A parallel circuit consists of two branches: one with impedance (4 + j3) Ω and another with (4 − j3) Ω connected to 100 V supply. Then
1. The total impedance is purely resistive.
2. The equivalent impedance magnitude is 4 Ω.
3. Total current drawn is 25 A.
4. Power factor of circuit is unity.
Which of the above statements is/are correct?
Answer & step-by-step solution
The equivalent resistance is not simply R; it is obtained after adding the two admittances. Complex-conjugate multiplication gives |4+j3|²=25. Adding admittances cancels ±j3/25 exactly. Key relation: Yeq = 1/(4+j3)+1/(4−j3)=8/25=0.32 S; Zeq=1/Yeq=3.125 Ω; I=V/Zeq=32 A; PF=cos0°=1.
Exam focus: For parallel complex branches, convert Z to Y first; it prevents sign and denominator mistakes. Common trap: Directly adding parallel impedances or assuming the equivalent resistance equals the real part of each branch.
Statement-wise Verification
- Statement 1 - Correct.
The total impedance is purely resistive.
The imaginary parts cancel in admittance because the branch impedances are complex conjugates. - Statement 2 - Incorrect.
The equivalent impedance magnitude is 4 Ω.
The equivalent magnitude is 3.125 Ω, not 4 Ω. - Statement 3 - Incorrect.
Total current drawn is 25 A.
I = 100/3.125 = 32 A, not 25 A. - Statement 4 - Correct.
Power factor of circuit is unity.
Equivalent impedance is purely real, so the phase angle is zero and power factor is 1.
Core Concept
Parallel AC impedances should be combined through admittance. For conjugate branches R+jX and R−jX, their susceptances cancel, leaving a real conductance. The equivalent resistance is not simply R; it is obtained after adding the two admittances.
Formula / Key Relationship
Yeq = 1/(4+j3)+1/(4−j3)=8/25=0.32 S; Zeq=1/Yeq=3.125 Ω; I=V/Zeq=32 A; PF=cos0°=1.
Step-by-Step Check
Complex-conjugate multiplication gives |4+j3|²=25. Adding admittances cancels ±j3/25 exactly. Numerical current and PF then follow directly, confirming B.
Why the Other Options Are Wrong
Option A is incorrect because it includes incorrect statement(s) 2. Option C is incorrect because it includes incorrect statement(s) 2, 3 and omits correct statement(s) 1, 4. Option D is incorrect because it includes incorrect statement(s) 2, 3.
Answer-Key Verification Note
The uploaded provisional key marks A . After independent technical verification, the defensible answer is B . This discrepancy is stated explicitly rather than silently copying the provisional key.
Exam Shortcut / Approach
For parallel complex branches, convert Z to Y first; it prevents sign and denominator mistakes.
Common Mistake
Directly adding parallel impedances or assuming the equivalent resistance equals the real part of each branch.
Quick Revision
Conjugate Impedances in Parallel is linked with Parallel Complex Impedances, AC Circuits and Basic Electrical Engineering. For parallel complex branches, convert Z to Y first; it prevents sign and denominator mistakes.
एक समानांतर परिपथ में दो शाखाएँ हैं: एक की प्रतिबाधा (4 + j3) Ω और दूसरी की (4 − j3) Ω, जो 100 V आपूर्ति से जुड़ी हैं। तब
1. कुल प्रतिबाधा पूर्णतः प्रतिरोधी है।
2. समतुल्य प्रतिबाधा का परिमाण 4 Ω है।
3. खींची गई कुल धारा 25 A है।
4. परिपथ का पावर फैक्टर इकाई है।
उपरोक्त में से कौन-सा/से कथन सही है/हैं?
उत्तर और चरणबद्ध हल
B — केवल 1 और 4। Yeq = 1/(4+j3) + 1/(4−j3) = 8/25 = 0.32 S (काल्पनिक भाग रद्द)। Zeq = 3.125 Ω (पूर्णतः प्रतिरोधी, PF = 1), I = 100/3.125 = 32 A। इसलिए 4 Ω और 25 A गलत हैं।
विस्तृत हल (मूल भाषा में)
Statement-wise Verification
- Statement 1 - Correct.
The total impedance is purely resistive.
The imaginary parts cancel in admittance because the branch impedances are complex conjugates. - Statement 2 - Incorrect.
The equivalent impedance magnitude is 4 Ω.
The equivalent magnitude is 3.125 Ω, not 4 Ω. - Statement 3 - Incorrect.
Total current drawn is 25 A.
I = 100/3.125 = 32 A, not 25 A. - Statement 4 - Correct.
Power factor of circuit is unity.
Equivalent impedance is purely real, so the phase angle is zero and power factor is 1.
Core Concept
Parallel AC impedances should be combined through admittance. For conjugate branches R+jX and R−jX, their susceptances cancel, leaving a real conductance. The equivalent resistance is not simply R; it is obtained after adding the two admittances.
Formula / Key Relationship
Yeq = 1/(4+j3)+1/(4−j3)=8/25=0.32 S; Zeq=1/Yeq=3.125 Ω; I=V/Zeq=32 A; PF=cos0°=1.
Step-by-Step Check
Complex-conjugate multiplication gives |4+j3|²=25. Adding admittances cancels ±j3/25 exactly. Numerical current and PF then follow directly, confirming B.
Why the Other Options Are Wrong
Option A is incorrect because it includes incorrect statement(s) 2. Option C is incorrect because it includes incorrect statement(s) 2, 3 and omits correct statement(s) 1, 4. Option D is incorrect because it includes incorrect statement(s) 2, 3.
Answer-Key Verification Note
The uploaded provisional key marks A . After independent technical verification, the defensible answer is B . This discrepancy is stated explicitly rather than silently copying the provisional key.
Exam Shortcut / Approach
For parallel complex branches, convert Z to Y first; it prevents sign and denominator mistakes.
Common Mistake
Directly adding parallel impedances or assuming the equivalent resistance equals the real part of each branch.
Quick Revision
Conjugate Impedances in Parallel is linked with Parallel Complex Impedances, AC Circuits and Basic Electrical Engineering. For parallel complex branches, convert Z to Y first; it prevents sign and denominator mistakes.
એક સમાંતર પરિપથમાં બે શાખાઓ છે: એકનો અવબાધ (4 + j3) Ω અને બીજાનો (4 − j3) Ω, જે 100 V પુરવઠા સાથે જોડાયેલ છે. તો
1. કુલ અવબાધ સંપૂર્ણ અવરોધક છે.
2. સમતુલ્ય અવબાધનું માન 4 Ω છે.
3. ખેંચાતો કુલ પ્રવાહ 25 A છે.
4. પરિપથનો પાવર ફેક્ટર એકમ છે.
ઉપરોક્તમાંથી કયું/કયાં વિધાન સાચું/સાચાં છે?
જવાબ અને પગલાવાર ઉકેલ
B — ફક્ત 1 અને 4. Yeq = 1/(4+j3) + 1/(4−j3) = 8/25 = 0.32 S (કાલ્પનિક ભાગ રદ). Zeq = 3.125 Ω (સંપૂર્ણ અવરોધક, PF = 1), I = 100/3.125 = 32 A. તેથી 4 Ω અને 25 A ખોટાં છે.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Statement-wise Verification
- Statement 1 - Correct.
The total impedance is purely resistive.
The imaginary parts cancel in admittance because the branch impedances are complex conjugates. - Statement 2 - Incorrect.
The equivalent impedance magnitude is 4 Ω.
The equivalent magnitude is 3.125 Ω, not 4 Ω. - Statement 3 - Incorrect.
Total current drawn is 25 A.
I = 100/3.125 = 32 A, not 25 A. - Statement 4 - Correct.
Power factor of circuit is unity.
Equivalent impedance is purely real, so the phase angle is zero and power factor is 1.
Core Concept
Parallel AC impedances should be combined through admittance. For conjugate branches R+jX and R−jX, their susceptances cancel, leaving a real conductance. The equivalent resistance is not simply R; it is obtained after adding the two admittances.
Formula / Key Relationship
Yeq = 1/(4+j3)+1/(4−j3)=8/25=0.32 S; Zeq=1/Yeq=3.125 Ω; I=V/Zeq=32 A; PF=cos0°=1.
Step-by-Step Check
Complex-conjugate multiplication gives |4+j3|²=25. Adding admittances cancels ±j3/25 exactly. Numerical current and PF then follow directly, confirming B.
Why the Other Options Are Wrong
Option A is incorrect because it includes incorrect statement(s) 2. Option C is incorrect because it includes incorrect statement(s) 2, 3 and omits correct statement(s) 1, 4. Option D is incorrect because it includes incorrect statement(s) 2, 3.
Answer-Key Verification Note
The uploaded provisional key marks A . After independent technical verification, the defensible answer is B . This discrepancy is stated explicitly rather than silently copying the provisional key.
Exam Shortcut / Approach
For parallel complex branches, convert Z to Y first; it prevents sign and denominator mistakes.
Common Mistake
Directly adding parallel impedances or assuming the equivalent resistance equals the real part of each branch.
Quick Revision
Conjugate Impedances in Parallel is linked with Parallel Complex Impedances, AC Circuits and Basic Electrical Engineering. For parallel complex branches, convert Z to Y first; it prevents sign and denominator mistakes.
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