The voltage across a 11 Ω resistor in the circuit shown below is:

F1 Vinanti Engineering 11.01.23 D8

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SSC JE Electrical 15 Nov 2022 Shift 2 Official Paper
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  1. \(\frac{1}{2} \mathrm{~V}\)
  2. 11 V
  3. 10 V
  4. \(-\frac{1}{2} \mathrm{~V}\)

Answer (Detailed Solution Below)

Option 4 : \(-\frac{1}{2} \mathrm{~V}\)
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Detailed Solution

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Concept

According to KVL, the algebraic sum of the total voltage drop in a closed loop is equal to zero.

Σ V = 0

Calculation

F1 Vinanti Engineering 11.01.23 D9

In parallel, the voltage remains the same.

Hence, the voltage across 4Ω is 10 V.

Applying KVL as per loop direction:

\(-19V_1-V_1-10=0\)

\(V_1=\space -\frac{1}{2} \mathrm{~V}\)

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