Related Question
Norton equivalent resistance for the given network is
R = (R1 || R2) + R3
R = (4 || 8) + 2 = (4 x 8) ⁄ (4 + 8) + 2 = 5.6Ω
Norton equivalent resistance = 5.6Ω
Related Question
In the given circuit the resistance R9 and R8 are Parallel with Resistance R10 therefore
(R9 + R8) || R10
= {(4 + 2) × 12} ⁄ {(4 + 2) + 12}
= (6 × 12) ⁄ (6 + 12)
RA = 4Ω
Now resistance RA and R7 is parallel with Resistance R6
∴ {(4 + 8) × 6} ⁄ {(4 + 8) + 6}
= (12 × 6) ⁄ (12 + 6)
RB = 4Ω
Now the circuit becomes as shown in the figure
Resistance RB and R7 is parallel with Resistance R4
{(4 + 8) × 6} ⁄ {(4 + 8) + 6}
= (12 × 6) ⁄ (12 + 6)
RC = 4Ω
Now resistance RC and R3 is parallel with Resistance R2
{(4 + 2) × 6} ⁄ {(4 + 2) + 6}
RD = 3
Now our final circuit becomes as shown in the figure
Therefore the equivalent resistance is
R1 + RD
Req = 4 + 3 = 7Ω
Kirchhoff’s Current Law (KCI ) or Kirchhoff’s Junction Rule. This law is based on the conservation of charge and may be stated as under:
The algebraic sum of the currents meeting at a junction in an electrical circuit is zero.
An algebraic sum is one in which the sign of the quantity is taken into account. For example, consider four conductors carrying currents I1, I2, I3, & I4 and meeting at point O as shown in Fig
If we take the signs of currents flowing towards point O as positive, then currents flowing away from point O will be assigned negative sign. Thus, applying Kirchhoff’s current law to the junction O we have,
(I1) + ( I2) + (−I3) + (−I4) = 0 or(I1) + ( I2) = (−I3) + (−I4)
i.e., Sum of incoming currents = Sum of outgoing currents.
Therefore, Kirchhoff’s current law may also be stated as under:
The sum of currents flowing towards any junction in an electrical circuit is equal to the sum of currents flowing away from that junction. Kirchhoff’s current law is rightly called the junction rule.
Kirchhoff’s current law is true because electric current is merely the flow of free electrons and they cannot accumulate at any point in the circuit. This is in accordance with the law of conservation of charge. Hence, Kirchhoff’s current law is based on the law of conservation of charge.
Current through the 10Ω resistance
I1 = V/R = 20/10 = 2A
I1 = 2A
Now current through the 20Ω resistance
I2 = V − (-10)/R = 20 + 10/30 = 1.5 A
I2 = 1.5 A
Determine the resistance RN of the network as seen from the network terminals. (Its value is the same as that of Rth).
RN = (4Ω || 8Ω) + 5Ω = (4 × 8)/(4 + 8) + 5
RN = 7.66
The value I for the current used in Norton’s Theorem is found by determining the open circuit voltage at the terminals AB and dividing it by the Norton resistance r.
According to voltage Division Rule
VAB = V1R3 ⁄ (R1 + R3)
= 24 × 8 ⁄ (4 + 8)
VAB = 16 V
Now Norton Current IN is
IN = VAB ⁄ RN
IN = 16 ⁄ 7.66
IN = 2.08
The Farad is the practical and the Sl unit of capacitance. The unit, named after Michael Faraday (1791-1867), was first suggested by Latimer Clark in 1867. The capacitor has a capacitance of 1 farad when a charge of 1 coulomb raises the potential between its plates to 1 volt.
The S.I unit of Inductance is Henry.
The S.I unit of resistance is OHM.
The S.I unit of Reluctance is amp-turns/Weber or Henry−1