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Ohm’s Law and Series Resistance

Find the value of current I in given fig.

Original question circuit, cleanly cropped: 8 V source with 2 Ω and 6 Ω series resistors.

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Answer & Explanation

Correct AnswerB. 1 A

Quick Explanation

B — 1 A. The 2 Ω and 6 Ω resistors are in series, so the 8 V source drives 8/(2 + 6) = 1 A.

Formula / Key Relation

R_total = 2 + 6 = 8 Ω

I = V/R_total = 8/8 = 1 A

Detailed Explanation

Correct answer: B — 1 A

Read the circuit connection

The figure shows an 8 V source, a 2 Ω resistor and a 6 Ω resistor in one closed loop. There is no branch at which the current can divide, so both resistors carry the same current I. They are therefore connected in series.

Step 1: Find the total resistance. Series resistances add:

R_total = R₁ + R₂

R_total = 2 + 6 = 8 Ω

Step 2: Apply Ohm’s law to the complete loop. Use the source voltage across the total resistance:

I = V/R_total

I = 8 V/8 Ω = 1 A

Step 3: State the result and direction. The source’s positive terminal is at the top. Conventional current passes through the 2 Ω resistor from left to right, matching the arrow shown. Hence I = 1 A, option B.

The source voltage is shared by the two resistors: the 2 Ω resistor has a 2 V drop and the 6 Ω resistor has a 6 V drop. Each resistor carries 1 A; each does not receive the full 8 V. This distinction between equal series current and divided voltage is the essential circuit concept.

Quick Trick

2+6=8 Ω and source is 8 V → current is immediately 1 A.

Why Other Options Are Wrong

6 A, 2 A and 8 A would produce total resistor drops of 48 V, 16 V and 64 V respectively across the 8 Ω series combination, which cannot match the 8 V source. Only 1 A satisfies KVL.

Common Mistake

Using only the 2 Ω or only the 6 Ω resistor when computing current instead of adding both series resistances.

Exam Tip

One loop, no branches → series resistance first, then one Ohm’s-law step.

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