Two resistors of 6 Ω and 3 Ω are connected in parallel across 12 V supply. Then
1. Equivalent resistance is 2 Ω.
2. Total current is 6 A.
3. Current through 3 Ω resistor is 4 A.
4. Power consumed is 72 W.
Which of the above statements is/are correct?
Answer & step-by-step solution
Equivalent resistance is found from reciprocal addition, total current is the sum of branch currents, and total real power is the sum of branch powers. For 6 Ω || 3 Ω, Req=(6×3)/(6+3)=2 Ω. Total current=12/2=6 A. Key relation: 1/R_eq=Σ(1/R_i); I_i=V/R_i; P_total=V I_total.
Exam focus: Solve branch currents first when the same supply voltage appears across every resistor. Common trap: Adding parallel resistances directly is incorrect.
Statement-wise Verification
- Statement 1 - Correct.
Equivalent resistance is 2 Ω.
Equivalent resistance is found from reciprocal addition, total current is the sum of branch currents, and total real power is the sum of branch powers. - Statement 2 - Correct.
Total current is 6 A.
Total current=12/2=6 A. - Statement 3 - Correct.
Current through 3 Ω resistor is 4 A.
The 3 Ω branch current=12/3=4 A. - Statement 4 - Correct.
Power consumed is 72 W.
Total power=12×6=72 W.
Core Concept
Parallel branches share the same voltage. Equivalent resistance is found from reciprocal addition, total current is the sum of branch currents, and total real power is the sum of branch powers.
Formula / Key Relationship
1/R_eq=Σ(1/R_i); I_i=V/R_i; P_total=V I_total.
Step-by-Step Check
For 6 Ω || 3 Ω, Req=(6×3)/(6+3)=2 Ω. Total current=12/2=6 A. The 3 Ω branch current=12/3=4 A. Total power=12×6=72 W.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 3, 4. Option B is incorrect because it omits correct statement(s) 4. Option C is incorrect because it omits correct statement(s) 1, 3.
Exam Shortcut / Approach
Solve branch currents first when the same supply voltage appears across every resistor.
Common Mistake
Adding parallel resistances directly is incorrect.
Quick Revision
Parallel Resistor Current and Power is linked with Parallel Resistance, DC Circuits and Basic Electrical Engineering. Solve branch currents first when the same supply voltage appears across every resistor.
6 Ω और 3 Ω के दो प्रतिरोधक 12 V सप्लाई के पार समांतर में जुड़े हैं। तब
1. समतुल्य प्रतिरोध 2 Ω है।
2. कुल करंट 6 A है।
3. 3 Ω प्रतिरोधक से करंट 4 A है।
4. खपत शक्ति 72 W है।
उपरोक्त में से कौन-सा/से कथन सही है/हैं?
उत्तर और चरणबद्ध हल
6 Ω || 3 Ω के लिए Req=(6×3)/(6+3)=2 Ω, कुल करंट=12/2=6 A, 3 Ω से करंट=12/3=4 A और P=12×6=72 W। मुख्य संबंध: 1/R_eq=Σ(1/R_i); I_i=V/R_i; P=V·I।
परीक्षा फोकस: समांतर में हर प्रतिरोधक पर समान वोल्टेज होता है। सामान्य गलती: समांतर प्रतिरोधों को सीधे जोड़ना।
विस्तृत हल (मूल भाषा में)
Statement-wise Verification
- Statement 1 - Correct.
Equivalent resistance is 2 Ω.
Equivalent resistance is found from reciprocal addition, total current is the sum of branch currents, and total real power is the sum of branch powers. - Statement 2 - Correct.
Total current is 6 A.
Total current=12/2=6 A. - Statement 3 - Correct.
Current through 3 Ω resistor is 4 A.
The 3 Ω branch current=12/3=4 A. - Statement 4 - Correct.
Power consumed is 72 W.
Total power=12×6=72 W.
Core Concept
Parallel branches share the same voltage. Equivalent resistance is found from reciprocal addition, total current is the sum of branch currents, and total real power is the sum of branch powers.
Formula / Key Relationship
1/R_eq=Σ(1/R_i); I_i=V/R_i; P_total=V I_total.
Step-by-Step Check
For 6 Ω || 3 Ω, Req=(6×3)/(6+3)=2 Ω. Total current=12/2=6 A. The 3 Ω branch current=12/3=4 A. Total power=12×6=72 W.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 3, 4. Option B is incorrect because it omits correct statement(s) 4. Option C is incorrect because it omits correct statement(s) 1, 3.
Exam Shortcut / Approach
Solve branch currents first when the same supply voltage appears across every resistor.
Common Mistake
Adding parallel resistances directly is incorrect.
Quick Revision
Parallel Resistor Current and Power is linked with Parallel Resistance, DC Circuits and Basic Electrical Engineering. Solve branch currents first when the same supply voltage appears across every resistor.
6 Ω અને 3 Ω ના બે અવરોધકો 12 V સપ્લાય સાથે સમાંતરમાં જોડેલા છે. તો
1. સમતુલ્ય અવરોધ 2 Ω છે.
2. કુલ કરંટ 6 A છે.
3. 3 Ω અવરોધકમાંથી કરંટ 4 A છે.
4. વપરાયેલ પાવર 72 W છે.
ઉપરોક્તમાંથી કયું/કયા વિધાન સાચું/સાચા છે?
જવાબ અને પગલાવાર ઉકેલ
6 Ω || 3 Ω માટે Req=(6×3)/(6+3)=2 Ω, કુલ કરંટ=12/2=6 A, 3 Ω માંથી કરંટ=12/3=4 A અને P=12×6=72 W. મુખ્ય સંબંધ: 1/R_eq=Σ(1/R_i); I_i=V/R_i; P=V·I.
પરીક્ષા મુદ્દો: સમાંતરમાં દરેક અવરોધક પર સમાન વોલ્ટેજ હોય છે. સામાન્ય ભૂલ: સમાંતર અવરોધોનો સીધો સરવાળો કરવો.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Statement-wise Verification
- Statement 1 - Correct.
Equivalent resistance is 2 Ω.
Equivalent resistance is found from reciprocal addition, total current is the sum of branch currents, and total real power is the sum of branch powers. - Statement 2 - Correct.
Total current is 6 A.
Total current=12/2=6 A. - Statement 3 - Correct.
Current through 3 Ω resistor is 4 A.
The 3 Ω branch current=12/3=4 A. - Statement 4 - Correct.
Power consumed is 72 W.
Total power=12×6=72 W.
Core Concept
Parallel branches share the same voltage. Equivalent resistance is found from reciprocal addition, total current is the sum of branch currents, and total real power is the sum of branch powers.
Formula / Key Relationship
1/R_eq=Σ(1/R_i); I_i=V/R_i; P_total=V I_total.
Step-by-Step Check
For 6 Ω || 3 Ω, Req=(6×3)/(6+3)=2 Ω. Total current=12/2=6 A. The 3 Ω branch current=12/3=4 A. Total power=12×6=72 W.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 3, 4. Option B is incorrect because it omits correct statement(s) 4. Option C is incorrect because it omits correct statement(s) 1, 3.
Exam Shortcut / Approach
Solve branch currents first when the same supply voltage appears across every resistor.
Common Mistake
Adding parallel resistances directly is incorrect.
Quick Revision
Parallel Resistor Current and Power is linked with Parallel Resistance, DC Circuits and Basic Electrical Engineering. Solve branch currents first when the same supply voltage appears across every resistor.
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