Electronics - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - True or False (Q.No. 1)
1.
In the given circuit, the Thevenin voltage equals 6 V.
Discussion:
5 comments Page 1 of 1.
Abhi said:
1 decade ago
Current flowing in loop 1 is,
-200I-600I-400I+12 = 0.
This implies, I=0.01A
Vth= 0.01A * 600 Ohms = 6V.
Hence the correct option is [A]
-200I-600I-400I+12 = 0.
This implies, I=0.01A
Vth= 0.01A * 600 Ohms = 6V.
Hence the correct option is [A]
Seema said:
8 years ago
I agree @Abhi.
Mustefa said:
7 years ago
Please explain the answer.
Vivek said:
7 years ago
Hi, Please explain this answer.
Eutrxa said:
5 years ago
When getting the Thevenin voltage the RL is deemed as open, and since it is open then no current would flow into the 800-ohm resistor making the circuit a series connection between the 400, 200 and 600-ohm resistors.
The Thevenin voltage is equal to the voltage drop in the 600-ohm resistor since the connection is in series we can use the voltage divider formula Vth = (12V) (600) / (600+400+200) = 6V.
The Thevenin voltage is equal to the voltage drop in the 600-ohm resistor since the connection is in series we can use the voltage divider formula Vth = (12V) (600) / (600+400+200) = 6V.
(1)
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