SSC JE Electrical Engineering 23 Jan 2018 Morning Paper

Question : 116 of 200

Marks: +1, -0

The open circuit voltage across the load terminals is 30 V. The terminal voltage drops to 20 V, when the load of 15 ohms is connected across the open circuited terminal. What is the internal resistance (in ohms) of the source?

Validate

Solution:

According to ohm’s law

R=V/I

∴15=20/I⇒I=4/3A

Now applying KVL in loop 1 we get

30+4/3r_s+20=0

⇒r_s=30/4\Ω=7.5\Ω

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