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Two resistors of 6Ω and 9Ω are connected in series to a 120 V source. The power consumed by 6Ω resistor is
Options
(a) 384Ω
(b) 616Ω
(c) 1500Ω
(d) 1800Ω
Correct Answer:
384Ω
Explanation:
Equivalent resistance in series = 6 + 9 = 15 Ω
Current flow in circuit, i = V / R = 120 / 15 = 8 A
Voltage in 6Ω resistor, V = iR = 8 × 6 = 48 V
Power consumed by 6Ω resistor,
P = V² / R = (48 × 48) / 6 = 8 × 48 = 384 W
Related Questions: - In R-L-C series circuit, the potential differences across each element is 20 V.
- Two straight wires each 10cm long are parallel to one another and seperated by 2 cm.
- Light propagates rectilinearly because of its
- If a wire is stretched to four times its length, then the specific resistance
- Bohr’s postulates quantizes
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In R-L-C series circuit, the potential differences across each element is 20 V.
- Two straight wires each 10cm long are parallel to one another and seperated by 2 cm.
- Light propagates rectilinearly because of its
- If a wire is stretched to four times its length, then the specific resistance
- Bohr’s postulates quantizes
Topics: Current Electricity (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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