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 1 of 3.
Damchi said:
8 years ago
Loop1: V =i1(R1+R2) - i2R2.
=> 21=40i1-35i2 ---------1.
Loop2: - i1R1+i2(R2+R3) =0.
=> - 35i1+49i2=0.
=> i1=(49/35)i2=(7/5)i2-------2.
Solving 1 and 2,
i2 = 1 A
i1 = 1.4 A.
Current through R2 = i1-i2
= 0.4 A = 400 mA.
=> 21=40i1-35i2 ---------1.
Loop2: - i1R1+i2(R2+R3) =0.
=> - 35i1+49i2=0.
=> i1=(49/35)i2=(7/5)i2-------2.
Solving 1 and 2,
i2 = 1 A
i1 = 1.4 A.
Current through R2 = i1-i2
= 0.4 A = 400 mA.
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)
Yazan Osrof said:
5 years ago
Total current = 1.4 amp.
R2 parallel with R3 = 10 ohms,
the ratio of V on 10 ohms = 0.667 * 21 = 14 volt ( The voltage is divided by the series resistors )
Then Return the circuit as it was before;
I = V/R.
= 14/R2 = 14/35 =0.4 Amp == 400 m Amp.
R2 parallel with R3 = 10 ohms,
the ratio of V on 10 ohms = 0.667 * 21 = 14 volt ( The voltage is divided by the series resistors )
Then Return the circuit as it was before;
I = V/R.
= 14/R2 = 14/35 =0.4 Amp == 400 m Amp.
(1)
Red said:
5 years ago
In a circuit, R1 and R2 are connected in series having a resistance of 5 ohms and 10 ohms respectively. The voltage across resistor R1 is equal to 4 V. Find the current passing through R2 and the voltage across the same resistor.
Please help.
Please help.
Swastik said:
1 decade ago
R2 & R3 are in parallel and then in series with R1
Solving, Eq Resistance = 15ohm
Then V(given)=21 V
So I=V/R=21/15 = 1.4Amp
Current across R2(By Current Divider Rule Formula)
= 1.4 x 14(R3)/(35+14)= 0.4Amp = 400 mA
Solving, Eq Resistance = 15ohm
Then V(given)=21 V
So I=V/R=21/15 = 1.4Amp
Current across R2(By Current Divider Rule Formula)
= 1.4 x 14(R3)/(35+14)= 0.4Amp = 400 mA
(1)
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.
Manish said:
8 years ago
If a series"parallel circuit has all 30-ohm resistors, what is the total resistance when R1 is in series with a parallel circuit consisting of R2 and R3?
* 45
* 50
* 60
* 90
Please give me the answer.
* 45
* 50
* 60
* 90
Please give me the answer.
Red said:
5 years ago
A parallel circuit has three branches of 6 ohms, 12 ohms, and 20 ohms. If A of current flows through the 12-ohm branch, what is the current flow through the 20-ohm branch?
Red said:
5 years ago
A lamp rated at 120 V, 100 W is to be used on a 220 V line. What size of the resistor should be connected in series with the lamp for correct operation?
Red said:
5 years ago
A store lightning system consists of 25 lamps of 100 watts each connected in parallel with 120 V sources. What is the total resistance of this system?
(1)
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