0.225+0.675 \]

\[ P_1+P_2=0.90\ \mathrm{W} \]

The two results agree. All the electrical power supplied is accounted for by the power dissipated in the resistors under our idealised assumptions.

05How would you test this experimentally?

Suppose you want to investigate whether equal current really means equal power.

A simple circuit with two resistors gives you a way to test the prediction.

The original measurement idea

Start by connecting the positive terminal of the supply to a switch, then an ammeter, followed by the 10 Ω resistor and the 30 Ω resistor. Return the circuit to the negative terminal of the supply.

The ammeter is connected in series because it measures the current passing through the circuit.

Suppose you measure the current before the first resistor, then move the ammeter to the section between the resistors. Switch off the supply before changing the connections.

The expected readings are identical.

Would that establish that both resistors dissipate equal power?

No. The experiment has only established that the current is equal. It hasn’t measured the voltage across either resistor.

An ammeter alone cannot establish the power dissipated by a component.

Revise the measurement

Add a voltmeter across each resistor.

A voltmeter is connected in parallel with the component whose potential difference you want to measure. Its positive lead should be connected to the end at higher potential, and its negative lead to the other end.

For an ideal voltmeter, the resistance is infinite, so it draws no current and does not affect the circuit. A real voltmeter has a very high, but finite, input resistance.

The improved investigation now measures both quantities needed for power:

\[ P_1=I_1V_1 \]

\[ P_2=I_2V_2 \]

Use resistors rated at 2 W or higher, so the predicted dissipation is comfortably within their ratings.

Predicted results

For the 6.0 V supply and the 10 Ω and 30 Ω resistors, the ideal model predicts:

These are theoretical predictions, not claimed experimental measurements.

A real investigation may produce slightly different readings because of resistor tolerances, meter uncertainty, connecting-wire resistance, and changes in resistance as components warm up.

The revised experiment addresses the original flaw: it measures both current and voltage, rather than assuming that identical current readings establish identical power.

06The contrasting case: what changes in parallel?

Here’s where students often make an overgeneralisation.

They learn that the larger resistor dissipates more power in series, then assume that a larger resistor must always dissipate more power.

That is incorrect.

Consider a 10 Ω resistor and a 30 Ω resistor connected in parallel across an ideal 6.0 V supply.

Both resistors connect across the same two points, so they have the same potential difference.

Now use Ohm’s law to find each branch current.

For the 10 Ω resistor:

\[ I_1=\frac{6.0}{10}=0.60\ \mathrm{A} \]

For the 30 Ω resistor:

\[ I_2=\frac{6.0}{30}=0.20\ \mathrm{A} \]

The currents are different because the resistors have different resistances.

Now calculate power using \(P=V^2/R\).

For the 10 Ω resistor:

\[ P_1=\frac{6.0^2}{10}=3.6\ \mathrm{W} \]

For the 30 Ω resistor:

\[ P_2=\frac{6.0^2}{30}=1.2\ \mathrm{W} \]

The 10 Ω resistor dissipates three times as much power as the 30 Ω resistor.

The result is the opposite of the series comparison.

Why?

Because the quantities held constant are different.

The important lesson isn’t simply that larger resistance means more power in series and less in parallel.

It’s that you must identify which electrical quantity is shared before choosing a power equation.

For a more involved application involving brightness and mixed series-parallel networks, see How to Compare Bulb Brightness in Series-Parallel Circuits.

07Worked examples

08Questions and solutions

Question 1

An ideal 9.0 V battery is connected to two resistors, 6.0 Ω and 12 Ω, in series.

Calculate:

a) The current through each resistor.

b) The potential difference across each resistor.

c) The power dissipated by each resistor.

d) Verify that the total power dissipated equals the power supplied by the battery.

Solution 1

The 12 Ω resistor dissipates 3.0 W, which is twice the power dissipated by the 6.0 Ω resistor. Both resistors carry the same 0.50 A current.

a) Current through each resistor

There is only one path, so:

\[ I_1=I_2=I \]

Apply the loop rule:

\[ 9.0=I(6.0+12) \]

\[ I=\frac{9.0}{18} \]

\[ \boxed{I=0.50\ \mathrm{A}} \]

b) Potential difference across each resistor

For the 6.0 Ω resistor:

\[ V_1=IR_1 \]

\[ V_1=0.50\times6.0 \]

\[ \boxed{V_1=3.0\ \mathrm{V}} \]

For the 12 Ω resistor:

\[ V_2=IR_2 \]

\[ V_2=0.50\times12 \]

\[ \boxed{V_2=6.0\ \mathrm{V}} \]

The voltage drops add to 9.0 V, as required by Kirchhoff’s voltage law.

c) Power dissipated

For the 6.0 Ω resistor:

\[ P_1=I^2R_1 \]

\[ P_1=0.50^2\times6.0 \]

\[ \boxed{P_1=1.5\ \mathrm{W}} \]

For the 12 Ω resistor:

\[ P_2=I^2R_2 \]

\[ P_2=0.50^2\times12 \]

\[ \boxed{P_2=3.0\ \mathrm{W}} \]

d) Power conservation

The power supplied by the battery is:

\[ P_s=V_sI \]

\[ P_s=9.0\times0.50=4.5\ \mathrm{W} \]

The total resistor power is:

\[ P_1+P_2=1.5+3.0=4.5\ \mathrm{W} \]

The two values agree.

The larger resistor dissipates more power because both resistors carry the same current, but the 12 Ω resistor has twice the resistance.

Question 2

A battery has an electromotive force (emf) of 18 V and an internal resistance of 2.0 Ω. It is connected to two external resistors, 4.0 Ω and 8.0 Ω, in series.

Calculate:

a) The circuit current.

b) The terminal voltage of the battery.

c) The power dissipated by each external resistor and by the battery’s internal resistance.

d) Explain why the external resistors still have a 2:1 power ratio, even though the terminal voltage is less than the battery’s emf.

Solution 2

The circuit current is approximately 1.29 A. The 8.0 Ω external resistor dissipates twice the power of the 4.0 Ω resistor, while the internal resistance also dissipates electrical power.

a) Circuit current

All three resistances are in series. The same current passes through the internal resistance and both external resistors.

Apply Kirchhoff’s voltage law:

\[ 18-I(2+4+8)=0 \]

Therefore:

\[ I=\frac{18}{14} \]

\[ \boxed{I=1.29\ \mathrm{A}} \]

b) Terminal voltage

The terminal voltage is the potential difference across the external circuit.

It is the emf minus the voltage drop across the internal resistance:

\[ V_{\mathrm{terminal}}=\varepsilon-Ir \]

where \(\varepsilon\) is the battery’s emf and \(r\) is its internal resistance.

Substituting:

\[ V_{\mathrm{terminal}} =18-\frac{18}{14}\times2 \]

\[ \boxed{V_{\mathrm{terminal}}=15.43\ \mathrm{V}} \]

The terminal voltage is lower than the emf because some energy is transferred inside the battery itself.

c) Power dissipated

For the 4.0 Ω resistor:

\[ P_1=I^2R_1 \]

\[ P_1=\left(\frac{18}{14}\right)^2\times4 \]

\[ \boxed{P_1=6.61\ \mathrm{W}} \]

For the 8.0 Ω resistor:

\[ P_2=I^2R_2 \]

\[ P_2=\left(\frac{18}{14}\right)^2\times8 \]

\[ \boxed{P_2=13.22\ \mathrm{W}} \]

For the internal resistance:

\[ P_r=I^2r \]

\[ P_r=\left(\frac{18}{14}\right)^2\times2 \]

\[ \boxed{P_r=3.31\ \mathrm{W}} \]

d) Explain the power ratio

The two external resistors carry the same current because they are connected in series.

Therefore:

\[ \frac{P_2}{P_1} =\frac{I^2R_2}{I^2R_1} \]

\[ =\frac{8}{4} \]

\[ \boxed{\frac{P_2}{P_1}=2} \]

The internal resistance reduces the current through the entire circuit, but it does not change the fact that the two external resistors share the same current.

As a final check, the total power supplied by the emf is:

\[ P_{\mathrm{source}}=\varepsilon I \]

\[ =18\times\frac{18}{14} =23.14\ \mathrm{W} \]

The total power dissipated is:

\[ 6.61+13.22+3.31=23.14\ \mathrm{W} \]

Energy conservation is satisfied.

Question 3

A 12.0 V ideal power supply is connected to two resistors, 6.0 Ω and 12 Ω, in parallel.

Calculate:

a) The current through each resistor.

b) The power dissipated by each resistor.

c) The total current supplied by the battery and the total power supplied.

d) Explain why the smaller resistor dissipates more power, even though the larger resistor has twice the resistance.

Solution 3

The 6.0 Ω resistor dissipates 24 W, twice the power of the 12 Ω resistor. This happens because parallel resistors share voltage, not current.

a) Current through each resistor

Both resistors are connected across the same two nodes, so each experiences the full 12.0 V.

For the 6.0 Ω resistor:

\[ I_1=\frac{V}{R_1} \]

\[ I_1=\frac{12.0}{6.0} \]

\[ \boxed{I_1=2.0\ \mathrm{A}} \]

For the 12 Ω resistor:

\[ I_2=\frac{12.0}{12} \]

\[ \boxed{I_2=1.0\ \mathrm{A}} \]

b) Power dissipated

Use \(P=V^2/R\), because the voltage is common.

For the 6.0 Ω resistor:

\[ P_1=\frac{12.0^2}{6.0} \]

\[ \boxed{P_1=24\ \mathrm{W}} \]

For the 12 Ω resistor:

\[ P_2=\frac{12.0^2}{12} \]

\[ \boxed{P_2=12\ \mathrm{W}} \]

c) Total current and power

Apply Kirchhoff’s current law at the junction where the current splits:

\[ I_s=I_1+I_2 \]

\[ I_s=2.0+1.0 \]

\[ \boxed{I_s=3.0\ \mathrm{A}} \]

The total power supplied is:

\[ P_s=V_sI_s \]

\[ P_s=12.0\times3.0 \]

\[ \boxed{P_s=36\ \mathrm{W}} \]

The total power dissipated is:

\[ P_1+P_2=24+12=36\ \mathrm{W} \]

Again, energy is conserved.

d) Explain the difference

The tempting mistake is to use \(P=I^2R\) and assume both resistors have the same current.

They don’t.

The 6.0 Ω resistor carries twice the current of the 12 Ω resistor. The larger current more than compensates for its smaller resistance.

Since the voltage is equal, the correct comparison is:

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

For parallel resistors, the smaller resistance dissipates more power when the potential difference is the same.

Question 4

A 12.0 V ideal supply is connected to a 4.0 Ω fixed resistor and a filament lamp in series.

The lamp has:

• A resistance of 2.0 Ω immediately after the switch is closed, while the filament is cold.
• A steady operating resistance of 6.0 Ω after the filament heats up.

Treat these resistance values as given for the two operating states.

Calculate the current and power dissipated by each component in both states.

Explain how the lamp’s power can increase even though the circuit current decreases.

Solution 4

The lamp’s power increases from 8.0 W to 8.64 W as it heats up, even though the current decreases from 2.0 A to 1.2 A. Its resistance increases enough to outweigh the reduction in current.

Cold state

Step 1: Calculate the total resistance.

\[ R_{\mathrm{total}}=4.0+2.0 \]

\[ R_{\mathrm{total}}=6.0\ \Omega \]

Step 2: Calculate the current.

\[ I=\frac{12.0}{6.0} \]

\[ \boxed{I=2.0\ \mathrm{A}} \]

Step 3: Calculate the power of each component.

For the fixed resistor:

\[ P_R=I^2R \]

\[ P_R=2.0^2\times4.0 \]

\[ \boxed{P_R=16\ \mathrm{W}} \]

For the cold lamp:

\[ P_L=I^2R_L \]

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

\[ \boxed{P_L=8.0\ \mathrm{W}} \]

Hot operating state

Step 4: Recalculate the total resistance.

The lamp’s resistance has increased to 6.0 Ω.

\[ R_{\mathrm{total}}=4.0+6.0 \]

\[ R_{\mathrm{total}}=10.0\ \Omega \]

Step 5: Calculate the new current.

\[ I=\frac{12.0}{10.0} \]

\[ \boxed{I=1.2\ \mathrm{A}} \]

Step 6: Calculate the new power values.

For the fixed resistor:

\[ P_R=1.2^2\times4.0 \]

\[ \boxed{P_R=5.76\ \mathrm{W}} \]

For the hot lamp:

\[ P_L=1.2^2\times6.0 \]

\[ \boxed{P_L=8.64\ \mathrm{W}} \]

Explain the result

Compare the two operating states:

The current is the same through both components at any given instant because they are in series.

However, the current is not the same between the cold and hot states.

This distinction matters. The lamp’s resistance increases as its filament temperature rises, changing the circuit current and the power distribution.

The tempting prediction is that decreasing current must mean decreasing power everywhere. But \(P=I^2R\) contains two variables. A sufficiently large increase in resistance can outweigh a decrease in current.

This is also why using a filament lamp’s cold resistance to predict its steady operating power can give an incorrect result.

Question 5

Consider a circuit containing an ideal 12.0 V supply, a 4.0 Ω resistor A, and a parallel combination of:

• Resistor B, with a fixed resistance of 8.0 Ω.
• An adjustable resistor X, with resistance \(R_X\).

Resistor A is connected in series with the entire parallel combination.

Find the value of \(R_X\) for which resistor A and resistor B dissipate equal power.

Then determine:

a) Which resistor dissipates more power when \(R_X\) is smaller than this value?

b) Which resistor dissipates more power when \(R_X\) is larger than this value?

c) Whether the critical value of \(R_X\) would change if the supply voltage were increased to 24.0 V.

Assume all resistors are ideal and ohmic, and the supply voltage is constant.

Solution 5

The two resistors dissipate equal power when:

\[ \boxed{R_X=8(1+\sqrt{2})\ \Omega} \]

\[ \boxed{R_X\approx19.3\ \Omega} \]

For smaller values of \(R_X\), resistor A dissipates more power. For larger values, resistor B dissipates more power. The critical resistance does not depend on the supply voltage.

This question is harder because resistor A and resistor B do not carry the same current.

They are not directly in series with each other. Resistor A carries the total circuit current, while the current splits between B and X.

Step 1: Apply conservation of charge at the junction.

Let:

• \(I_A\) be the current through resistor A.
• \(I_B\) be the current through resistor B.
• \(I_X\) be the current through resistor X.

Kirchhoff’s current law gives:

\[ I_A=I_B+I_X \]

Therefore, we cannot simply compare the resistance values of A and B using \(P=I^2R\) with a common current.

Step 2: Apply conservation of energy around the loop.

Let \(V_A\) be the voltage across resistor A, and \(V_P\) be the voltage across the parallel combination.

The loop rule gives:

\[ 12.0=V_A+V_P \]

Both B and X connect across the same two nodes, so:

\[ V_B=V_X=V_P \]

Step 3: Express the two powers.

We want resistor A and resistor B to dissipate equal power.

For resistor A:

\[ P_A=\frac{V_A^2}{R_A} \]

Since \(R_A=4.0\ \Omega\):

\[ P_A=\frac{V_A^2}{4} \]

For resistor B:

\[ P_B=\frac{V_P^2}{8} \]

Set the powers equal:

\[ \frac{V_A^2}{4}=\frac{V_P^2}{8} \]

Rearrange:

\[ 2V_A^2=V_P^2 \]

Taking the positive square root, since both voltage drops are positive in the direction of conventional current:

\[ V_P=\sqrt{2}V_A \]

Step 4: Find the required voltage division.

Substitute into the loop equation:

\[ 12=V_A+\sqrt{2}V_A \]

\[ 12=V_A(1+\sqrt{2}) \]

Therefore:

\[ V_A=\frac{12}{1+\sqrt{2}} \]

And:

\[ V_P=\frac{12\sqrt{2}}{1+\sqrt{2}} \]

Step 5: Determine the equivalent resistance of the parallel section.

Resistor A and the parallel combination are in series with each other, so they carry the same current.

Using Ohm’s law:

\[ I_A=\frac{V_A}{4} \]

The equivalent resistance of the parallel section is:

\[ R_P=\frac{V_P}{I_A} \]

Substitute the expressions for voltage and current:

\[ R_P=\frac{V_P}{V_A/4} \]

Since \(V_P=\sqrt{2}V_A\):

\[ R_P=4\sqrt{2} \]

\[ \boxed{R_P\approx5.66\ \Omega} \]

Step 6: Calculate the adjustable resistance.

Now use the parallel resistance relationship:

\[ R_P=\frac{R_BR_X}{R_B+R_X} \]

Substitute the known values:

\[ 4\sqrt{2}=\frac{8R_X}{8+R_X} \]

Multiply both sides:

\[ 4\sqrt{2}(8+R_X)=8R_X \]

Expand:

\[ 32\sqrt{2}+4\sqrt{2}R_X=8R_X \]

Rearrange:

\[ 32\sqrt{2}=R_X(8-4\sqrt{2}) \]

Therefore:

\[ R_X=\frac{32\sqrt{2}}{8-4\sqrt{2}} \]

Simplifying:

\[ \boxed{R_X=8(1+\sqrt{2})\ \Omega} \]

\[ \boxed{R_X\approx19.3\ \Omega} \]

a) What happens below 19.3 Ω?

Resistor A dissipates more power.

When \(R_X\) decreases, the equivalent resistance of the parallel section decreases. The voltage across the parallel section also decreases relative to the voltage across A.

Resistor B consequently receives a smaller voltage drop, and its power decreases relative to A’s power.

b) What happens above 19.3 Ω?

Resistor B dissipates more power.

As \(R_X\) increases, the equivalent resistance of the parallel section increases. More of the supply voltage is dropped across that section.

Eventually, resistor B dissipates more power than A.

At the limiting case where X becomes an open circuit, only A and B remain in series. The current through them is then equal, and B has twice the resistance of A. Consequently, B dissipates twice the power.

c) Does increasing the supply voltage change the critical resistance?

No.

The condition for equal power was:

\[ \frac{V_A^2}{4}=\frac{V_P^2}{8} \]

The resulting voltage ratio depends only on the resistance values.

If the supply voltage doubles, both voltage drops double at the equality point. Both resistor powers increase by a factor of four, but their powers remain equal.

The critical value of \(R_X\) therefore remains approximately 19.3 Ω.

The deeper lesson is that power comparisons depend on the circuit’s current and voltage relationships, not just on resistance values. A resistor’s position in a network determines which conservation relationships apply, and those relationships determine the power distribution.

08The decision process to remember

When comparing power in a circuit, use this order:

1. Identify every junction. Apply Kirchhoff’s current law to establish where current is shared or split.
2. Identify complete loops. Apply Kirchhoff’s voltage law to establish the voltage drops.
3. Work out which components actually share the same current or potential difference.
4. Apply Ohm’s law to find any missing electrical quantities.
5. Use the most appropriate power equation.
6. Check that the total power supplied equals the total power dissipated, accounting for any internal resistance.

For two fixed resistors in series, the decision is particularly simple:

\[ I_1=I_2 \]

Therefore:

\[ \frac{P_1}{P_2}=\frac{R_1}{R_2} \]

The larger resistance dissipates more power.

But don’t carry that shortcut into a parallel circuit or a mixed network without checking the connections first.

The next useful step is learning to track current through branching circuits, where one component may carry the total current while other components carry only part of it. That’s where the junction rule becomes just as important as the loop rule.“`