Electronics - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 24)
24.
When 50 V is applied to four series resistors, 100 µA flows. If R1 = 12 k
, R2 = 47 k, and R3 = 57 k, what is the value of R4?
, R2 = 47 k, and R3 = 57 k, what is the value of R4?38.4 k
3.84 k
384 k
3.84 M
Discussion:
7 comments Page 1 of 1.
ASWATHY said:
10 years ago
Rtotal = R1+R2+R3+R4.
= 12+47+57+R4.
= 116+R4.
V = IR.
R = V/I.
= 50/100*10^6.
= 500K.
500K = 116K+R4.
R4 = 384K.
= 12+47+57+R4.
= 116+R4.
V = IR.
R = V/I.
= 50/100*10^6.
= 500K.
500K = 116K+R4.
R4 = 384K.
(1)
Ramdas said:
1 decade ago
V=I(R1+R2+R3+R4).
V=50 I=100*10^(-6) R1=12K R2=47K R3=57K R4 = X.
50= 100*10^(-6) . (12K+47K+57K+X).
50= 100*10^(-6) . (116K+X).
50/(100*10^(-6)) = (116+X).
500K = 116K+X.
X = 384K.
V=50 I=100*10^(-6) R1=12K R2=47K R3=57K R4 = X.
50= 100*10^(-6) . (12K+47K+57K+X).
50= 100*10^(-6) . (116K+X).
50/(100*10^(-6)) = (116+X).
500K = 116K+X.
X = 384K.
Venki said:
1 decade ago
V=I(R1+R2+R3+R4)
V=50
I=100*10^-6
R1=12k,R2=47k,R3=57k
So 50/100*10-6=12k+47k+57k+R4
500k=116k+R4
Then R4=500k-116k
R4=384k
V=50
I=100*10^-6
R1=12k,R2=47k,R3=57k
So 50/100*10-6=12k+47k+57k+R4
500k=116k+R4
Then R4=500k-116k
R4=384k
Vikas said:
1 decade ago
R=V/I.
Hence 50/100*10^-6=500.
But r= (r1+r2+r3) +r4.
or 500= (r1+r2+r3) +r4.
or 500= (116) +r4.
Hence r4=384.
Hence 50/100*10^-6=500.
But r= (r1+r2+r3) +r4.
or 500= (r1+r2+r3) +r4.
or 500= (116) +r4.
Hence r4=384.
Sreeyush Sudhakaran said:
1 decade ago
116k+R4=(50/(100x10^-6))
116k+R4=500K
R4=500k-116k=384k
116k+R4=500K
R4=500k-116k=384k
Naseem ul aziz said:
1 decade ago
rt=r1+r2+r3+r4
rt=50/0.0001=500k ohm
500k-12k-47k-57k=r4=384k ohm
rt=50/0.0001=500k ohm
500k-12k-47k-57k=r4=384k ohm
Vishnu said:
1 decade ago
v=i(r1+r2+r3+r4)
v=50
i=100*10^-6
r1=12k,r2=47k,r3=57k
so 50/100*10-6=12k+47k+57k+r4
500k=116k+r4
then r4=500k-116k
r4=384k
v=50
i=100*10^-6
r1=12k,r2=47k,r3=57k
so 50/100*10-6=12k+47k+57k+r4
500k=116k+r4
then r4=500k-116k
r4=384k
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