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Voltage/current dividers Trivial to understand, but still very useful. Use in combination with equivalent resistances to quickly find a particular voltage or current in a circuit. Voltage divider VS + – R1 iS R2 R3 + vR2 – Want to know vR2. Easily solved with KCL, KVL, & equivalent resistances. = = = = + + + + That’s it. The total voltage across a series string is divided among the resistors according to a simple ratio. EE 201 voltage/current dividers – 1 vR1 + – R1 VS + – R2 R3 – + vR2 Easy to find the other voltages, too. = – vR3 + = + + + + Put in some values. + + 15 V – 5 kΩ – 15 kΩ 10 kΩ – EE 201 vR1 vR3 = + vR2 = + + + + ( )= . ( )= . ( )= – + = + + voltage/current dividers – 2 Current divider Same idea, but with current. IS R1 R2 iR2 R3 + vR – = = = = + + Want to know iR2. Easily solved with KCL, KVL, & equivalent resistances. + + The total is divided according a simple ratio determined by the resistors. EE 201 voltage/current dividers – 3 The other currents are just as easy. IS R1 iR1 R2 iR2 R3 = iR3 = + + + + Plug in some numbers. = 9 mA 3 k! 5 k! iR1 iR2 15 k! iR3 = = EE 201 + + + + + + ( )= ( )= ( )= voltage/current dividers – 4 Example 1 24 ! R1 R2 VS + – 12 V R3 48 ! R4 36 ! + In the circuit, find vR4. vR4 36 ! – R5 12 ! vR4 = vR3 Combine parallel combinations. R12 16 ! VS + – 12 V R5 EE 201 12 ! Use voltage divider. = 18 ! + R34 vR4 – = + + + + ( ) = . voltage/current dividers – 5 Example 2 IS Find the equivalent resistance in each branch. IS 0.5 A 0.5 A R2 120 ! R3 40 ! R5 iR3 60 ! R1 10 ! R13 40 ! R4 20 ! R45 R6 80 ! iR13 80 ! R6 80 ! In the circuit, find iR3. Use a current divider to find iR13. = = EE 201 + + = + + Referring back to the original circuit, iR13 divides between R2 and R3. ( . )= . = + + ( . )= . voltage/current dividers – 6 Example 3 R1 1 k! VS R2 + – 3 k! Without RL, = RL In the simple divider circuit at right, if RL is attached in parallel with R2, the voltage across R1 doubles. What is the value of RL? From the two expressions for v’R1 + + With RL, = + EE 201 + + = = = Then = = = = + = . voltage/current dividers – 7 Example 4 R1 IS 1 mA iR1 R2 iR2 R3 iR3 From the current divider equation, we know that the currents are proportional to the inverse of the resistance in each branch. : : = : For the simple current divider at right, design it (i.e. choose resistor values) so that the currents are in the ratio iR1:iR2 :iR3 = 1:2:4 and the total power dissipated in the resistors is 10 mW. The equivalent resistance of the three in parallel is = + + : + = + = = = . Therefore, : : = : : R2 = 2R3 and R1 = 2R2 (R1 = 4R3). = . Find Req, = Finally: R3 = 1.75Req = 17.5 kΩ; R2 = 2R3 = 35 kΩ; R1 = 2R2 = 70 kΩ EE 201 voltage/current dividers – 8