Question

Please give a detailed answer and with sketch of the circuit to support the step. Thanks.

Use Norton's theorem to find V. in the circuit below. 12 ΚΩ 2 ΚΩ 10 kΩ 36v () 24 ΚΩξ () 3 mA 1kΩξ ν.

Answer 1

eleng Norton equivalent Resistance for this all Independent Sources deactivated 12K 2 k lokr } 24kn Rith 12k, 24ks are in arall - Ri= 12kx 24k. (12k +241) R - 8kr circit becomes 2K 10k lok El Rin Series. all Resistance are in 80 Ron = R'+2k+1015 =(81 +2x +lok) m=20k

alculating Norton equivalent Current - 12k ess 10) 3 2415 36v Isc 3MA apply Source trans formation 2k Tols I-36 3mA Isc PIRK 320 24k 12k I. = 3mA Indarallel Zk Jok So 3MA Isc I= 3mA 8k opply Source transformation again – Sk 2k 21 lok In Series 3mA V=3 mA x OK V=24v

Node- lok lok 3mA v Isc & VG=or apply nodal analysis (KCL) at node – – VI-VG-24 V - VG + 3 mA = 0 Tok Tok 1-24 + v +3=0 1o 10 2V – 2,4+3=0 10. 8 -06 = 0 W = 3 Vol vi-va - Ise = 3 ore so ISCE lok or Isc= 0.3MA

so Norton equivalent circuit becomes TO Тce (1) -0.3mA 16 apply Current division Io = 20k - x0.3 mA (20k tlk) - 200 x 0.3mm Io - I mA so Vo= vo Toxik & Volt or vo= 2 2 Volt) Anser
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