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 1 of 2.
Karthik Netha said:
4 years ago
Nice explanation @Anurag Jaiswal.
(1)
Joseph said:
6 years ago
Can someone please make a figure about this question.
Syed said:
8 years ago
Nice @Manirose.
Jatin said:
9 years ago
Thanks to all for the great explanation.
(2)
M SALMAN KHAN said:
9 years ago
Thanks @Anurag.
(2)
A C Parmar said:
10 years ago
5.6 K Ohm = 10 V.
4.7 K Ohm = ?
4.7*10/5.6 = 8.3928571 V.
4.7 K Ohm = ?
4.7*10/5.6 = 8.3928571 V.
(3)
Jagadesh said:
10 years ago
Voltage in 4.7 kv = 4.7 k*i.
i = Voltage in 5.6k/5.6k.
i = 10/5.6 k A.
v in 4.7 k = 4.7 k*10/5.6 k.
= 8.39 V.
i = Voltage in 5.6k/5.6k.
i = 10/5.6 k A.
v in 4.7 k = 4.7 k*10/5.6 k.
= 8.39 V.
(1)
Prince manju said:
10 years ago
Because in series circuit current through the both the resistors is same so that we can multiply the 5.6k ohm resistor's current with 4.7 K resistor.
Vinod said:
10 years ago
v1/r1 = v2/r2.
10/5.6 = x/4.7.
x = 8.39.
10/5.6 = x/4.7.
x = 8.39.
(1)
Ravi said:
1 decade ago
I is same in series:
So, I = v1/r1 = 10/5.6 = 1.8.
v2 = I*r2 = 1.8*4.7 = 8.4.
So, I = v1/r1 = 10/5.6 = 1.8.
v2 = I*r2 = 1.8*4.7 = 8.4.
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