Current through 3 Ω resistor is 0.8 ampere, then potential drop through 4 Ω resistor is (a) 9.6 V (b) 2.6 V (c) 4.8 V (d) 1.2 V

Answer: (c). P.d. across the parallel pair: V = 0.8 × 3 = 2.4 V. Current in 6 Ω: 2.4/6 = 0.4 A, so total current = 0.8 + 0.4 = 1.2 A. Drop across 4 Ω: 1.2 × 4 = 4.8 V So the answer is (c) 4.8 V.

Current through 3 Ω resistor is 0.8 ampere, then potential drop… | NEET 1993 (AIPMT)

Q208NEET 1993 · AIPMTEasy

Kirchhoff's Laws & Measuring Instruments

Current through $3\ \Omega$ resistor is $0.8$ ampere, then potential drop through $4\ \Omega$ resistor is

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Answer: (c)

P.d. across the parallel pair: $V = 0.8 \times 3 = 2.4\text{ V}$ .
Current in $6\ \Omega$ : $\dfrac{2.4}{6} = 0.4\text{ A}$ , so total current $= 0.8 + 0.4 = 1.2\text{ A}$ .
Drop across $4\ \Omega$ : $1.2 \times 4 = 4.8\text{ V}$
So the answer is (c) 4.8 V.

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