20.5.4 P.d. Across Any Two Points in DC Circuit
Figure 20.24 shows part of a large circuit. Points A, B and C are nodes. We wish to determine the p.d. across points A and D. In order to achieve that, let us start from point A (or point D), move through R₁, R₂, the battery and finally stop at D (or A).
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| Figure 20.24 Path from A to D through R₁, R₂, and a battery to determine the potential difference. |
After arriving at point D, we would have passed through three voltages:
(a) potential fall $−I_1R_1$
(b) potential fall $−I_2R_2$
(c) potential rise $+E_3$
Then the p.d. $\Delta V_{AD}$ between A and D is given by
$\Delta V_{AD}=V_D-V_A$
$= (−I_1R_1) + (−I_2R_2) + (+E_3)$
$= −I_1R_1 − I_2R_2 + E_3$
Note that if we had followed the path DCBA, the p.d. ΔVDA would be
$\Delta V_{DA}=V_A-V_D$
$= −I_1R_1 + I_2R_2 - E_3$
$=-\Delta V_{AD}$
20.5.5 Earthing or Grounding a Point in DC Circuit
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| Figure 20.25 Point X is connected to earth, so its potential is taken as zero (Vₓ = 0). |
Suppose a point X in a circuit is connected directly to earth by a conductor, as shown in Figure 20.25. Then the potential at X is taken to be zero. In other words, potential $V_X = 0$. Refer to the circuit shown in Figure 20.24. We have
$V_D − V_A = −I_1R_1 − I_2R_2 + E_3$
If point A is earthed, then $V_A = 0$. The potential at point D relative to A will become
$V_D = −I_1R_1 − I_2R_2 + E_3$
EXAMPLE 20.6
Refer to the circuit shown in Figure 20.26. Neglect the internal resistance of the batteries.
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| Figure 20.26 A multi-loop circuit with resistors and ideal batteries (internal resistance neglected), used to analyze currents and potential differences using Kirchhoff’s laws. |
(a) Determine the electric potential at point G relative to (i) point B (ii) point C.
(b) Determine the electric potential at G if (i) B is earthed (ii) C is earthed.
Answer
Step 1: Draw the current direction at each resistor. Label each current.
Step 2: Draw a ‘voltage’ arrow beside each component in the circuit.
The circuit should look like the one shown in Figure 20.27.
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| Figure 20.27 A circuit diagram showing voltage arrows across each component, indicating the direction of potential differences. |
Step 3: Apply Kirchhoff’s current law to the currents at node B. We get
$(+I_1) + (+I_2) + (−I_3) = 0$
$I_1 + I_2 = I_3$ .......... (1)
Step 4: Move round loop ABEFA. Apply Kirchhoff’s voltage law to all the voltages encountered in the closed loop. We get
$(−I_1(2)) + (−6) + (+I_2(3)) + (−I_1(1)) + (+3) = 0$
$−3I_1 + 3I_2 = 3$
$−I_1 + I_2 = 1$ .......... (2)
Step 5: Move round loop BCDEB. Apply Kirchhoff’s voltage law to all the voltages encountered in the closed loop. We get
$(−I_3(2)) + (+9) + (−I_3(1)) + (−I_2(3)) + (+6) = 0$
$−3I_2 − 3I_3 = −15$
$I_2 + I_3 = 5$ .......... (3)
(1) + (2),
$2I_2 = I_3 + 1$
$2I_2 − I_3 = 1$ .......... (4)
(3) + (4),
$3I_2 = 6$
$I_2 = +2 \ A$
From (3),
$2 + I_3 = 5$
$I_3 = +3 \ A$
From (1),
$I_1 + (+2) = +3$
$I_1 = +1 \ A$
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| Figure 20.28 A section of a circuit between points B and G showing two voltage changes encountered when moving from B to G. |
(a) (i) Refer to the circuit between B and G, as shown in Figure 20.28. Move from B to G. We will encounter two voltages. They produce the p.d. $\Delta V_{BG}$ given by
$\Delta V_{BG}= V_G-V_B$
$\Delta V_{BG} = (+I_1(2)) + (−3)$
$= (+1)(2) − 3 = −1 \ V$
$V_G = V_B − 1$
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| Figure 20.29 A section of a circuit between points C and G showing two voltage changes encountered when moving from C to G. |
(ii) Refer to the circuit between C and G, as shown in Figure 20.29. Move from C to G. We will encounter three voltages. They produce the p.d.$\Delta V_{CG}$ given by
$\Delta V_{CG}= V_G-V_C$
$\Delta V_{CG} = (+I_3(2)) + (+I_1(2)) + (−3)$
$= (+3)(2) + (+1)(2) − 3 = +5 \ V$
$V_G = V_C + 5$
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| Figure 20.30 A graph showing how the electric potential changes as we move from point C to point B and then to point G. |
Figure 20.30 shows how the electric potential changes as we move from point C to point B and then to point G.
(b) (i) If B is earthed, then $V_B= 0$. We get
$V_G = V_B − 1 = −1 \ V$
(ii) If C is earthed, then $V_C= 0$. We get
$V_G = V_C + 5 = +5 \ V$
EXAMPLE 20.7
Refer to the circuit shown in Figure 20.31. Neglect the internal resistance of the batteries.
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| Figure 30.31 A circuit diagram with multiple loops and ideal batteries, where internal resistance is neglected. |
Determine
(a) the current that flows through each component in the circuit
(b) the electric potential at point A relative to point B, which is earthed.
Answer
(a) Refer to the circuit shown in Figure 20.32.
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| Figure 20.32 A multi-loop circuit showing current directions $I_1$, $I_2$ and $I_3$ in different branches, used for analysis using Kirchhoff’s laws. |
Apply Kirchhoff’s current law to the currents at node C. We get
$−I_1 − I_2 + I_3 = 0$ .......... (1)
Apply Kirchhoff’s voltage law to the closed loop CADC. We get
$(−I_1(3)) + (−2) + (−I_1(1)) + (+3) + (−I_3(2)) = 0$
$−4I_1 − 2I_3 = −1$ .......... (2)
Apply Kirchhoff’s voltage law to the closed loop CDBC. We get
$(+I_3(2)) + (+1) + (+I_2(2)) + (−2) + (+I_2(1)) = 0$
$3I_2 + 2I_3 = 1$ .......... (3)
(2) + (3),
$−4I_1 + 3I_2 = 0$
$I_2 = \frac{4}{3} I_1$
(1) × 2,
$−2I_1 − 2I_2 + 2I_3 = 0$ .......... (4)
(2) + (4),
$−6I_1 − 2I_2 = −1$
$6I_1 = −2I_2 + 1$
$= −2\left( \frac{4}{3}I_1 \right) + 1$
$6I_1 + \frac{8}{3}I_1 = 1$
$I_1 = \frac{3}{26} \ A$
$= +0.115 \ A ≈ +0.12 \ A$
$I_2 = \frac{4}{3} \times \frac{3}{26} = \frac{4}{26} \ A$
$= +0.154 \ A ≈ +0.15 \ A$
$I_3=I_1+I_2$
$= \frac{7}{26}\ A = +0.27 \ A$
Note: The calculations would have been neater and probably quicker if we had used matrices and determinants to solve this problem.
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| Figure 20.33 A section of the circuit used to determine the electric potential at point A relative to point B by moving from B to C and then to A, encountering four voltage changes. |
(b) To determine the electric potential at A relative to B, consider ther part of the circuit as shown in Figure 20.33. Starting from B, move to C and finally to A (Note: we can also take the path BDA). We will encounter four voltages.
$\Delta V_{BA}=V_A-V_B$
$=(-2)+(+I_2(+1))+(-I_3(3))+(-2)$
$=-2+(+0.154)-3(+0.115)-2=-4.2$
$V_A=-4.2+V_B$
Since point B is earted, we have $V_B=0$. Hence
$V_A=-4.2 \ V$
Application of Potential Difference (P.d.) Across Any Two Points in a DC Circuit and Earthing (Grounding) a Point
1. Potential Difference Across Any Two Points
In a DC circuit, the potential difference (p.d.) between any two points represents the work done per unit charge in moving a charge between those points. It can be determined by analyzing the circuit using Kirchhoff’s Voltage Law (KVL), which states that the algebraic sum of all voltages around a closed loop is zero.
By moving from one point to another in a circuit, we encounter voltage rises (from batteries) and voltage drops (across resistors). The net change gives the potential difference between the two points.
2. Application in Circuit Analysis
Understanding the potential difference between two points helps in:
- Determining current flow in different branches
- Analyzing multi-loop circuits
- Verifying Kirchhoff’s laws
3. Earthing (Grounding) a Point in a DC Circuit
Earthing or grounding refers to assigning a reference point in the circuit a potential of zero volts. This simplifies circuit analysis by providing a common reference for measuring all other voltages.
Once a point is grounded:
- All voltages are measured relative to this point
- Circuit calculations become simpler and more systematic
- It improves safety in practical electrical systems
Conclusion
The potential difference between any two points in a DC circuit represents the energy change per unit charge and can be determined using Kirchhoff’s laws. Assigning a ground (zero potential) simplifies circuit analysis by providing a common reference point. Together, these concepts make it easier to understand voltage distribution and analyze complex DC circuits accurately.
Frequently Asked Questions (FAQ)
1. What is potential difference (P.d.) in a DC circuit?
Potential difference is the work done per unit charge to move a charge between two points in a circuit.
2. How do you find the potential difference between two points?
It can be determined by applying Kirchhoff’s Voltage Law and summing all voltage rises and drops along a path between the two points.
3. Why is grounding important in a DC circuit?
Grounding provides a reference point of zero potential, making it easier to measure and analyze voltages in the circuit.
4. Does grounding affect current flow?
In most circuit analysis, grounding does not change current flow; it only sets a reference level for voltage.
5. What happens if a circuit has no ground?
The circuit can still function, but voltage measurements become more complex without a common reference point.
6. Can any point in a circuit be grounded?
Yes, any point can be chosen as ground, depending on convenience for analysis.
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