This lecture continues the discussion on circuit analysis, focusing on the concept of an open circuit. It contrasts the theoretical ideal case with the practical reality and explains how very high resistance components can behave like open circuits in parallel configurations.
Ideal Open Circuit
An ideal open circuit is a theoretical construct. It assumes a break in the path with infinite resistance, leading to specific (and impossible) electrical characteristics. Key characteristics:
- Resistance (R): ∞ ohms (infinite)
- Current (I): 0 amperes
- Voltage (V): ∞ volts (infinite)
Violation of KCL and Conservation of Charge
An ideal open circuit violates Kirchhoff's Current Law (KCL) and the law of conservation of charge. Consider a current source (I) connected to a single path that becomes an open circuit (i1 = 0A).
- KCL states: I = i1.
- Since I ≠ 0 and i1 = 0, KCL is violated.
- This implies that charges are being destroyed, which is impossible as charge can neither be created nor destroyed. Therefore, an ideal open circuit cannot exist in a real, functioning circuit.
- Note on Terminology: A "circuit" implies a closed loop. An "open circuit" refers to a path within a closed circuit that has broken or is of extremely high resistance, not an open network.
Practical Open Circuit
In reality, a break in a circuit (like two separated wires or a faulty switch) has a finite voltage across it, even though no current flows. Key characteristics:
- Resistance (R): ∞ ohms (practically infinite)
- Current (I): 0 amperes
- Voltage (V): Finite value. The voltage across the open-circuit terminals is equal to the voltage present at that point with no load.
Example: A 50V source supplying 10A.
- Point A (from the source) is at 50V. Point B is at 0V.
- The air between A and B is the "open circuit" (infinite R).
- Current through the air is 0A.
- The voltage across the open circuit is 50V (V_A - V_B = 50V - 0V = 50V), which is a finite value.
High Resistance as an Open Circuit
When two resistors are connected in parallel, the current primarily flows through the path of least resistance. Understanding this requires knowledge of Engineering Circuit Analysis: Voltage, Current, and Resistance Basics.
Scenario: A 10V source, a 2 MΩ resistor (R1), and a 2 Ω resistor (R2) in parallel.
-
Calculate Equivalent Resistance (Req): Req = (R1 · R2) / (R1 + R2) ≈ (2 MΩ · 2 Ω) / (2 MΩ) ≈ 2 Ω
-
Calculate Total Current (I): I = V / Req = 10V / 2 Ω = 5A
-
Apply Current Divider Rule to find i1 (through 2 MΩ) and i2 (through 2 Ω):
- i1 = I · R2 / (R1 + R2) = 5A · 2 Ω / (2 MΩ + 2 Ω) ≈ 5 · 10⁻⁶ A (microamps) = ~0A
- i2 = I · R1 / (R1 + R2) = 5A · 2 MΩ / (2 MΩ + 2 Ω) ≈ 5A
Conclusion:
- The 2 MΩ resistor conducts a negligible current (~0A).
- The 2 Ω resistor conducts almost all the current (5A).
- Therefore, a very high resistance connected in parallel with a very low resistance will effectively act as an open circuit. For a deeper understanding of these configurations, refer to the Complete Guide to Series, Parallel, and Combined Electrical Circuits.
in the last lecture we understood the concept of short-circuit and we discussed two cases of short circuit the
first case was ideal short-circuit case and the second case was practical short circuit case and in case of ideal short
circuit resistance of the path is 0 ohms voltage across the path is 0 volts and the current through the path is equal to
infinite amperes but in case of practical short circuit resistance and voltage both are 0 but the current is
not infinite but it is finite and now in this lecture we are going to understand the concept of open circuit and like
previous lecture we will understand the ideal case of open circuit and the practical case of open circuit and we
will begin our discussion with the ideal case of open circuit in case of open circuit resistance of the path has
infinite this means you will have a path having the infinite resistance all you can see there will be no part like in
this case if you see these two points then you will find there is no path there is no resistance
having the infinite value but here we are having the path which is air so here is the path and here is having the
resistance which is of very high value so the air will act as the path which is having a resistance close to infinity
and when this happens it is very clear that current will be equal to 0 because you are calculating current let's say
current is high one you are calculating this current through this part which is the part developed due to air
and it is having the resistance it is having the resistance equal to infinity and therefore this current will face the
obstruction or you can say the restriction in moving this path and this current will become zero so current high
one is zero through the path provided by the air and therefore current is zero and if we talk about the voltage then it
is infinite in case of ideal open circuit it is infinite because resistance is infinite and if you
calculate the drop if you consider a very small current very very small current flowing through this path he
will find voltages going to be infinity so ideally the open circuit should have the resistance of the path equal to
infinite homes current flowing through the path will be zero amperes and the voltage across the path is equal to
infinite volts and here in this case we are violating K CL KCl is violated in this case and now we will understand how
KCl is violated in this case and how the violation of KCl implies the violation of law of conservation of charge in this
case we are having a current source and the current source is providing the current equal to I amperes and therefore
current I will enter this point and you can see that current i1 equal to zero ampere is leaving this point and
according to the rule current I should be equal to i1 I is not equal to zero but hi 1 is equal to zero this means I
will not be equal to i1 and this is like destroying the charges I will explain how it is like destroying the charges
you are having the current which is not equal to zero this means charges are moving to produce this
but after this point the current is becoming zero and there is no parallel path so the current high should be same
as high one there is no parallel path and therefore it is like destroying the charges in this particular case and we
know from the law of conservation of charge that charge can neither be created and nor they can be destroyed
but here we are destroying the charges so this particular case is not possible practically because here we are
violating KCl and this means we are violating the law of conservation of charge and this is all for the ideal
case of open circuit and now we will move on to the practical case of open circuit but before that I want to clear
one point by a circuit we mean a network which is closed like in this case you can see that the network is closed but
we are using become open circuit so how can a circuit be open when circuit itself means closed Network so don't
confuse yourself this clearly means that we are having a path in the circuit which is having the infinite resistance
now let's discuss the practical case of open circuit in this case again we are having the resistance equal to infinite
ohms current equal to zero amperes but the voltage is not infinite it is finite let's understand how the voltage is
equal to finite in this case we are having a 50 volt source and the current it is providing is 10 amperes to the
connection of two resistances like this and if we consider this point and this point then you can see that we are
having the air as the path this means the resistance between these two points is equal to infinite ohms and this means
the current through this path is equal to zero ampere now we will calculate the voltage across
the part we are having the voltage source equal to 50 volts and if we consider this point at zero volt
potential then this point will have the potential equal to 50 volts then in this scenario you will get 50 minus 0 equal
to 50 volts has the voltage at this point is having the 50 volts potential this point will also have the same
potential this point will also have the same 50 volts potential at this point and this point will have the same 50
volts potential there is no drop there is no resistance the wire is simply connected this makes this points
potential equal to 50 volts similarly this point will have the potential equal to zero volts now when you calculate the
potential difference across the two points you will find it is equal to 50 volts minus zero volt which will give
you 50 volts as the voltage which is not infinite it is a finite value so in practical case of open circuit voltage
is finite this is all for the practical case of open circuit and now we will take one example in which we will
understand when two resistances are connected in parallel in which one resistance is having the very high value
as compared to the other resistance then the resistance with the high value will act as open circuit viable attacked as
open circuit because the current will choose this part not this part very low current will flow through the part
having the high resistance because the current is having the option of these two paths and it will choose the path
with low resistance let's try to understand this by calculating the current in different paths first we will
calculate the current in this path let's say the current is high then we will calculate current
I won which is the current through this resistance and then current i2 which is the current through this resistance we
can redraw this circuit by considering a single resistance which is the equivalent resistance and we will
understand how to calculate the equivalent resistance in the coming presentations but for now simply
understand that when two resistances are connected in parallel then the equivalent resistance will be equal to
r1 multiplied to r2 divided by r1 plus r2 r1 is equal to two megaohms r2 is equal to 2 ohms so we will have
the equivalent resistance equal to 2 mega ohms or we can write 2 multiplied to 10 raise to power 6 ohms multiplied
to 2 ohms divided by 2 multiplied to 10 raised to power 6 ohms plus 2 ohms when you solve it you will get 4 multiplied
to 10 raised to power 6 in the numerator and in denominator you will get two point zero zero zero zero zero 2
multiplied to 10 raised to power 6 now this you can write 2 multiplied to 10 raised to power 6 so when you solve it
you will get equivalent resistance is equal to 2 ohms we have calculated the equivalent resistance and we know the
voltage is equal to 10 volts therefore current I will be equal to 10 volts divided by 2 ohms and this is equal to 5
amperes so current I is equal to 5 amperes and now we will calculate i1 and i2 using the current divider rule we
will understand what is current divider rule in the coming presentations but now we will directly use the current divider
rule we will calculate i1 forced hi 1 will be equal to the main current which is high
five ampere multiplied to the other resistance we are calculating current i1 through two mega ohm resistance but we
need to multiply main current by the other resistance which is two ohms and then we will divide it by the sum of two
resistances so we have two mega ohms plus 2 ohms in the denominator and we just saw that 2 multiplied to 10 raised
to power 6 plus 2 is nearly equal to 2 multiplied to 10 raised to power 6 so we can write I 1 is equal to 5 multiplied
to 2 which is equal to 10 10 divided by 2 multiplied to 10 raised to power 6 ampere when you solve it he will get I 1
is equal to 5 multiplied to 10 raised to power minus 6 ampere and this current is very low nearly equal to zero ampere so
current i1 is nearly equal to zero ampere and when you calculate i2 using the same current divider rule it will be
equal to 5 ampere multiplied to the other resistance which is 2 mega ohm 2 multiplied to 10 raised to power 6 and
then in denominator we will have the sum of 2 resistances which is nearly equal to 2 multiplied to 10 raised to power 6
this and this will cancel this implies I 2 is equal to 5 ampere main current was equal to 5 ampere and we are getting I 2
equal to 5 ampere so all the current which we are getting from the voltage source is passing through the 2 ohm
resistance and nearly zero ampere current is passing through 2 mega ohm resistance and therefore we can say that
this will act as the open circuit because zero current is flowing through this path so whenever you have a very
high resistance connected in parallel with a low resistance then the path with the high resistance will act as open
circuit so this is all for this lecture if you have any doubt um ask in the comment section I will end this lecture
here soon the next one [Applause] [Music]
Yes, when a very high resistance is placed in parallel with a very low resistance, it effectively acts as an open circuit. For instance, a 2 MΩ resistor in parallel with a 2 Ω resistor from a 10V source conducts negligible current (microamps), while the 2 Ω resistor carries almost all the current.
Use the current divider rule: current through the high resistor equals total current times the low resistance divided by the sum of both resistances. For a 5A total current, a 2 MΩ resistor gets about 5A × 2Ω / (2MΩ + 2Ω) ≈ 5 microamps, effectively zero.
Ideal open circuits help with theoretical understanding but can lead to contradictions like violating KCL. Practical open circuits reflect real-world behavior where voltages are finite and currents are zero, enabling accurate analysis of faulty switches, broken wires, or high-impedance components.
An ideal open circuit is a theoretical concept assuming a break with infinite resistance, resulting in zero current but infinite voltage. It violates Kirchhoff's Current Law and the conservation of charge because it would imply charges are destroyed. Thus, it cannot exist in a real, functioning circuit.
In a practical open circuit, the resistance is effectively infinite, so current is zero, but the voltage across the break is finite. For example, a 50V source with an air gap between its terminals has 50V across the gap but no current flows through the air.
Because the 2 Ω resistor offers a much lower resistance path, most of the current (5A) flows through it. The current divider rule shows that only a tiny fraction, determined by the ratio of the low resistance to the sum, goes through the high resistor.
It represents the voltage that would be present across the open terminals if no load were connected. For example, if a wire is disconnected at a point in a 50V circuit, the voltage between the two ends is 50V, even though no current flows across the break.
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