Equivalent Circuits
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Equivalent Circuits Samantha R. Summerson 2 September, 2009 1 Equivalent Resistance Recall from last time: π1 π 1 + π£1 − π + π2 + + π£2 π£ππ (π‘) ± − π 2 π£ππ’π‘ (π‘) − − Figure 1: Example circuit. Last time we determined that π£ππ’π‘ = π 2 π£ππ . π 1 + π 2 This relation is called voltage divider. In general, for a series of resistors, π π π£π = ∑ π£ππ . π π The voltage across resistor π π is the input voltage scaled by a its resistance, π π , and divided by the sum of all the resistances in the series (including π π ). What about resistors in parallel? Can we ο¬nd an equivalent resistance? From KVL we know that π2 π1 π πππ (π‘) + + π£1 ↑ − − 1 + π 1 π£1 − π 2 π£1 = π£2 , π 1 π1 = π 2 π2 . From KCL, we know πππ = π1 + π2 , ⇒ π1 = πππ − π2 . Using this value of π1 in our KVL equation, we get π 1 (πππ − π2 ) ⇒ π 1 πππ π 1 ⇒ πππ π 1 + π 2 = π 2 π 2 , = (π 1 + π 2 )π2 , = π2 . (1) This last equation (1) is an expression for the current divider. When there are parallel resistors in a circuit, current is divided between the resistors (as opposed to series resistors where voltage is divided). In general, ππ = π π πππ . π π + π π Note that the numerator of the divider is the resistance of the other resistor. Plugging in the value of π2 into our π£ − π relation for π 2 , we get π£2 = π 2 π2 π 1 π 2 πππ π 1 + π 2 = Thus, the equivalent resistance is π ππ = π 1 π 2 . π 1 + π 2 In shorthand, we write π ππ = (π 1 β£β£π 2 ). Note that our expression for π ππ can also be written in terms of the reciprocals of the resistances: ( )−1 1 1 π ππ = + . π 1 π 2 The reciprocal of resistance, π 1 , is called conductance. This quantity is measured in siemens. We therefore conclude that for parallel resistors the conductances add, whereas for series resistors the resistances add. In general when we want to measure current or voltage, we will need to connect some additional element to the circuit. Often this additional element behaves as a resistor. In this case, we want the value of πππ ππ π‘ππππ to aο¬ect the circuit behavior as little as possible. Consider the following circuit: The resistor, π πΏ , represents the load resistance of the external element added to the circuit. Using our current divider rule, we know π ππ = = π πΏ π 2 π πΏ + π 2 (π πΏ β£β£π 2 ) Now that we have two resistors in series, we can use the voltage divider rule: 2 π π 1 π2 + π£1 − π1 + + + π£ππ’π‘ π£ππ (π‘) ± − π 2 − π πΏ − π 1 π1 + π£1 − π π£πΏ + + π2 + π£ππ (π‘) ± π ππ − π£ππ’π‘ (π‘) − − π£ππ’π‘ = π ππ π£ππ π 1 + π ππ = π πΏ π 2 π πΏ +π 2 π 2 π 1 + π π πΏπΏ+π 2 = π πΏ π 2 π£ππ π 1 (π πΏ + π 2 ) + π πΏ π 2 π£ππ Since we are using π πΏ to measure, say, π£ππ’π‘ , we need to pick π πΏ such that ( ) ( ) π 1 π πΏ π 2 π£ππ ≈ π£ππ , π 1 + π 2 π 1 (π πΏ + π 2 ) + π πΏ π 2 ) ( π 2 π£ππ . ≈ π 1 π 2 π πΏ + π 1 + π 1 As π πΏ → ∞, we see that the right-hand side approaches the left-hand side. In order the two sides to be approximately the same, we want π 1 π 2 << π 1 + π 2 . π πΏ This is called the no load condition. Example. Find the equivalent resistance of the circuit shown in Fig.2. π ππ = ((π 2 β£β£π 3 ) + π 4 β£β£π 1 ) Example. Find the equivalent circuit for the circuit in Fig.3. The KCL equation is π1 + π = π2 , and the KVL equation is π£1 + π£ − π£ππ = 0. 3 π 2 π 3 π 1 π 4 Figure 2: Example - ο¬nd the equivalent resistance. π1 π π 1 π + π£1 − π2 + + + π£ππ (π‘) ± π 2 − ←π£ − − Figure 3: Example - what circuit would a terminal see? Using the π£ − π relation for π 2 and the KCL equation, π£ ⇒ π1 ⇒π£ ( ) π 2 ⇒ 1+ π£ π 1 ⇒π£ = π 2 (π1 + π) π£1 = π 1 π£ππ − π£ = π 1 ( ) π£ππ − π£ = π 2 +π π 1 ) ( π£ππ = π 2 +π π 1 π 2 π 1 π 2 = π£ππ + π π 1 + π 2 π 1 + π 2 This looks like a current divider!!! 4
