Electrical Engineering - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 11)
11.
Two resistors are in series: a 5.6 k
resistor and a 4.7 k resistor. The voltage drop across the 5.6 k resistor is 10 V. The voltage across the 4.7 k resistor is
resistor and a 4.7 k resistor. The voltage drop across the 5.6 k resistor is 10 V. The voltage across the 4.7 k resistor isDiscussion:
17 comments Page 2 of 2.
Sushama said:
1 decade ago
R1+R2 = 5.6+4.7 = 10.3 K.
I = V/10.3.
V1 = 10V = 5.6XV/10.3.
V = 10X10.3/5.6 = 18.39.
Voltage across 4.7 K = 8.394.
I = V/10.3.
V1 = 10V = 5.6XV/10.3.
V = 10X10.3/5.6 = 18.39.
Voltage across 4.7 K = 8.394.
Manirose said:
1 decade ago
The easy method:
V1/r1 = V2/r2.
So,
5.7/10 = 4.7/v2.
1.785*1000 = v2/4.7.
0.001785*4.7 = v2 .
V2 = 8.394.
V1/r1 = V2/r2.
So,
5.7/10 = 4.7/v2.
1.785*1000 = v2/4.7.
0.001785*4.7 = v2 .
V2 = 8.394.
Rakesh yadav said:
1 decade ago
Current flowing through all resistance in series is same.
In this question value of one resistor and voltage is given so we can find out the value of current.
10/(5.6*10^3) = 0.001785.
Voltage across 4.7 k ohm is V = IR : 4.7*10^3 * 0.001785 = 8.394.
In this question value of one resistor and voltage is given so we can find out the value of current.
10/(5.6*10^3) = 0.001785.
Voltage across 4.7 k ohm is V = IR : 4.7*10^3 * 0.001785 = 8.394.
Anand said:
1 decade ago
Nice explanation Anurag
Anurag jaiswal said:
1 decade ago
current will be same for both the register because it is in series,
so if we consider current across the registers is I,
Soby I=V/R;
I=10/(5.6*1000)=0.001786A
So Voltage across 4.7 kohm register:
V=0.001786*(4.7*1000)=8.394V
so if we consider current across the registers is I,
Soby I=V/R;
I=10/(5.6*1000)=0.001786A
So Voltage across 4.7 kohm register:
V=0.001786*(4.7*1000)=8.394V
Raju said:
1 decade ago
Thank you monali
Monalisa said:
1 decade ago
current across 5.6 kohm register =10/5.6*1000=0.00178 A
so,voltage across 4.7 Kohm =.00178*4.7=8.3A
so,voltage across 4.7 Kohm =.00178*4.7=8.3A
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