How to Solve a Circuit with a Shorted Resistor

An ideal 12 V supply feeds a 4 Ω resistor, followed by a 6 Ω resistor. You connect a plain wire directly across the two terminals of the 6 Ω resistor.

What happens to the current through the 6 Ω resistor? Does it increase, decrease, or stay the same?

It becomes zero. The wire forces both terminals of the resistor to the same electrical potential, leaving no potential difference to drive current through it. Current still flows through the circuit, but it travels through the bypass wire instead.

The 4 Ω resistor is still in the circuit, so it limits the total current to:

\[ I=\frac{12}{4}=3\text{ A} \]

This is the key to solving a circuit with a shorted resistor: the short-circuit wire carries the current that would otherwise have flowed through the bypassed resistor, while the rest of the circuit still determines how much current flows overall.

01Why does a shorted resistor have zero current?

Start by looking at the connections, rather than adding up the resistances.

A resistor is short-circuited when a wire provides a direct connection between its two terminals. In an ideal circuit, that wire has zero resistance.

Imagine the resistor has terminals A and B.

The wire connects A directly to B. Since an ideal wire has no resistance, there is no potential difference between these two points.

They are at the same electrical potential. In circuit analysis, we call them the same node.

The resistor is connected between those exact same points, so it must also have zero potential difference across it.

Mathematically:

\[ V_R=0\text{ V} \]

Now apply Ohm’s law:

\[ V=IR \]

Rearranging for current:

\[ I_R=\frac{V_R}{R} \]

For a finite resistor:

\[ I_R=\frac{0}{R}=0\text{ A} \]

The resistor carries no current.

Notice something important: zero current through the resistor does not mean zero current through the circuit.

The wire can still carry a substantial current. In fact, it carries the current that would otherwise have passed through the resistor’s branch.

This distinction is what makes short-circuit questions easier once you recognise the connections.

02Apply conservation laws before reducing resistance

A common mistake is to immediately remove the shorted resistor and calculate the equivalent resistance.

That shortcut can work, but only after you have identified which components are actually bypassed.

Start with two fundamental circuit principles.

Kirchhoff’s current law: conservation at junctions

Kirchhoff’s current law (KCL) states that the total current entering a junction equals the total current leaving it.

This follows from conservation of electric charge.

Consider the junction where current reaches the two parallel paths: the resistor and the bypass wire.

Let:

• \(I_{\text{in}}\) be the current entering the junction.
• \(I_R\) be the current through the resistor.
• \(I_w\) be the current through the bypass wire.

KCL gives:

\[ I_{\text{in}}=I_R+I_w \]

We already established that the shorted resistor carries zero current.

Therefore:

\[ I_{\text{in}}=0+I_w \]

So:

\[ \boxed{I_w=I_{\text{in}}} \]

All the incoming current travels through the bypass wire in the ideal-circuit model.

At the second junction, the current leaves the wire and continues through the rest of the circuit.

Kirchhoff’s voltage law: conservation around loops

Kirchhoff’s voltage law (KVL) states that the sum of all potential differences around a closed loop is zero.

This follows from conservation of energy.

For a loop containing a power supply, a series resistor, and a shorted resistor, write:

\[ \mathcal E-IR_s-V_R=0 \]

where:

• \(\mathcal E\) is the supply’s electromotive force (emf), in volts.
• \(I\) is the current through the series resistor, in amperes.
• \(R_s\) is the series resistance, in ohms.
• \(V_R\) is the potential difference across the shorted resistor.

Since \(V_R=0\):

\[ \boxed{\mathcal E=IR_s} \]

The remaining series resistance determines the circuit current.

Sign convention: Take conventional current as flowing from the positive supply terminal through the external circuit. Moving through a resistor in the direction of current gives a potential drop of \(-IR\). Moving through the supply from its negative to positive terminal gives a rise of \(+\mathcal E\).

The correct solving order

Use this sequence whenever you encounter a shorted resistor.

The order matters. Conservation laws establish what the circuit is doing before you simplify its resistance.

03How could you experimentally check the result?

Suppose you wanted to test whether the resistor really carries zero current when a wire bypasses it.

The question is not simply whether current flows in the circuit. A main-line ammeter can answer that.

You need to establish which path carries the current.

Setting up the measurement

Consider the 12 V supply, 4 Ω series resistor, and 6 Ω resistor with an ideal bypass wire.

Use a low-voltage, current-limited supply. The 4 Ω resistor remains in series to limit the current.

The current travels from the supply, through the 4 Ω resistor, and reaches junction A. It can then pass through either the 6 Ω resistor or the bypass wire before the paths rejoin at B.

To investigate the current distribution, use:

• One ammeter in the main circuit before the junction.
• One ammeter in series with the 6 Ω resistor.
• One ammeter in series with the bypass wire.
• One voltmeter connected across the 6 Ω resistor.

What should the measurements show?

For the ideal-circuit model, the predicted results are:

The first measurement design, using only the main-line ammeter, would show that 3 A flows through the circuit.

However, it would not establish how that current divides between the two branches.

Adding the branch ammeters and voltmeter addresses this limitation. The predicted pattern is zero current through the resistor, nearly all the current through the bypass path, and zero potential difference across the resistor.

In an actual experiment, the readings will depend on the resistance of the wire, contacts, and measuring instruments. The ideal prediction is a model, not a claim that a real wire has exactly zero resistance.

A small estimate: what if the wire has resistance?

Suppose the bypass path, including its contacts, has a resistance of \(0.10\ \Omega\). This is an illustrative assumed value, not a universal value for a wire.

The 6 Ω resistor and the bypass path are now parallel:

\[ R_p=\frac{6\times0.10}{6+0.10} \]

\[ R_p=0.0984\ \Omega \]

Including the 4 Ω series resistor:

\[ R_{\text{total}}=4+0.0984 \]

\[ I_{\text{total}}=\frac{12}{4.0984} =2.928\text{ A} \]

The voltage across the parallel section is:

\[ V_p=2.928\times0.0984 \]

\[ V_p=0.288\text{ V} \]

The current through the 6 Ω resistor is therefore:

\[ I_6=\frac{0.288}{6} =0.048\text{ A} \]

The bypass-wire current is:

\[ I_w=\frac{0.288}{0.10} =2.88\text{ A} \]

Check KCL:

\[ 2.928=0.048+2.88 \]

The result makes sense. Most of the current travels through the lower-resistance bypass path, but the resistor still carries a small current.

This is why we use the ideal-short approximation in ordinary circuit questions unless the wire’s resistance is specifically provided.

05A short can affect more than one resistor

Here is a particularly important complication.

A wire connected across one resistor can also short-circuit other resistors if they share the same two connection points.

Consider a circuit with:

• An ideal 18 V supply.
• A 2 Ω resistor in series with the rest of the circuit.
• A 6 Ω resistor in the upper parallel branch.
• A 4 Ω resistor in the lower parallel branch.
• An ideal wire connected directly across the 4 Ω resistor.

Both resistors connect between the same two parallel rails.

The wire connects those rails directly, making them the same electrical node.

Consequently, both resistors have zero potential difference across them.

Calculate the current

Both parallel resistors have zero voltage:

\[ V_6=V_4=0 \]

Therefore:

\[ I_6=\frac{0}{6}=0\text{ A} \]

\[ I_4=\frac{0}{4}=0\text{ A} \]

The 2 Ω resistor is the only remaining resistance limiting the supply current.

Using KVL:

\[ 18-I(2)-0=0 \]

\[ \boxed{I=9\text{ A}} \]

KCL tells us that all 9 A must pass through the bypass wire.

The result is:

• Current through the 2 Ω resistor: 9 A.
• Current through the 6 Ω resistor: 0 A.
• Current through the 4 Ω resistor: 0 A.
• Current through the bypass wire: 9 A.

This is a useful distinction: a short across a resistor does not necessarily affect only that resistor. You must check whether the wire also connects the two terminals of other components.

There is another important boundary case.

If the 2 Ω series resistor were replaced by an ideal wire, the ideal supply would be connected directly across an ideal short circuit.

The ideal model would then have no finite solution for the current. You cannot calculate a meaningful finite current using Ohm’s law because the total resistance would be zero.

In a real circuit, the source’s internal resistance and the resistance of the connecting wires limit the current.

07Short-circuited versus open-circuited resistors

These two situations are easy to confuse because both can produce zero current through a resistor.

However, they are physically different.

The most important distinction is the voltage.

A shorted resistor has zero potential difference because its terminals are connected directly.

An open-circuited resistor can have a potential difference across its terminals, but current cannot pass through the broken path.

Zero current does not automatically mean an open circuit. Always inspect the connections and voltage.

For a shorted resistor, removing the resistor while leaving the bypass wire in place makes no difference to an ideal circuit.

Removing the bypass wire, however, can completely change the circuit.

08Questions and solutions

Question 1

An ideal 12.0 V supply is connected to a 5.0 Ω resistor in series with a 10.0 Ω resistor.

An ideal wire is connected directly across the 10.0 Ω resistor.

Calculate:

(a) The total current supplied by the battery.

(b) The current through the 10.0 Ω resistor.

(c) The current through the bypass wire.

(d) The potential difference across both resistors.

Solution 1

The total current is 2.4 A, the 10.0 Ω resistor carries zero current, and the bypass wire carries 2.4 A.

The wire connects both terminals of the 10.0 Ω resistor, making them equipotential.

Step 1: Calculate the voltage across the shorted resistor.

\[ V_{10}=0\text{ V} \]

Step 2: Calculate its current using Ohm’s law.

\[ I_{10}=\frac{0}{10.0} \]

\[ \boxed{I_{10}=0\text{ A}} \]

Step 3: Apply KVL to the complete circuit.

Only the 5.0 Ω resistor has a potential drop.

\[ 12.0-I(5.0)-0=0 \]

\[ I=\frac{12.0}{5.0} \]

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

Step 4: Apply KCL to calculate the bypass-wire current.

\[ I_{\text{total}}=I_{10}+I_w \]

\[ 2.4=0+I_w \]

\[ \boxed{I_w=2.4\text{ A}} \]

Step 5: Calculate the potential difference across the 5.0 Ω resistor.

\[ V_5=IR \]

\[ V_5=2.4\times5.0 \]

\[ \boxed{V_5=12.0\text{ V}} \]

The 10.0 Ω resistor has zero voltage, while the 5.0 Ω resistor accounts for the entire supply voltage.

Question 2

An ideal 24.0 V supply feeds a 4.0 Ω resistor in series with a parallel network.

The upper branch contains an 8.0 Ω resistor.

The lower branch contains an unknown resistor \(X\) in series with a 5.0 Ω resistor. An ideal wire is connected directly across the 5.0 Ω resistor.

An ammeter in the main circuit measures a total current of 4.0 A.

Calculate:

(a) The potential difference across the parallel network.

(b) The current in each parallel branch.

(c) The resistance \(X\).

(d) The current through the bypass wire.

(e) The current and potential difference across the 5.0 Ω resistor.

Solution 2

The unknown resistance is \(2.67\ \Omega\). The upper branch carries 1.0 A, the lower branch and bypass wire carry 3.0 A, and the 5.0 Ω resistor carries zero current.

Step 1: Identify the shorted resistor.

The 5.0 Ω resistor is bypassed, so:

\[ V_5=0\text{ V} \]

\[ I_5=0\text{ A} \]

The lower branch therefore has an effective resistance of \(X\).

Step 2: Calculate the voltage across the 4.0 Ω resistor.

The total supply current is 4.0 A.

\[ V_4=IR \]

\[ V_4=4.0\times4.0 \]

\[ V_4=16.0\text{ V} \]

Step 3: Apply KVL to find the parallel-network voltage.

\[ V_p=24.0-16.0 \]

\[ \boxed{V_p=8.0\text{ V}} \]

Step 4: Calculate the upper branch current.

Using Ohm’s law:

\[ I_8=\frac{V_p}{R_8} \]

\[ I_8=\frac{8.0}{8.0} \]

\[ \boxed{I_8=1.0\text{ A}} \]

Step 5: Apply KCL to find the lower branch current.

\[ I_{\text{total}}=I_8+I_{\text{lower}} \]

\[ 4.0=1.0+I_{\text{lower}} \]

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

Step 6: Calculate the unknown resistance.

The 5.0 Ω resistor has zero voltage, so the entire 8.0 V across the lower branch is across \(X\).

Using Ohm’s law:

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

\[ X=\frac{8.0}{3.0} \]

\[ \boxed{X=2.67\ \Omega} \]

Step 7: Find the bypass-wire current.

Since the 5.0 Ω resistor carries zero current, all 3.0 A in the lower branch passes through the bypass wire.

\[ \boxed{I_w=3.0\text{ A}} \]

The results satisfy KCL:

\[ 4.0=1.0+3.0 \]

They also satisfy KVL:

\[ 16.0+8.0=24.0\text{ V} \]

The important reasoning step was recognising that the unknown resistor remained in the lower branch, even though the 5.0 Ω resistor did not contribute to its resistance.

Question 3

An ideal supply with an unknown emf feeds a 2.0 Ω series resistor, followed by two parallel branches.

The upper branch contains an 8.0 Ω resistor.

The lower branch contains a 3.0 Ω resistor in series with a 6.0 Ω resistor. An ideal wire is connected directly across the 6.0 Ω resistor.

An ammeter placed in the bypass wire reads 2.0 A.

Calculate:

(a) The potential difference across the parallel network.

(b) The current through the 8.0 Ω resistor.

(c) The total supply current.

(d) The emf of the supply.

(e) The potential difference across each resistor.

Explain why the current through the 3.0 Ω resistor is not zero, even though the 6.0 Ω resistor is shorted.

Solution 3

The supply emf is 11.5 V, the total current is 2.75 A, and the upper branch carries 0.75 A. The 6.0 Ω resistor has zero current, but the 3.0 Ω resistor carries 2.0 A.

Step 1: Determine the current through the lower branch.

The ammeter measures 2.0 A in the bypass wire.

Since the 6.0 Ω resistor is shorted, its current is zero.

Therefore, all the lower-branch current passes through the 3.0 Ω resistor and the bypass wire.

\[ \boxed{I_3=2.0\text{ A}} \]

Step 2: Calculate the voltage across the 3.0 Ω resistor.

\[ V_3=IR \]

\[ V_3=2.0\times3.0 \]

\[ \boxed{V_3=6.0\text{ V}} \]

The 6.0 Ω resistor has zero voltage, so the entire lower-branch potential difference is 6.0 V.

Both parallel branches share this voltage.

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

Step 3: Calculate the current through the upper 8.0 Ω resistor.

\[ I_8=\frac{V_p}{R_8} \]

\[ I_8=\frac{6.0}{8.0} \]

\[ \boxed{I_8=0.75\text{ A}} \]

Step 4: Apply KCL to calculate the total current.

The total current is the sum of the two branch currents.

\[ I_{\text{total}}=I_8+I_3 \]

\[ I_{\text{total}}=0.75+2.0 \]

\[ \boxed{I_{\text{total}}=2.75\text{ A}} \]

Step 5: Calculate the voltage across the 2.0 Ω series resistor.

\[ V_2=IR \]

\[ V_2=2.75\times2.0 \]

\[ \boxed{V_2=5.5\text{ V}} \]

Step 6: Apply KVL to find the supply emf.

The supply voltage equals the voltage drop across the series resistor plus the voltage across the parallel network.

\[ \mathcal E=V_2+V_p \]

\[ \mathcal E=5.5+6.0 \]

\[ \boxed{\mathcal E=11.5\text{ V}} \]

Step 7: Summarise the potential differences.

The 3.0 Ω resistor still carries current because it is not bypassed. It remains in the lower branch, in series with the short-circuit path.

The 6.0 Ω resistor alone has been bypassed.

Question 4

An ideal 12.0 V supply is connected to a 3.0 Ω resistor in series with a 6.0 Ω resistor.

An ideal wire is connected directly across the 6.0 Ω resistor.

A student argues:

“The 6.0 Ω resistor carries zero current, so it behaves like an open circuit. Removing the bypass wire should make no difference.”

Evaluate this claim.

Calculate the total current and voltage across the 6.0 Ω resistor in each of the following situations:

(a) The original circuit, with the bypass wire connected.

(b) The 6.0 Ω resistor is physically removed, but the bypass wire remains connected.

(c) The bypass wire is removed, but the 6.0 Ω resistor remains connected.

Solution 4

The student’s claim is incorrect. A shorted resistor is not an open circuit. With the bypass wire connected, the total current is 4.0 A. Removing the resistor while retaining the wire changes nothing, but removing the wire makes the two resistors a series circuit with a current of 1.33 A.

Step 1: Analyse the original circuit.

The bypass wire connects both terminals of the 6.0 Ω resistor.

Therefore:

\[ V_6=0\text{ V} \]

\[ I_6=0\text{ A} \]

The 3.0 Ω resistor is the only resistance limiting the circuit current.

\[ I=\frac{12.0}{3.0} \]

\[ \boxed{I=4.0\text{ A}} \]

The bypass wire carries 4.0 A.

Step 2: Remove the 6.0 Ω resistor but leave the wire.

The wire still connects the same two points.

The circuit therefore behaves exactly as it did before.

\[ \boxed{I=4.0\text{ A}} \]

The voltage across the removed resistor is no longer defined as a component voltage, but the two connection points remain at the same potential.

This demonstrates that the resistor itself was not contributing to the circuit’s behaviour.

Step 3: Remove the bypass wire but leave the 6.0 Ω resistor.

Now the circuit has a single current path through both resistors.

They are connected in series.

The total resistance becomes:

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

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

The new circuit current is:

\[ I=\frac{12.0}{9.0} \]

\[ \boxed{I=1.33\text{ A}} \]

Step 4: Calculate the new voltage across the 6.0 Ω resistor.

\[ V_6=IR_6 \]

\[ V_6=\frac{12.0}{9.0}\times6.0 \]

\[ \boxed{V_6=8.0\text{ V}} \]

The voltage across the 3.0 Ω resistor is:

\[ V_3=\frac{12.0}{9.0}\times3.0 \]

\[ \boxed{V_3=4.0\text{ V}} \]

KVL confirms:

\[ 8.0+4.0=12.0\text{ V} \]

The comparison is important:

The resistor having zero current does not mean the circuit path is open. The wire is still providing a complete path for current.

Question 5

A cell has an emf of 24.0 V and an internal resistance of 2.0 Ω.

Its external circuit contains a 4.0 Ω resistor in series with a parallel network containing:

• A 6.0 Ω resistor.
• A 3.0 Ω resistor.

An ideal switch connects directly between the two rails of the parallel network. When closed, the switch provides a short-circuit path across both resistor branches.

Calculate the following for both switch positions:

(a) The total current supplied by the cell.

(b) The terminal voltage of the cell.

(c) The current through each resistor and the switch.

(d) The power dissipated by each external resistor.

(e) Explain why opening the switch increases the cell’s terminal voltage, even though the total circuit current decreases.

Assume all resistors are ohmic, the connecting wires and closed switch are ideal, and the internal resistance remains constant.

Solution 5

When the switch is closed, the total current is 4.0 A and the terminal voltage is 16.0 V. Both parallel resistors carry zero current.

When the switch is opened, the total current decreases to 3.0 A, but the terminal voltage increases to 18.0 V. The 6.0 Ω and 3.0 Ω resistors then carry 1.0 A and 2.0 A respectively.

The reason is that the cell has internal resistance. Its terminal voltage changes as the current changes.

Step 1: Analyse the closed-switch condition.

When the switch is closed, it connects nodes A and B directly.

Both nodes are at the same potential.

Therefore:

\[ V_6=V_3=0\text{ V} \]

The current through both parallel resistors is zero.

The switch carries the current instead.

The only resistances limiting the total current are the cell’s internal resistance and the external 4.0 Ω resistor.

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

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

Step 2: Calculate the total current when the switch is closed.

Apply Ohm’s law using the cell’s emf and total circuit resistance:

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

\[ I=\frac{24.0}{6.0} \]

\[ \boxed{I=4.0\text{ A}} \]

The switch carries 4.0 A.

Step 3: Calculate the terminal voltage when the switch is closed.

The terminal voltage is the potential difference across the cell’s external terminals.

It is calculated using:

\[ V_t=\mathcal E-Ir \]

where \(r\) is the cell’s internal resistance.

Substitute:

\[ V_t=24.0-(4.0\times2.0) \]

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

The entire external terminal voltage is across the 4.0 Ω resistor because the parallel network has zero voltage.

Check:

\[ V_4=IR \]

\[ V_4=4.0\times4.0 \]

\[ V_4=16.0\text{ V} \]

Step 4: Calculate the power dissipated when the switch is closed.

Power is the rate at which electrical energy is transferred.

For a resistor:

\[ P=I^2R \]

The 4.0 Ω resistor dissipates:

\[ P_4=4.0^2\times4.0 \]

\[ \boxed{P_4=64\text{ W}} \]

Both parallel resistors have zero current and zero voltage.

Therefore:

\[ \boxed{P_6=P_3=0\text{ W}} \]

The closed ideal switch also dissipates zero power.

Step 5: Analyse the open-switch condition.

Opening the switch removes the bypass path.

The 6.0 Ω and 3.0 Ω resistors are now connected in parallel normally.

Calculate their equivalent resistance:

\[ R_p=\frac{6.0\times3.0}{6.0+3.0} \]

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

The external resistance is:

\[ R_{\text{external}}=4.0+2.0 \]

\[ R_{\text{external}}=6.0\ \Omega \]

Including the internal resistance:

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

\[ \boxed{R_{\text{total}}=8.0\ \Omega} \]

Step 6: Calculate the new total current.

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

\[ I=\frac{24.0}{8.0} \]

\[ \boxed{I=3.0\text{ A}} \]

Notice that the total current has decreased from 4.0 A to 3.0 A.

Step 7: Calculate the new terminal voltage.

\[ V_t=\mathcal E-Ir \]

\[ V_t=24.0-(3.0\times2.0) \]

\[ \boxed{V_t=18.0\text{ V}} \]

The terminal voltage has increased from 16.0 V to 18.0 V.

This is not a contradiction.

When the switch opens, the total current decreases. Less voltage is lost across the cell’s internal resistance.

With the switch closed, the internal voltage drop was:

\[ V_r=4.0\times2.0=8.0\text{ V} \]

With the switch open, it becomes:

\[ V_r=3.0\times2.0=6.0\text{ V} \]

The reduction in internal voltage drop allows the terminal voltage to increase.

Step 8: Calculate the voltage across the external 4.0 Ω resistor.

\[ V_4=IR \]

\[ V_4=3.0\times4.0 \]

\[ \boxed{V_4=12.0\text{ V}} \]

The voltage across the parallel network is:

\[ V_p=18.0-12.0 \]

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

Step 9: Calculate the individual branch currents.

For the 6.0 Ω resistor:

\[ I_6=\frac{6.0}{6.0} \]

\[ \boxed{I_6=1.0\text{ A}} \]

For the 3.0 Ω resistor:

\[ I_3=\frac{6.0}{3.0} \]

\[ \boxed{I_3=2.0\text{ A}} \]

The switch is open, so:

\[ \boxed{I_{\text{switch}}=0\text{ A}} \]

KCL confirms:

\[ I_{\text{total}}=I_6+I_3 \]

\[ 3.0=1.0+2.0 \]

Step 10: Calculate the power dissipated in each external resistor.

For the 4.0 Ω resistor:

\[ P_4=I^2R \]

\[ P_4=3.0^2\times4.0 \]

\[ \boxed{P_4=36\text{ W}} \]

For the 6.0 Ω resistor:

\[ P_6=1.0^2\times6.0 \]

\[ \boxed{P_6=6\text{ W}} \]

For the 3.0 Ω resistor:

\[ P_3=2.0^2\times3.0 \]

\[ \boxed{P_3=12\text{ W}} \]

The total external power is:

\[ P_{\text{external}}=36+6+12 \]

\[ \boxed{P_{\text{external}}=54\text{ W}} \]

The results for both switch positions are:

The interesting result is that opening the short circuit reduces the total current but increases the cell’s terminal voltage.

The cell’s emf stays constant. Its terminal voltage changes because the voltage lost inside the cell depends on the current through its internal resistance.

This is a useful reminder that you cannot always treat a battery’s terminal voltage as fixed when internal resistance is included.

09Where to go next

Once you can identify a shorted resistor, the remaining circuit is usually a standard series, parallel, or combination-circuit problem.

The key is to preserve the correct connections:

• A shorted resistor has zero potential difference and zero current in the ideal model.
• The bypass wire can carry the current that would otherwise have passed through the resistor.
• Other resistors connected between the same two shorted nodes may also be bypassed.
• Components in series with the shorted section remain part of the circuit.

For more practice with the remaining network, see How to Solve Combination Circuits Without Losing Track of Current and How to Find Branch Currents in a Three-Branch Parallel Circuit.

The most reliable habit is to identify the nodes first, apply KCL and KVL second, and simplify the resistances last.