CIE AS/A Level Physics 10.2 Kirchhoff’s laws Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 10.2 Kirchhoff’s laws Study Notes – New Syllabus
CIE AS/A Level Physics 10.2 Kirchhoff’s laws Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- recall Kirchhoff’s first law and understand that it is a consequence of conservation of charge
- recall Kirchhoff’s second law and understand that it is a consequence of conservation of energy
- derive, using Kirchhoff’s laws, a formula for the combined resistance of two or more resistors in series
- use the formula for the combined resistance of two or more resistors in series
- derive, using Kirchhoff’s laws, a formula for the combined resistance of two or more resistors in parallel
- use the formula for the combined resistance of two or more resistors in parallel
- use Kirchhoff’s laws to solve simple circuit problems
Kirchhoff’s First Law (Current Law — KCL)
Statement: At any junction (or node) in an electrical circuit, the sum of currents entering the junction equals the sum of currents leaving the junction.
\( \mathrm{\sum I_{in} = \sum I_{out}} \)
Explanation:
- Electric current is the flow of charge per unit time.
- When charge flows into a junction, it cannot disappear or accumulate — it must flow out again.
- This law applies to both direct current (DC) and alternating current (AC) circuits.
Origin (Conservation of Charge):
- The total charge entering a junction per second equals the total charge leaving it per second.
- Hence, no charge is lost or created at a junction — charge is conserved.
Kirchhoff’s First Law is a direct consequence of the law of conservation of electric charge — charge can neither be created nor destroyed within an electrical circuit.
Example
At a junction, three wires meet. The currents are \( \mathrm{I_1 = 4 \, A} \) entering, \( \mathrm{I_2 = 2.5 \, A} \) entering, and \( \mathrm{I_3} \) leaving. Find the value and direction of \( \mathrm{I_3.} \)
▶️ Answer / Explanation
By Kirchhoff’s First Law:
\( \mathrm{I_1 + I_2 = I_3} \)
\( \mathrm{I_3 = 4.0 + 2.5 = 6.5 \, A.} \)
Therefore, a current of 6.5 A leaves the junction.
Example
At a junction, a current of \( \mathrm{8.0 \, A} \) enters and splits into two branches carrying \( \mathrm{3.0 \, A} \) and \( \mathrm{I_2.} \) Find \( \mathrm{I_2.} \)
▶️ Answer / Explanation
Applying Kirchhoff’s First Law:
\( \mathrm{I_{in} = I_{out}} \Rightarrow 8.0 = 3.0 + I_2 \)
\( \mathrm{I_2 = 5.0 \, A.} \)
Hence, 5.0 A flows out through the second branch.
Kirchhoff’s Second Law (Voltage Law — KVL)
Statement: In any closed loop of an electrical circuit, the sum of the electromotive forces (e.m.f.s) is equal to the sum of the potential drops (voltage drops) across all resistive elements.
\( \mathrm{\sum \varepsilon = \sum IR} \)
Explanation:
- As a charge moves around a closed circuit, it gains energy from sources (cells, batteries) and loses energy through resistors or other components.
- The total energy gained per unit charge (sum of e.m.f.s) equals the total energy lost per unit charge (sum of potential differences).
Origin (Conservation of Energy):
- Energy supplied by sources = Energy used in resistors and other devices.
- No energy is created or lost overall — it is only converted from electrical to other forms (heat, light, etc.).
Kirchhoff’s Second Law is a direct consequence of the law of conservation of energy — total energy gained by charges in a circuit equals total energy lost per cycle of flow.
Example
A circuit contains a battery of e.m.f. \( \mathrm{12 \, V} \) and two resistors of \( \mathrm{2.0 \, \Omega} \) and \( \mathrm{4.0 \, \Omega} \) in series. Find the current in the circuit and the potential difference across each resistor.
▶️ Answer / Explanation
Total resistance \( \mathrm{R_T = 2.0 + 4.0 = 6.0 \, \Omega.} \)
Current: \( \mathrm{I = \dfrac{\varepsilon}{R_T} = \dfrac{12}{6} = 2.0 \, A.} \)
Potential drops:
- \( \mathrm{V_1 = IR_1 = 2.0 \times 2.0 = 4.0 \, V.} \)
- \( \mathrm{V_2 = IR_2 = 2.0 \times 4.0 = 8.0 \, V.} \)
Check Kirchhoff’s Second Law:
\( \mathrm{E.m.f. = V_1 + V_2 \Rightarrow 12 = 4 + 8} \)
Law satisfied — energy gained equals energy lost.
Example
A battery of e.m.f. \( \mathrm{9.0 \, V} \) and internal resistance \( \mathrm{1.0 \, \Omega} \) is connected to an external resistor of \( \mathrm{4.0 \, \Omega.} \) Find the current and terminal potential difference.
▶️ Answer / Explanation
Total resistance: \( \mathrm{R_T = r + R = 1.0 + 4.0 = 5.0 \, \Omega.} \)
Current: \( \mathrm{I = \dfrac{\varepsilon}{R_T} = \dfrac{9.0}{5.0} = 1.8 \, A.} \)
Potential drop inside battery: \( \mathrm{V_{internal} = Ir = 1.8 \times 1.0 = 1.8 \, V.} \)
Terminal potential difference: \( \mathrm{V = \varepsilon – Ir = 9.0 – 1.8 = 7.2 \, V.} \)
Check Kirchhoff’s Second Law:
\( \mathrm{\varepsilon = V + Ir \Rightarrow 9.0 = 7.2 + 1.8} \)
Law verified — total energy supplied equals total energy lost.
Derivation — Combined Resistance of Resistors in Series (using Kirchhoff’s laws)
Setup: Two resistors \( \mathrm{R_1} \) and \( \mathrm{R_2} \) are connected in series with a source of e.m.f. \( \mathrm{\varepsilon} \). The same current \( \mathrm{I} \) flows through both resistors because series connection gives a single current path.
Apply Kirchhoff’s second law (loop/energy law):
Around the loop the sum of e.m.f.s equals the sum of potential drops:
\( \mathrm{\varepsilon = V_{R_1} + V_{R_2}} \)
Use Ohm’s law for each resistor (\( \mathrm{V = IR} \)):
\( \mathrm{\varepsilon = I R_1 + I R_2 = I (R_1 + R_2)} \)
Define the equivalent (combined) resistance \( \mathrm{R_{eq}} \) as the single resistance that would draw the same current from the same source, so \( \mathrm{\varepsilon = I R_{eq}} \).
Comparing the two expressions for \( \mathrm{\varepsilon} \) gives:
\( \mathrm{I R_{eq} = I (R_1 + R_2) \ \Rightarrow \ R_{eq} = R_1 + R_2} \)
Generalisation for \( \mathrm{n} \) resistors in series:
\( \mathrm{R_{eq} = R_1 + R_2 + R_3 + \dots + R_n = \sum_{k=1}^{n} R_k} \)
Key point: In series the same current flows through each resistor and the total (combined) resistance is the arithmetic sum of the individual resistances.
Example
A battery of e.m.f. \( \mathrm{12.0 \ V} \) is connected to two resistors in series: \( \mathrm{R_1 = 4.0 \ \Omega} \) and \( \mathrm{R_2 = 6.0 \ \Omega} \). Find
(a) the combined resistance,
(b) the current in the circuit, and
(c) the potential difference across each resistor.
▶️ Answer / Explanation
(a) Combined resistance:
\( \mathrm{R_{eq} = R_1 + R_2 = 4.0 + 6.0 = 10.0 \ \Omega.} \)
(b) Current (using \( \mathrm{I = \dfrac{\varepsilon}{R_{eq}}} \)):
\( \mathrm{I = \dfrac{12.0}{10.0} = 1.20 \ A.} \)
(c) Potential differences (using \( \mathrm{V = IR} \)):
Across \( \mathrm{R_1: \ V_1 = I R_1 = 1.20 \times 4.0 = 4.80 \ V.} \)
Across \( \mathrm{R_2: \ V_2 = I R_2 = 1.20 \times 6.0 = 7.20 \ V.} \)
Check: \( \mathrm{V_1 + V_2 = 4.80 + 7.20 = 12.0 \ V = \varepsilon} \) (Kirchhoff’s second law satisfied).
Example
Three resistors \( \mathrm{R_1 = 2.0 \ \Omega, \ R_2 = 3.0 \ \Omega, \ R_3 = 5.0 \ \Omega} \) are connected in series across a \( \mathrm{24 \ V} \) supply. Find
(a) the equivalent resistance,
(b) the circuit current, and
(c) the power dissipated in \( \mathrm{R_3} \).
▶️ Answer / Explanation
(a) Equivalent resistance:
\( \mathrm{R_{eq} = 2.0 + 3.0 + 5.0 = 10.0 \ \Omega.} \)
(b) Circuit current:
\( \mathrm{I = \dfrac{\varepsilon}{R_{eq}} = \dfrac{24}{10.0} = 2.40 \ A.} \)
(c) Power dissipated in \( \mathrm{R_3} \):
First find \( \mathrm{V_3 = I R_3 = 2.40 \times 5.0 = 12.0 \ V.} \)
Then power \( \mathrm{P_3 = V_3 I = 12.0 \times 2.40 = 28.8 \ W.} \)
(Alternatively \( \mathrm{P_3 = I^2 R_3 = (2.40)^2 \times 5.0 = 5.76 \times 5.0 = 28.8 \ W} \).)
Use the Formula for the Combined Resistance of Two or More Resistors in Series
Formula:
\( \mathrm{R_{eq} = R_1 + R_2 + R_3 + \dots + R_n} \)
This equation is used whenever resistors are connected **in series**, meaning they are joined end-to-end and the same current flows through each one.
Key Points When Using the Formula:
- The same current \( \mathrm{I} \) flows through all series components.
- The total potential difference across the combination is the sum of the potential differences across each resistor: \( \mathrm{V = V_1 + V_2 + V_3 + \dots} \).
- The combined resistance always increases when resistors are added in series — total resistance is greater than any individual resistance.
Example
Three resistors of \( \mathrm{10 \, \Omega} \), \( \mathrm{15 \, \Omega} \), and \( \mathrm{25 \, \Omega} \) are connected in series. Calculate their total (equivalent) resistance.
▶️ Answer / Explanation
Using the series formula:
\( \mathrm{R_{eq} = R_1 + R_2 + R_3 = 10 + 15 + 25 = 50 \, \Omega.} \)
Therefore, the total resistance of the combination is \( \mathrm{50 \, \Omega.} \)
Interpretation: The total resistance is simply the sum of the individual resistances, as the current must pass through each resistor one after another.
Example
A battery of e.m.f. \( \mathrm{9.0 \, V} \) is connected in series with two resistors: \( \mathrm{R_1 = 1.5 \, \Omega} \) and \( \mathrm{R_2 = 3.5 \, \Omega.} \) Find:
- (a) The total resistance
- (b) The current in the circuit
- (c) The potential difference across each resistor
▶️ Answer / Explanation
(a) Total resistance:
\( \mathrm{R_{eq} = R_1 + R_2 = 1.5 + 3.5 = 5.0 \, \Omega.} \)
(b) Current in the circuit:
Using \( \mathrm{I = \dfrac{\varepsilon}{R_{eq}}} \): \( \mathrm{I = \dfrac{9.0}{5.0} = 1.8 \, A.} \)
(c) Voltage across each resistor:
\( \mathrm{V_1 = I R_1 = 1.8 \times 1.5 = 2.7 \, V.} \)
\( \mathrm{V_2 = I R_2 = 1.8 \times 3.5 = 6.3 \, V.} \)
Check (Kirchhoff’s second law):
\( \mathrm{V_1 + V_2 = 2.7 + 6.3 = 9.0 \, V = \varepsilon.} \)
Law satisfied — total voltage across resistors equals source voltage.
Example
In Example 2, a third resistor of \( \mathrm{R_3 = 2.0 \, \Omega} \) is added in series. Find the new total resistance and current.
▶️ Answer / Explanation
New total resistance:
\( \mathrm{R_{eq} = 1.5 + 3.5 + 2.0 = 7.0 \, \Omega.} \)
New current:
\( \mathrm{I = \dfrac{\varepsilon}{R_{eq}} = \dfrac{9.0}{7.0} = 1.29 \, A.} \)
Interpretation: Adding another resistor in series increases the total resistance, thereby decreasing the current through the circuit.
Derivation — Combined Resistance of Resistors in Parallel (Using Kirchhoff’s Laws)
Setup:
Consider two resistors, \( \mathrm{R_1} \) and \( \mathrm{R_2} \), connected in parallel across a voltage source of e.m.f. \( \mathrm{V} \).
In a parallel circuit:
- The potential difference \( \mathrm{V} \) across each resistor is the same.
- The total current from the source \( \mathrm{I} \) splits into \( \mathrm{I_1} \) and \( \mathrm{I_2} \), where \( \mathrm{I = I_1 + I_2.} \)
Step 1 — Apply Ohm’s Law to each branch:
\( \mathrm{I_1 = \dfrac{V}{R_1}} \quad \text{and} \quad \mathrm{I_2 = \dfrac{V}{R_2}} \)
Step 2 — Apply Kirchhoff’s First Law at the junction:
\( \mathrm{I = I_1 + I_2} \)
Substitute the expressions for \( \mathrm{I_1} \) and \( \mathrm{I_2:} \)
\( \mathrm{I = \dfrac{V}{R_1} + \dfrac{V}{R_2}} \)
Factor out \( \mathrm{V} \):
\( \mathrm{I = V \left( \dfrac{1}{R_1} + \dfrac{1}{R_2} \right)} \)
Step 3 — Define equivalent resistance \( \mathrm{R_{eq}} \):
The equivalent resistance is defined by \( \mathrm{I = \dfrac{V}{R_{eq}}.} \)
Substitute this into the above equation:
\( \mathrm{\dfrac{V}{R_{eq}} = V \left( \dfrac{1}{R_1} + \dfrac{1}{R_2} \right)} \)
Cancel \( \mathrm{V} \) (since \( \mathrm{V \neq 0} \)):
\( \mathrm{\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}} \)
Generalization for n resistors in parallel:
\( \mathrm{\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dots + \dfrac{1}{R_n}} \)
Key Idea:
In a parallel circuit, the reciprocal of the total (combined) resistance equals the sum of the reciprocals of individual resistances. This arises from Kirchhoff’s first law (conservation of charge — current splitting at junctions) and Ohm’s law.
Important Note:
- The total resistance of resistors in parallel is always less than the smallest individual resistance.
- Adding more resistors in parallel decreases the total resistance, as more current paths become available.
Example
Two resistors \( \mathrm{R_1 = 6.0 \, \Omega} \) and \( \mathrm{R_2 = 3.0 \, \Omega} \) are connected in parallel. Calculate their combined resistance.
▶️ Answer / Explanation
Using the formula:
\( \mathrm{\dfrac{1}{R_{eq}} = \dfrac{1}{6.0} + \dfrac{1}{3.0}} \)
\( \mathrm{\dfrac{1}{R_{eq}} = \dfrac{1 + 2}{6} = \dfrac{3}{6} = \dfrac{1}{2}} \)
\( \mathrm{R_{eq} = 2.0 \, \Omega.} \)
Hence, the total resistance of the two resistors in parallel is \( \mathrm{2.0 \, \Omega.} \)
Use the Formula for the Combined Resistance of Two or More Resistors in Parallel
Formula:
\( \mathrm{\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dots + \dfrac{1}{R_n}} \)
Key Concepts:
- Each resistor in parallel has the same potential difference across it.
- The total current from the source is the sum of the currents through each resistor: \( \mathrm{I = I_1 + I_2 + I_3 + \dots} \).
- The combined resistance \( \mathrm{R_{eq}} \) is always less than the smallest individual resistance because more current paths reduce total opposition.
Example
Two resistors, \( \mathrm{R_1 = 4.0 \, \Omega} \) and \( \mathrm{R_2 = 6.0 \, \Omega} \), are connected in parallel. Find the total resistance.
▶️ Answer / Explanation
Using the formula:
\( \mathrm{\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{1}{4.0} + \dfrac{1}{6.0}} \)
\( \mathrm{\dfrac{1}{R_{eq}} = \dfrac{3 + 2}{12} = \dfrac{5}{12}} \)
\( \mathrm{R_{eq} = \dfrac{12}{5} = 2.4 \, \Omega.} \)
Hence, the total resistance is \( \mathrm{2.4 \, \Omega.} \)
Check: \( \mathrm{R_{eq}} \) is less than the smallest resistor (4 Ω)
Example
Three resistors \( \mathrm{R_1 = 6 \, \Omega} \), \( \mathrm{R_2 = 3 \, \Omega} \), and \( \mathrm{R_3 = 2 \, \Omega} \) are connected in parallel. Find the combined resistance.
▶️ Answer / Explanation
Using the formula:
\( \mathrm{\dfrac{1}{R_{eq}} = \dfrac{1}{6} + \dfrac{1}{3} + \dfrac{1}{2}} \)
Find the LCM of denominators (6): \( \mathrm{\dfrac{1}{6} + \dfrac{2}{6} + \dfrac{3}{6} = \dfrac{6}{6}} \)
\( \mathrm{\dfrac{1}{R_{eq}} = 1.0 \ \Rightarrow \ R_{eq} = 1.0 \, \Omega.} \)
Hence, the total resistance is \( \mathrm{1.0 \, \Omega.} \)
Check: \( \mathrm{R_{eq}} \) is smaller than the smallest resistor (2 Ω)
Example
A \( \mathrm{12.0 \, V} \) battery is connected across two resistors in parallel: \( \mathrm{R_1 = 8.0 \, \Omega} \) and \( \mathrm{R_2 = 12.0 \, \Omega.} \) Find:
- (a) The combined resistance,
- (b) The total current from the battery,
- (c) The current through each resistor.
▶️ Answer / Explanation
(a) Combined resistance:
\( \mathrm{\dfrac{1}{R_{eq}} = \dfrac{1}{8.0} + \dfrac{1}{12.0}} = \dfrac{3 + 2}{24} = \dfrac{5}{24}} \)
\( \mathrm{R_{eq} = \dfrac{24}{5} = 4.8 \, \Omega.} \)
(b) Total current from the battery:
\( \mathrm{I = \dfrac{V}{R_{eq}} = \dfrac{12.0}{4.8} = 2.5 \, A.} \)
(c) Currents in each branch:
\( \mathrm{I_1 = \dfrac{V}{R_1} = \dfrac{12.0}{8.0} = 1.5 \, A.} \)
\( \mathrm{I_2 = \dfrac{V}{R_2} = \dfrac{12.0}{12.0} = 1.0 \, A.} \)
Check using Kirchhoff’s First Law: \( \mathrm{I = I_1 + I_2 = 1.5 + 1.0 = 2.5 \, A.} \)
Hence, the circuit obeys charge conservation, and the total resistance gives correct current division.
Example
Three identical resistors, each of resistance \( \mathrm{12 \, \Omega} \), are connected in parallel. Find the combined resistance.
▶️ Answer / Explanation
For identical resistors, the formula simplifies to:
\( \mathrm{\dfrac{1}{R_{eq}} = n \times \dfrac{1}{R}} \ \Rightarrow \ \mathrm{R_{eq} = \dfrac{R}{n}} \)
Here \( \mathrm{R = 12 \, \Omega} \), \( \mathrm{n = 3} \):
\( \mathrm{R_{eq} = \dfrac{12}{3} = 4.0 \, \Omega.} \)
Hence, the total resistance is \( \mathrm{4.0 \, \Omega.} \)
Note: When identical resistors are connected in parallel, the total resistance is the resistance of one divided by the number of resistors.
Using Kirchhoff’s Laws to Solve Simple Circuit Problems
Overview:
Kirchhoff’s two circuit laws — the Current Law (KCL) and the Voltage Law (KVL) — are powerful tools for analysing complex circuits containing multiple loops and junctions. They allow us to determine unknown currents, voltages, and resistances in any part of a circuit.
Recap of the Laws:
| Law | Statement | Conservation Principle |
|---|---|---|
| Kirchhoff’s First Law (KCL) | The total current entering a junction equals the total current leaving it. | Conservation of charge |
| Kirchhoff’s Second Law (KVL) | In any closed loop, the sum of e.m.f.s equals the sum of potential drops. | Conservation of energy |
Step-by-Step Method for Solving Circuits Using Kirchhoff’s Laws:
- Label all currents in the circuit (assign directions — even if wrong, the correct sign will emerge in the solution).
- Apply KCL at junctions to relate the currents.
- Apply KVL to each independent loop to relate voltages, resistances, and currents.
- Use Ohm’s law \( \mathrm{V = IR} \) for resistive elements in each loop.
- Solve simultaneous equations for unknown currents or voltages.
Example
In the circuit below:
- Battery 1: \( \mathrm{\varepsilon_1 = 6.0 \, V} \), internal resistance negligible
- Battery 2: \( \mathrm{\varepsilon_2 = 12.0 \, V} \), internal resistance negligible
- Resistors: \( \mathrm{R_1 = 2.0 \, \Omega} \), \( \mathrm{R_2 = 3.0 \, \Omega} \), \( \mathrm{R_3 = 4.0 \, \Omega} \)
Find the current in each branch using Kirchhoff’s laws.
▶️ Answer / Explanation
Step 1 — Label Currents:
- \( \mathrm{I_1} \): current from \( \varepsilon_1 \) through \( R_1 \)
- \( \mathrm{I_2} \): current from \( \varepsilon_2 \) through \( R_3 \)
- \( \mathrm{I_3} \): current through the middle resistor \( R_2 \)
At the central junction, by KCL: \( \mathrm{I_3 = I_1 – I_2} \) (assuming directions are consistent).
Step 2 — Apply KVL to Loop 1 (left loop):
\( \mathrm{6.0 – 2I_1 – 3I_3 = 0} \)
Step 3 — Apply KVL to Loop 2 (right loop):
\( \mathrm{12.0 – 4I_2 – 3I_3 = 0} \)
Step 4 — Substitute \( \mathrm{I_3 = I_1 – I_2} \) into both equations:
Equation (1): \( \mathrm{6 = 2I_1 + 3(I_1 – I_2)} \ \Rightarrow \ \mathrm{6 = 5I_1 – 3I_2} \)
Equation (2): \( \mathrm{12 = 4I_2 + 3(I_1 – I_2)} \ \Rightarrow \ \mathrm{12 = 3I_1 + I_2} \)
Step 5 — Solve Simultaneous Equations:
- From (1): \( \mathrm{5I_1 – 3I_2 = 6} \)
- From (2): \( \mathrm{3I_1 + I_2 = 12} \)
Multiply (2) by 3 to eliminate \( \mathrm{I_2} \): \( \mathrm{9I_1 + 3I_2 = 36} \)
Add to (1): \( \mathrm{14I_1 = 42 \Rightarrow I_1 = 3.0 \, A.} \)
Substitute into (2): \( \mathrm{3(3.0) + I_2 = 12 \Rightarrow I_2 = 3.0 \, A.} \)
Then \( \mathrm{I_3 = I_1 – I_2 = 0.} \)
Results:
- \( \mathrm{I_1 = 3.0 \, A} \)
- \( \mathrm{I_2 = 3.0 \, A} \)
- \( \mathrm{I_3 = 0 \, A} \)
Interpretation: Both batteries supply equal currents through their branches, and no current flows through the central resistor since both sides are at the same potential.
Example
A battery of e.m.f. \( \mathrm{9.0 \, V} \) and internal resistance \( \mathrm{1.0 \, \Omega} \) is connected to a resistor \( \mathrm{4.0 \, \Omega.} \) Find:
- (a) The current in the circuit
- (b) The terminal potential difference
▶️ Answer / Explanation
Step 1 — Apply KVL around the loop:
\( \mathrm{\varepsilon – I r – I R = 0} \)
Substitute values: \( \mathrm{9.0 – I(1.0) – I(4.0) = 0} \Rightarrow 9.0 = 5I \)
Step 2 — Solve for \( \mathrm{I} \):
\( \mathrm{I = 1.8 \, A.} \)
Step 3 — Find terminal potential difference:
\( \mathrm{V = \varepsilon – I r = 9.0 – 1.8(1.0) = 7.2 \, V.} \)
Results:
- Current \( \mathrm{I = 1.8 \, A} \)
- Terminal voltage \( \mathrm{V = 7.2 \, V} \)
Check: The potential drop across internal resistance (1.8 V) plus the external resistor (7.2 V) equals total e.m.f. (9.0 V) Kirchhoff’s second law verified.
Example
In the circuit below, a 12 V battery is connected to two resistors in parallel: \( \mathrm{R_1 = 6 \, \Omega} \) and \( \mathrm{R_2 = 3 \, \Omega.} \) Find the current through each resistor and verify Kirchhoff’s first law at the junction.
▶️ Answer / Explanation
Step 1 — Currents in each branch:
\( \mathrm{I_1 = \dfrac{V}{R_1} = \dfrac{12}{6} = 2.0 \, A.} \)
\( \mathrm{I_2 = \dfrac{V}{R_2} = \dfrac{12}{3} = 4.0 \, A.} \)
Step 2 — Total current from the battery:
\( \mathrm{I = I_1 + I_2 = 2.0 + 4.0 = 6.0 \, A.} \)
Verification:
At the junction: total current entering = total current leaving. \( \mathrm{6.0 \, A = 2.0 + 4.0 \, A} \) Kirchhoff’s first law holds.