SSC JE Electrical Engineering 23 Jan 2018 Morning Paper

Question : 115 of 200

Marks: +1, -0

Determine the potential difference (in V) between nodes B and D.

Solution:

Applying KVL, in loop 1 we get

V_D + 4×6 - 5× 6 - V_B = 0

N_D - V_B = 30 - 24

∴ V_D -V_B = 6 V

∴ V_\BD =- 6 V

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