Internal Resistance in Electric Circuits: Simple Circuit, Terminal Voltage, and Solved Examples

INTERNAL RESISTANCE OF SOURCES

20.2.1 Internal Resistance

1. Normally, when current flows through a source of electrical energy, the current experiences some opposition. This opposition to the current within the source is known as the internal resistance of the source od electrical energy.

Figure 20.4 A schematic diagram of a battery which possesses internal resistance r and emf E

2. Figure 20.4 shows a schematic diagram of a battery which possesses internal resistance r and emf E.

20.2.2 Effect of Internal Resistance on Terminal Potential Difference

1. Terminal Potential Difference

When no current flows through a battery, the voltage existing across the two terminals of the baterry is equal to the emf E of the battery, as shown in Figure 20.4. However, when a current flows through the battery, which normally has internal resistance, the voltage across the terminals becomes $V_T$ which is less than E, as shown in Figure 20.5.

Figure 20.5 Figure 20.5. When a current flows through a battery with internal resistance, the terminal voltage $V_T$ is less than the electromotive force E due to the voltage drop inside the battery.

2. The voltage $V_T (< E)$ is known as the terminal p.d.

3. Example: without current, E = 1.5 V, but with current flowing through the battery, $V_T=1.3 \ V$. It looks like 0.2 V has been 'lost' due to the presence of internal resistance in the battery.

20.3 A Simple Circuit

20.3.1 A Simple Complete Circuit

Figure 20.6 When the circuit is closed, current flows through the battery and its internal resistance. As a result, part of the emf is used to overcome the internal resistance, causing the terminal voltage $V_T$ to be less than the emf E.

1. A simple circuit

Figure 20.6 shows a simple complete electric circuit. It is composed of two parts:

(a) a source of electrical energy of emf E
(b) an ohmic resistor of resistance R

A current I flows through the circuit.

2. Equation for a complete circuit

Refer to the circuit above. Suppose the source of electrical energy is a battery which produces an electrical power output $P_s$. This power is converted to heat which is dissipated at a rate of $P_r$ by the internal resistance r, and at a rate of $P_R$ by the external resistance R. 

According to the law of conservation of energy, we have

$\boxed{P_{total} = P_{internal}+ P_{external}}$

Thus:

$IE = I^2r + I^2R$

Factorizing:

$\boxed{E = I(r + R)}$

2. Terminal Potential Difference

From the equation of the circuit:

$E = Ir + IR$

Since $V_T= IR$, we can write:

$E = Ir + V_T$

where  $V_T= IR$ is the terminal p.d. which exist across the battery when a current flows in the circuit. Hence, terminal p.d. is given by

$\boxed{V_T = E − Ir}$

and is determined by

• the current I in the circuit and
• the internal resistance r of the battery

20.3.2 Determination of the Internal Resistance of a Battery

We can use the simple circuit described above to determine the internal resistance of a battery.

Figure 20.7. An electrical circuit set up with a resistor R and a rheostat. The rheostat is adjusted to allow a small current to flow through the circuit.

Method

1. Set up the circuit as shown in Figure 20.7. R is a resistor. Adjust the rheostat until a low current is flowing in the circuit.
2. Record the terminal p.d. V using the voltmeter and the current I using the ammeter.
3. Repeat the procedure above using several different values of V and record the corresponding values of I.
4. Plot a graph of V against I.

Result

We will get a line with a negative gradient for the V–I graph, as shown in Figure 20.8.

Figure 20.8 A V–I graph showing a straight line with a negative gradient, indicating that the current decreases as the potential difference increases.

Theory

$V = E − Ir$

or

$V = (−r)I + E$

$y = −mx + c$

This equation is equivalent to y = −mx + c if we let: $y = V$; $x = I$; $m = r$; $c = E$, Hence the internal resistance of the battery is given by: 

r = gradient of the line

Example 20.1: Internal Resistance and Current (Step-by-Step Solution)

Figure 20.9. (a) An electrical circuit containing a 10.0 Ω ohmic resistor with a switch that is initially open and then closed at time T. (b) A graph showing the voltmeter reading as a function of time, where the voltage is V before time T and changes when the switch is closed.

Refer to the circuit shown in Figure 20.9(a). R is an ohmic resistor of resistance 10.0 Ω. Initially, the switch is open. At time T the switch is closed. The graph shown the voltmeter reading is V before time T shown in Figure 20.9(b). Determine

(a) the current in the circuit after the switch is closed

(b) the internal resistance of the battery

Answer

Given:

• Resistance, $R = 10.0 \ \Omega$
• Emf of battery, $E = 6.0 \ V$ (before current flows)
• Terminal voltage, $V_T = 5.0 \ V$ (after current flows)

Concept:

When no current flows, the voltmeter reads the electromotive force (emf) of the battery:

$E = V$ (no current)

When current flows, the voltmeter reads the terminal potential difference:

$V_T = IR$

The relation between emf, terminal voltage, and internal resistance is:

$E = V_T + Ir$

(a) Current in the circuit

Using Ohm’s Law:

$V_T = IR$

Substitute values:

$5.0 = I(10.0)$

$I = 0.50 \ A$

Explanation: The terminal voltage is the voltage across the external resistor. Using Ohm’s Law, we can directly calculate the current flowing in the circuit.

(b) Internal resistance of the battery

Using the equation:

$E = V_T + Ir$

Substitute values:

$6.0 = 5.0 + (0.50)r$

$1.0 = 0.50r$

$r = 2.0 \ \Omega$

Explanation: The emf is greater than the terminal voltage because some energy is lost inside the battery due to internal resistance. This loss is represented by the term Ir.

Example 20.2

A resistor of 100 Ω is connected to a battery with an electromotive force (emf) of 12.0 V and an internal resistance of 1.0 Ω. Determine:

(a) the current in the circuit

(b) the terminal potential difference

(c) the electrical power generated by the battery

(d) the power dissipated by the resistor

(e) the power dissipated by the internal resistance

(f) the amount of heat dissipated by the 100 Ω resistor in one minute

Answer

Given:

• Emf, $E = 12.0 \ V$
• Resistance, $R = 100 \ \Omega$
• Internal resistance, $r = 1.0 \ \Omega$
• Time, $t = 60 \ s$

Concept:

For a complete circuit:

$I = \frac{E}{R + r}$

Terminal voltage:

$V = E − Ir$

Power:

$P = IV$, $P = I^2R$, $P = IE$

Heat energy:

$Q = Pt$

(a) Current in the circuit

$I = E / (R + r)$

$I = 12.0 / (100 + 1.0)$

$I = 12.0 / 101$

$I = 0.119 \ A ≈ 0.12 \ A$

(b) Terminal potential difference

$V = E − Ir$

$V = 12.0 − (0.119 × 1.0)$

$V = 11.9 \ V$

(c) Power generated by the battery

$P = IE$

$P = (0.119)(12.0)$

$P = 1.43 \ W$

(d) Power dissipated by the resistor

$P = I^2R$

$P = (0.119)^2 \times 100$

$P ≈ 1.42 \ W$

(e) Power dissipated by internal resistance

$P = I^2r$

$P = (0.119)^2 \times 1.0$

$P ≈ 0.014 \ W$

(f) Heat energy dissipated in 1 minute

$Q = Pt$

$Q = 1.42 \times 60$

$Q ≈ 85.2 \ J$

Explanation:

The total resistance in the circuit is the sum of external resistance and internal resistance. The battery supplies electrical energy, which is partly dissipated in the external resistor and partly inside the battery due to internal resistance. Most of the energy is converted into heat in the external resistor, while only a small portion is lost internally.

Example 20.3 Internal Resistance of a Battery and Unrealistic Results

A dry cell battery with an emf of 1.50 V is connected to a small light bulb with a resistance of 12.0 Ω. The bulb is observed to dissipate 0.90 W of power. Determine:

(a) The internal resistance of the battery

(b) What is unreasonable about the result

(c) Which assumptions may be incorrect or inconsistent

Answer

Given:

• Emf, $E = 1.50 \ V$
• Load resistance, $R = 12.0 \ \Omega$
• Power delivered to load, $P = 0.90 \ W$

Concept:

Power dissipated in the load resistor:

$P = I^2R$

So current is:

$I = \sqrt{\frac{P}{R}}$

Also, for a battery with internal resistance:

$E = I(R + r)$

From this, internal resistance is:

$r = \frac{E}{I} − R$

(a) Internal Resistance

Step 1: Find current

$I = \sqrt{\frac{0.90}{12.0}}$

$I ≈ 0.274 \ A$

Step 2: Calculate internal resistance

$r = (E/I) − R$

$r = (1.50/0.274) − 12.0$

$r ≈ 5.47 − 12.0$

$r ≈ −6.53 \ \Omega$

Explanation: The result gives a negative internal resistance, which is not physically possible for a normal battery.

(b) What is unreasonable?

The calculated internal resistance is negative. In real physical systems:

• Internal resistance must always be positive
• A negative value indicates an impossible situation

(c) Which assumptions are incorrect?

The inconsistency likely comes from unrealistic assumptions:

• The power delivered (0.90 W) is too large for a small 1.50 V dry cell
• Such a battery cannot sustain that current under normal conditions
• The internal resistance of real dry cells is not negligible and limits current

Conclusion:

This example shows the importance of checking whether given values are physically realistic. Even if calculations are correct mathematically, they must also make sense in real-world physics.

Example 20.4: Electric Shock and Internal Resistance

A person accidentally touches the terminals of a high-voltage power supply. The resistance of the human body between the hands is approximately 12.0 kΩ. The power supply provides a voltage of 15.0 kV and has an internal resistance of 1500 Ω. Determine:

(a) Sketch a circuit diagram representing the situation

(b) The current flowing through the person’s body

(c) The power dissipated in the body

(d) The required internal resistance to limit the current to 0.800 mA or less

(e) Whether increasing internal resistance affects the performance of the power supply

Answer

Given:

• Body resistance, $R = 12.0 \ k\Omega = 12,000 \ \Omega$
• Voltage, $V = 15.0 \ kV = 15,000 \ V$
• Internal resistance, $r = 1500 \ \Omega$
• Maximum safe current, $I_{max}= 0.800 \ mA = 0.0008 \ A$

Concept:

The total resistance in the circuit is:

$R_{total}= R + r$

Using Ohm’s Law:

$I = \frac{V}{R_{total}}$

Power dissipated in the body:

$P = I^2R$

(a) Circuit Diagram

The situation can be modeled as a simple series circuit consisting of a voltage source, its internal resistance, and the resistance of the human body.

(b) Current through the body

Step 1: Total resistance

$R_{total} = 12,000 + 1500 = 13,500 \ \Omega$

Step 2: Current

$I = 15,000 / 13,500$

$I ≈ 1.11 \ A$

Explanation: The current is extremely large due to the high voltage, making this situation very dangerous.

(c) Power dissipated in the body

$P = I^2R$

$P = (1.11)^2 \times 12,000$

$P ≈ 1.23 \times 12,000$

$P ≈ 14,760 \ W$

Explanation: A very large amount of power is converted into heat in the body, which can cause severe injury.

(d) Required internal resistance for safety

Step 1: Use Ohm’s Law

$I = \frac{V}{R + r_{new}}$

$0.0008 = \frac{15,000}{12,000 + r_{new}}$

Step 2: Solve for total resistance

$12,000 + r_{new} = 15,000/0.0008$

$12,000 + r_{new} = 18,750,000 \ \Omega$

Step 3: Find new internal resistance

$r_{new} = 18,750,000 − 12,000$

$r_{new} ≈ 18.74 \ M \Omega$

Explanation: A very large internal resistance is needed to limit the current to a safe level.

(e) Effect on power supply performance

Increasing the internal resistance significantly reduces the current supplied by the source.

Conclusion: Yes, this modification would reduce the effectiveness of the power supply, especially for devices requiring large currents. High internal resistance limits current flow, making the source unsuitable for low-resistance loads.

Example 20.5: Charging and Discharging a Car Battery

Figure 20.10 A person opens the car hood and checks the battery condition, illustrating a real-life application of internal resistance and current flow in an automobile electrical system.

A rechargeable car battery has an emf of 12.5 V and an internal resistance of 0.040 Ω. The battery is being charged with a current of 50 A. Determine:

(a) The terminal potential difference during charging

(b) The rate at which thermal energy is dissipated inside the battery

(c) The rate at which electrical energy is converted into chemical energy

(d) The terminal voltage and thermal power when the battery supplies the same current to a motor

Answer

Given:

• Emf, $E = 12.5 \ V$
• Internal resistance, $r = 0.040 \ \Omega$
• Current, $I = 50 \ A$

Concept:

Charging condition:

$V = E + Ir$

Discharging condition:

$V = E − Ir$

Thermal power loss:

$P_{heat} = I^2r$

Power converted to chemical energy:

$P_{chem} = IE$

(a) Terminal Voltage during Charging

$V = E + Ir$

$V = 12.5 + (50 \times 0.040)$

$V = 12.5 + 2.0$

$V = 14.5 \ V$

Explanation: During charging, extra voltage is required to overcome internal resistance and push current into the battery.

(b) Thermal Energy Dissipation

$P = I^2r$

$P = (50)^2 \times 0.040$

$P = 2500 \times 0.040$

$P = 100 \ W$

Explanation: This power is lost as heat due to internal resistance.

(c) Electrical to Chemical Energy Conversion

$P = IE$

$P = 50 \times 12.5$

$P = 625 \ W$

Explanation: This represents useful energy stored in the battery as chemical energy.

(d) When the battery supplies current (discharging)

Terminal voltage:

$V = E − Ir$

$V = 12.5 − (50 \times 0.040)$

$V = 12.5 − 2.0$

$V = 10.5 \ V$

Thermal power:

$P = I^2r$

$P = (50)^2 \times 0.040$

$P = 100 \ W$

Explanation: When discharging, the terminal voltage drops due to internal resistance, but heat loss remains the same because it depends on I²r.

Conclusion:

This example highlights the difference between charging and discharging processes. During charging, the terminal voltage is higher than the emf, while during discharging it is lower. In both cases, internal resistance causes energy loss in the form of heat.

Example 20.6: Charging a Battery and Heat Generation

An automobile battery with an emf of 11.5 V is being charged. During charging, its terminal voltage rises to 15.0 V while a current of 8.0 A flows into the battery. The battery has a mass of 18.0 kg and a specific heat capacity of 0.280 kcal/kg·°C. Assume that no heat is lost to the surroundings. Determine:(a) The internal resistance of the battery

(b) The power dissipated inside the battery

(c) The rate of temperature increase (in °C/min)      

Answer

Given:

• Emf, $E = 11.5 \ V$
• Terminal voltage during charging, $V = 15.0 \ V$
• Current, $I = 8.0 \ A$
• Mass, $m = 18.0 \ kg$
• Specific heat, $c = 0.280 \ kcal/kg^0C$

Concept:

During charging, the terminal voltage is greater than the emf:

$V = E + Ir$

Power dissipated inside the battery:

$P = I^2r$

Heat energy relation:

$Q = mc\Delta T$

Rate of temperature increase:

$P = mc \left(\frac{dT}{dt}\right)$

Conversion: 1 kcal = 4186 J

(a) Internal resistance

Using:

$V = E + Ir$

$15.0 = 11.5 + (8.0)r$

$3.5 = 8.0r$

r = 0.4375 Ω ≈ 0.44 Ω

Explanation: During charging, extra voltage is needed to overcome the internal resistance of the battery.

(b) Power dissipated inside the battery

$P = I^2r$

$P = (8.0)^2 \times 0.4375$

$P = 64 \times 0.4375$

P = 28.0 W

Explanation: This power is converted into heat inside the battery.

(c) Rate of temperature increase

Step 1: Convert specific heat to SI units

$c = 0.280 \times 4186 = 1172.08 J/kg^0C$

Step 2: Use formula

$P = mc \left(\frac{dT}{dt}\right)$

$28.0 = (18.0)(1172.08)\left(\frac{dT}{dt}\right)$

$28.0 = 21097.44\left(\frac{dT}{dt}\right)$

$\left(\frac{dT}{dt}\right) = 28.0/21097.44$

$\left(\frac{dT}{dt}\right)$ ≈ 0.00133 °C/s

Step 3: Convert to °C/min

$\left(\frac{dT}{dt}\right) ≈ 0.00133 \times 60$

$\left(\frac{dT}{dt}\right)$ ≈ 0.080 °C/min

Explanation: The temperature increases slowly because the battery has a large mass and heat capacity.

Conclusion:

During charging, the battery’s internal resistance causes energy loss in the form of heat. This leads to a gradual increase in temperature, although the rate is relatively small due to the battery’s thermal properties.

Example 20.7: Internal Resistance and Unreasonable Results

A small battery with an emf of 1.60 V and an internal resistance of 0.250 Ω is connected to an external circuit. It is observed that a current of 7.00 A flows through the circuit. Determine:

(a) The terminal voltage of the battery
(b) The resistance of the external load
(c) What is unreasonable about the result
(d) Which assumptions may be incorrect or inconsistent

Answer

Given:

• Emf, $E = 1.60 \ V$
• Internal resistance, $r = 0.250 \ \Omega$
• Current,$I = 7.00 \ A$

Concept:

When a battery supplies current, the terminal voltage is:

$V = E − Ir$

The external load resistance is given by Ohm’s Law:

$R = V / I$

(a) Terminal Voltage

$V = E − Ir$

$V = 1.60 − (7.00 \times 0.250)$

$V = 1.60 − 1.75$

V = −0.15 V

Explanation: The terminal voltage becomes negative, which is physically unrealistic for a battery delivering current in a normal circuit.

(b) Load Resistance

$R = V / I$

$R = (−0.15) / 7.00$

R ≈ −0.021 Ω

Explanation: A negative resistance is not possible for a passive load, indicating an inconsistency in the given situation.

(c) What is unreasonable?

The results show:

• Negative terminal voltage
• Negative load resistance

These are not physically meaningful in a normal electrical circuit. Therefore, the situation described is unrealistic.

(d) Which assumptions are unreasonable?

The main issue lies in the assumption that such a large current (7.00 A) can be drawn from a small battery with low emf.

• Small batteries cannot supply such high currents without significant voltage drop
• The internal resistance would prevent such a large current from flowing
• The battery would likely overheat or fail under these conditions

Conclusion:

This problem illustrates that not all mathematically obtained results are physically valid. It is important to check whether the assumptions and values used are realistic in practical situations.

Applications of Internal Resistance in Electric Circuits

Understanding internal resistance in electric circuits is important in many real-life applications. It helps explain how electrical energy is distributed, lost, and utilized in practical systems.

1. Battery Performance

Internal resistance affects how efficiently a battery delivers energy. A battery with high internal resistance produces a lower terminal voltage when current flows, reducing its effectiveness in powering devices.

2. Design of Simple Electric Circuits

In a simple circuit, both internal resistance and external resistance determine the total current. Engineers must consider internal resistance to ensure that circuits operate safely and efficiently.

3. Terminal Voltage in Devices

The terminal potential difference is the actual voltage supplied to a device. Due to internal resistance, this voltage is always less than the emf when current flows. This is important for devices that require stable voltage.

4. Power Loss in Electrical Systems

Some electrical energy is always lost as heat inside the source due to internal resistance. Minimizing this loss is important in designing efficient electrical systems such as power supplies and batteries.

5. Charging and Discharging Batteries

Internal resistance plays a key role during charging and discharging. It affects the rate at which energy is stored or delivered, and also influences heat generation inside the battery.

6. High Current Applications

In devices that require large currents, such as electric vehicles or heaters, internal resistance can cause significant energy loss and heating. Proper design is required to reduce these effects.

Conclusion: Internal Resistance in Electric Circuits

Internal resistance is an essential concept in understanding how real electric circuits operate. Unlike ideal sources, practical batteries always have internal resistance, which causes a drop in terminal voltage when current flows.

This voltage drop reduces the efficiency of energy transfer, since part of the electrical energy is dissipated as heat inside the source. As a result, the terminal potential difference is always less than the electromotive force (emf) during operation.

In simple electric circuits, the total current depends on both external resistance and internal resistance. Therefore, analyzing a complete circuit requires considering all sources of resistance.

Understanding internal resistance is important in designing efficient electrical systems, improving battery performance, and minimizing energy losses. It also helps explain real-world behavior in devices such as batteries, power supplies, and electronic equipment.

In summary: Internal resistance plays a key role in determining current, terminal voltage, power distribution, and overall efficiency in electric circuits.

Frequently Asked Questions (FAQ): Internal Resistance in Electric Circuits

1. What is internal resistance?

Internal resistance is the resistance inside a power source, such as a battery, that opposes the flow of electric current. It causes part of the electrical energy to be lost as heat within the source.

2. Why is terminal voltage less than emf?

When current flows in a circuit, some voltage is dropped across the internal resistance of the battery. As a result, the terminal voltage becomes lower than the electromotive force (emf).

3. What is the formula for terminal voltage?

The terminal voltage is given by the formula:

$V = E − Ir$

where E is the emf, I is the current, and r is the internal resistance.

4. How does internal resistance affect current?

Internal resistance reduces the total current in a circuit. The current is determined by the equation:

$I = E / (R + r)$

where R is the external resistance and r is the internal resistance. 

5. Can internal resistance be zero?

In real-life situations, internal resistance cannot be zero. However, in ideal models, it is sometimes assumed to be zero to simplify calculations.

6. Why is internal resistance important?

Internal resistance is important because it affects the efficiency of electrical systems, power loss, and the performance of batteries and electronic devices.

7. What happens if internal resistance is very large?

If the internal resistance is very large, the current in the circuit becomes very small, and the battery becomes inefficient in delivering power to external devices.

8. How can internal resistance be reduced?

Internal resistance can be reduced by improving the design of the battery, using better conductive materials, and minimizing impurities inside the source.

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Tip: Understanding internal resistance helps in solving many problems related to current, voltage, and power in electric circuits.

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