In the next tutorial we will look at Maximum Power Transfer. Thévenin's theorem and its dual, Norton's theorem, are widely used to make circuit analysis simpler and to study a circuit's initial-condition and steady-state response.

By using current division principle, the current flowing through the 20 Ω resistor will be, $$I_{20 \Omega} = 5 \lgroup \frac{\frac{40}{3}}{\frac{40}{3} + 20} \rgroup$$, $$I_{20 \Omega} = 5 \lgroup \frac{40}{100} \rgroup = 2A$$. As far as the load resistance, RL is concerned this single resistance, RS is the value of the resistance looking back into the network with all the current sources open circuited and IS is the short circuit current at the output terminals as shown below. For the given circuit, calculate the current flows through the 5Ω resistor using Norton’s theorem. Any collection of batteries and resistances with two terminals is electrically equivalent to an ideal current source i in parallel with a single resistor r.The value of r is the same as that in the Thevenin equivalent and the current i can be found by dividing the open circuit voltage by r. We use cookies to ensure that we give you the best experience on our website. Any active network can be represented as an equivalent current source with an impedance connected in parallel across it. In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. To determine internal resistance or Norton equivalent resistance RN of the network under consideration, remove the branch between A and B and also replace the voltage source by its internal resistance. Norton’s theorem says that a linear two-terminal electric circuit may be exchanged with a Norton equivalent circuit consisting of a current source, I N, in parallel with a Norton resistor R N.Where I N is the short-circuit current through the terminals load resistor and R N is equivalent resistance at the terminals when all the independent sources are turned off. In this post, you will learn the statement of Norton’s theorem for dc circuits with solved examples including limitations, and applications.

Norton’s theorem states that any linear complex electrical circuit can be reduced into a simple electric circuit with one current and resistance connected in parallel. Follow these steps in order to find the Norton’s equivalent circuit, when only the sources of independent type are present. If one resistance in the circuit is changed rapidly (load), then the Norton’s theorem can be used to perform calculations easily. We know that current select a path with low resistance and a short circuit path is considered as zero resistance. The value of the internal resistor Rs is found by calculating the total resistance at the terminals A and B giving us the following circuit. Now find the short circuit current which is Norton’s current of the network.

Norton’s Theorem Explanation. Norton’s Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load. Step 4 − Find Norton’s resistance RN by using the following formula.

Find the current through the short-circuited path by any circuit analysis method.

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Current I1 is the total current supplied by the source. We know that Norton’s resistance, RN is same as that of Thevenin’s resistance RTh.

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