Electrical Engineering - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 6)
6.
Three 10 k
resistors are connected in series. A 20 k resistor is connected in parallel across one of the 10 k resistors. The voltage source is 24 V. The total current in the circuit is
resistors are connected in series. A 20 k resistor is connected in parallel across one of the 10 k resistors. The voltage source is 24 V. The total current in the circuit is900 
A
A9 mA
90 mA
800 A
ADiscussion:
23 comments Page 2 of 3.
Aamir said:
10 years ago
In the given question, R1, R2 and R3 are in series and R4 is in parallel with any one. So for example R4||R3.
Now R1 and R2 in series 10+10 = 20 k and from R3 and R4 = 6.66 k, which is now in series with R1 and R2.
Then Rt is 26.66 k.
To find I = v/r.
I = 24v/26.66k = 0.900 mA = 900 uA.
Now R1 and R2 in series 10+10 = 20 k and from R3 and R4 = 6.66 k, which is now in series with R1 and R2.
Then Rt is 26.66 k.
To find I = v/r.
I = 24v/26.66k = 0.900 mA = 900 uA.
(3)
Pappish said:
10 years ago
Three 10k in series but one 10k in parallel with 20k = 10*20/30 = 6.66 k + 20 k = 26.66 k.
I = 24/26.66 k = 900 uA.
I = 24/26.66 k = 900 uA.
Mayur shankariya said:
9 years ago
@Uday.
Thanks for the explanation.
Thanks for the explanation.
Shashikant Chiranjivi said:
9 years ago
Thanks @Arun.
Sourabh said:
9 years ago
Nice explanations, thank you all.
Sanjay said:
9 years ago
Thanks for explanation @Trupti.
Rion said:
8 years ago
Here,
R = 20k||10k + 20k.
I = 24/R.
R = 20k||10k + 20k.
I = 24/R.
Thiyanesh said:
8 years ago
Thanks @Ali.
Rakesh said:
8 years ago
How to solve it? please explain me.
Anomii said:
7 years ago
Good explanation. Thanks.
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