Question

You are the expert designer searching for a simple circuit to provide 1/5 of a signal current I to a load resistance R.

Part a) (2 points): Suggest a solution using one resistor (draw the circuit diagram).

Part b) (3 points): What must be the value of the resistance?

Part c) (3 points): What is the input resistance of the resulting current divider?

Part d): For a particular value of R, you discover that the otherwise-best-available resistor is actually 10% too high, i.e., R’₁ = 1.1R₁ = 1.1R/4 . Suggest two circuit topologies using one additional resistor in each topology that will solve this problem.

Topology (i) (1 point)

Topology (ii) (1 point)

Part e): What is the value of the resistor required in each case in Part d)?

Topology (i) (3 point)

Topology (ii) (2 points)

Part f): What is the input resistance of the current divider in each case in Part d)?

Topology (i) (3 points)

Topology (ii) (2 points)

Answer 1

Topselogy : 1 Г 1 6 И , үт, 2 14 | «ko Det fj= # nadan R' — туды. К = : 2- те s be = 2 x Sc Ѕт 0,4y = Be - Е,

Topology : 2, I ( 9 AR) = (HR) Ax 2. IR = RAR? LIRE RAR? R?= 0;1R © : : Resistance (Topo 32J = SAR Resistence (Trebo - 2) = R © Tapologe. d Rina e te ART B 44 Roti Rin = R

Tapology : 2: Rin= 1.LR. || (Rt 01R) — 14 12 х 41 = 1 - 1 - 1 gos R = .. 10

| | © Resistente value of the 03