Introduction

The voltage drop when a power supply (in this case a battery) is placed under load can be modelled as the effect of an internal resistance.

The power supply has an internal EMF, which is the "no-load" (\( I_{\mathrm{L}} = 0 \)) voltage. Load current will cause a voltage drop across the internal resistance, lowering the voltage seen at the supply terminals.

The effect of the internal resistance may be modelled using voltage division, using the internal resistance, and the total resistance.


Calculations using Voltage Division

The general formula for supply voltage with internal resistance is as follows:

\( V_{\mathrm{s}} = \mathcal{E} \cdot \frac{R_{\mathrm{L}}}{R_{\mathrm{L}} + R_{\mathrm{Int}}} \)

where \( V_{\mathrm{s}} \) is the supply voltage seen at the supply terminals, \(\mathcal{E} \) is the internal EMF of the power supply, \( R_{\mathrm{L}} \) is the total load resistance, and \( R_{\mathrm{Int}} \) is the internal resistance of the power supply.

It is not possible to measure \( R_{\mathrm{Int}} \) directly, as it is not possible to separate it from the EMF generating mechanism (e.g. in a battery the electrolyte may contribute a resistive effect at the same time as it assists generating the EMF).

The theoretical method of measuring \( R_{\mathrm{Int}} \) is given by the formula below.

\( R_{\mathrm{Int}} = \frac{V_{\mathrm{oc}}}{I_{\mathrm{sc}}} \)

where \( R_{\mathrm{Int}} \) is the internal resistance, \( V_{\mathrm{oc}} \) is the open-curcuit voltage (no-load voltage), and \( I_{\mathrm{sc}} \) is the short-circuit current. This method is generally not used on real sources, since the short circuit current is often dangerously high. Further, many power supplies (especially current-limited power supplies) have a value of \( R_{\mathrm{Int}} \) that depends on the load current. Most methods of measuring \( R_{\mathrm{Int}} \) rely on measuring the voltage drop for a known load current.

The internal voltage drop is given in the equation below.

\( V_{\mathrm{Int}} = \mathcal{E} - V_{\mathrm{s}} \)

where \( V_{\mathrm{Int}} \) is the internal voltage drop, and \( \mathcal{E} \) and \( V_{\mathrm{s}} \) have the same meanings as above.


Calculations using Load Current

If the load current is known, the calculations can be simplified.

The general formula for supply voltage with internal resistance is as follows:

\( V_{\mathrm{s}} = \mathcal{E} - I_{\mathrm{L}} R_{\mathrm{Int}} \)

where \( V_{\mathrm{s}} \) is the supply voltage seen at the supply terminals, \(\mathcal{E} \) is the internal EMF of the power supply, \( I_{\mathrm{L}} \) is the load current, and \( R_{\mathrm{Int}} \) is the internal resistance of the power supply.

It is not possible to measure \( R_{\mathrm{Int}} \) directly, as it is not possible to separate it from the EMF generating mechanism (e.g. in a battery the electrolyte may contribute a resistive effect at the same time as it assists generating the EMF).

The theoretical method of measuring \( R_{\mathrm{Int}} \) is given by the formula below.

\( R_{\mathrm{Int}} = \frac{\mathcal{E} - V_{\mathrm{s}}}{I_{\mathrm{L}}} \)

where \( R_{\mathrm{Int}} \) is the internal resistance, and all other symbols have the same meanings as above.

\( V_{\mathrm{Int}} = I_{\mathrm{L}} R_{\mathrm{Int}} \)

where \( V_{\mathrm{Int}} \) is the internal voltage drop, and all other symbols have the same meanings as above.


Mixing and Matching Calculations

It is possible to mix and match calculations to get voltages, currents and resistances. As long as the method complies with Ohm's Law and Kirchhoff's Laws, it will give the correct results.


Question 1

A marked-up version of the circuit is shown below.


The objective is to find the following quantities:

  • The voltage drop across \( R_1 \): \( V_{R1} \).
  • The voltage drop across \( R_2 \): \( V_{R2} \).
  • The current through \( R_3 \): \( I_{R3} \).
  • The battery output voltage: \( V_{\mathrm{B}} \).


Solution

The load current \( I_{\mathrm{L}} \) needs to be calculated first. In order to do this, the total load resistance, \( R_{\mathrm{L}} \) needs to be calculated.

\( R_{\mathrm{L}} = R_4 \parallel (R_1 + R_2 \parallel R_3) \).

The value is calculated by plugging the values into the formula above.

\( R_{\mathrm{L}} = 15 \parallel (5 + 20 \parallel 20) = 15 \parallel (5 + 10) = 15 \parallel 15 = \) 7.5 Ω.

The load current \( I_{\mathrm{L}} \) is calculated from the load resistance \( R_{\mathrm{L}} \), internal resistance \( R_{\mathrm{Int}} \), and the internal EMF \( E \) of the battery.

\( I_{\mathrm{L}} = \frac{E}{R_{\mathrm{L}} + R_{\mathrm{Int}}} = \frac{80}{7.5 + 0.5} = \frac{80}{8} = \) 10 A.

The battery voltage \( V_{\mathrm{B}} \) is calculated using either Ohm's Law, or voltage division.

Ohm's Law: \( V_{\mathrm{B}} = E - I_{\mathrm{L}} R_{\mathrm{Int}} = 80 - 10 \cdot 0.5 = 80 - 5 = \) 75 V #.

Voltage Division: \( V_{\mathrm{B}} = E \cdot \frac{R_{\mathrm{L}}}{R_{\mathrm{L}} + R_{\mathrm{Int}}} = 80 \cdot \frac{7.5}{7.5 + 0.5} = 80 \cdot \frac{7.5}{8} = \) 75 V #.

The voltages across \( R_1 \) and \( R_2 \) can be calculated using voltage division.

\( V_{R1} = V_{\mathrm{B}} \cdot \frac{R_1}{R_1 + R_2 \parallel R_3} \).

Substituting the appropriate values yields the calculation below.

\( V_{R1} = 75 \cdot \frac{5}{5 + 20 \parallel 20} = 75 \cdot \frac{5}{15} = 75 \cdot \frac{1}{3} = \) 25 V #.

The voltage across \( R_2 \) must therefore be 50 V by Kirchhoff's Voltage Law.

\( V_{R2} = V_{\mathrm{B}} - V_{\mathrm{R1}} = 75 - 25 = \) 50 V #.

The current through \( R_3 \) is calculated using Ohm's Law. Remember that \( R_2 \) and \( R_3 \) have the same voltage drop across them.

\( I_{R3} = \frac{V_{R2}}{R_3} = \frac{50}{20} = \) 2.5 A #.


Question 2

A marked-up version of the circuit is shown below.


The objective is to find the following quantities:

  • The battery EMF: \( E \).

Given that the current through \( R_2 \) is 7.5 A.


Solution

The 150 Ω resistor \( R_3 \) has 7.5 A current flow through it. This instantly implies that the voltage drop across \( R_2 \) and \( R_3 \) (the resistors are in parallel) is 1125 V (\( 150 \cdot 7.5 \)).

The battery output voltage is the sum of the voltage drop of \( R_2 \) (or \( R_3 \)), and the voltage drop across \( R_4 \) \( V_{\mathrm{R4}} \).

\( V_{\mathrm{B}} = V_{R2} + V_{R4}.

The voltage drop across \( R_4 \) needs to be calculated. To this, the current needs calculation.

\( I_{R4} = I_{R2} + I_{R3} = 7.5 + \frac{1125}{25} = 7.5 + 45 = \) 52.5 A.

The battery voltage may then be calculated using Ohm's Law.

\( V_{\mathrm{B}} = V_{R2} + V_{R4} = 1125 + 52.5 \cdot 100 = 1125 + 5250 = \) 6375 V.

The battery EMF value is given using Ohm's Law.

\( E = V_{\mathrm{B}} + I_{\mathrm{L}} R_{\mathrm{T}} = 6375 + I_{\mathrm{L}} R_{\mathrm{Int}} \).

\( I_{\mathrm{L}} \) is the total load current (including \( R_3 \), \( R_5 \) and \( R_6 \)), and \( R_{\mathrm{Int}} \) is the internal resistance of the voltage source.

The total load current is calculated using Ohm's Law on the battery voltage and the branch current through \( R_3 \) and friends.

\( I_{\mathrm{L}} = I_{R4} + \frac{V_{\mathrm{B}}}{R_3 + R_5 \parallel R_6} = I_{R4} + \frac{V_{\mathrm{B}}}{R_7} \).

The resistance of \( R_3 \) and friends (\( R_7 \)) is calculated below.

\( R_7 = R_3 + R_5 \parallel R_6 = 50 + 200 \parallel 200 = \) 150 Ω.

The load current \( I_{\mathrm{L}} \) can then be calculated using Kirchhoff's Current Law and Ohm's Law.

\( I_{\mathrm{L}} = I_{R4} + \frac{V_{\mathrm{B}}}{R_7} = 52.5 + \frac{6375}{150} = 52.5 + 42.5 = \) 95 A (6 sf).

The internal EMF can now be calculated.

\( E = 6375 + I_{\mathrm{L}} R_{\mathrm{Int}} = 6375 + 95 \cdot 5 = 6375 + 475 = \) 6850 V (4 sf) #.


Question 3

A marked-up version of the circuit is shown below.


The objective is to find the following quantities:

  • The battery EMF: \( E \).
  • The voltage drop across \( R_6 \): \( V_{R6} \).

Given that the voltage drop across \( R_5 \) is 32 V.


Solution

The 10 Ω resistor \( R_5 \) has 32 V voltage drop across it. This instantly implies that the current through \( R_5 \) is 3.2 A. The current will also be the same through \( R_4 \) (same voltage drop and same resistance). The current through \( R_1 \) and \( R_7 \) will be 6.4 A.

This 6.4 A current will be referred to as \( I_{R1} \).

The battery output voltage is the sum of the voltage drops of \( R_1 \), \( R_5 \) (or \( R_4 \)), and \( R_7 \).

\( V_{\mathrm{B}} = V_{R1} + 32 + V_{R7} \).

The voltage drops across \( R_1 \) and \( R_7 \) need to be calculated using Ohm's Law.

\( V_{\mathrm{B}} = I_{R1} \cdot (R_1 + R_7) + 32 = 6.4 \cdot (8 + 24) + 32 = 204.8 + 32 = \) 236.8 V #.

The voltage across \( R_6 \) can now be calculated using voltage division.

\( V_{R6} = V_{\mathrm{B}} \cot \frac{R_6}{R_6 + R_2 \parallel R_3 + R_8 \parallel R_9} = 236.8 \cdot \frac{12}{12 + 5 \parallel 5 + 15 \parallel 15} = 236.8 \cdot \frac{12}{22} = \) 129.164 V (6 sf) or 129.1 V (4 sf) #.

In order to calculate the internal EMF of the battery, the total load current \( I_{\mathrm{L}} \) must be known. We already know that one of the branch currents is 6.4 A, we need to calculate the other. This other branch current is equal to the current through \( R_6 \).

\( I_{\mathrm{L}} = I_{R1} + I_{R6} = 6.4 + \frac{129.164}{12} = 6.4 + 10.7636 = \) 17.1636 A (6 sf).

The internal EMF can now be calculated.

\( E = 236.8 + I_{\mathrm{L}} R_{\mathrm{Int}} = 236.8 + 17.1636 \cdot 1.5 = 236.8 + 25.7454 = \) 262.5 V (4 sf) #.


Kētanga mutunga: Rāhina, 25 Haratua 2020, 11:34 AM