Electronics - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 8)
8.
With 21 V applied, if R1 = 5 ohms, R2 = 35 ohms, and R3 = 14 ohms, what is the current of R2 if R1 is series connected with parallel circuit R2 and R3?
Discussion:
22 comments Page 2 of 3.
Lester said:
9 years ago
Not getting, please explain the solution.
Kamran said:
9 years ago
Thanks @Pankaj.
P@$hoo said:
9 years ago
Nice explanation @Pankaj.
Happy said:
9 years ago
Thanks, @Pankaj. Your explanation is very nice.
Deepthi said:
9 years ago
Thanks @Pankaj.
Chinmay said:
10 years ago
I satisfied with your explanation.
Pankaj said:
1 decade ago
1st calculate the total Resistance.
Rt = R1+R2||R3.
= 5+(35*14)/35+14 = 15.
According to voltage divider:
Voltage across R1 is V1 = v*R1/Rt.
V1=21*5/15=7.
Then voltage across the parallel circuit is 21-7=14;
Then current across I2 = 14/R2 = 14/35 = 0.4 A or 400 mA.
Rt = R1+R2||R3.
= 5+(35*14)/35+14 = 15.
According to voltage divider:
Voltage across R1 is V1 = v*R1/Rt.
V1=21*5/15=7.
Then voltage across the parallel circuit is 21-7=14;
Then current across I2 = 14/R2 = 14/35 = 0.4 A or 400 mA.
(2)
Jhedd said:
1 decade ago
It is the total resistance.
James Adams said:
1 decade ago
Where did the 15 ohms come from?
Orange said:
1 decade ago
By applying mesh analysis at each loop.
40I1-35I2=21---->eqn(1).
-35I1+49I2=0----->eqn(2).
RESOLVING BOTH WE GET,
I1=1.1amp and I2= 1.5amp.
current at R2 is I1-I2 =====> 1.5-1.1=0.44amp=400mA.
40I1-35I2=21---->eqn(1).
-35I1+49I2=0----->eqn(2).
RESOLVING BOTH WE GET,
I1=1.1amp and I2= 1.5amp.
current at R2 is I1-I2 =====> 1.5-1.1=0.44amp=400mA.
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