Electronics - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 15)
15.
What are the minimum and maximum output voltages?
Discussion:
12 comments Page 1 of 2.
Akila said:
7 years ago
vout = v1 * (r2/r1+r2).
= 24 * (10k/15k),
= 16v.
= 24 * (10k/15k),
= 16v.
(2)
ABDULLAH AKHTAR said:
9 years ago
Total current = v/(r1+r2)
= 24/15 = 1.6A
Voltage drop across 10 kohm resistance is the max. voltage.
So, max.vol = 10*1.6 = 16v or min.volt = 0v.
= 24/15 = 1.6A
Voltage drop across 10 kohm resistance is the max. voltage.
So, max.vol = 10*1.6 = 16v or min.volt = 0v.
(1)
Naseem ul aziz said:
1 decade ago
r1 has voltage drop of 8v
so,vmin=0 volt
vmax=16 volt
so,vmin=0 volt
vmax=16 volt
Venu said:
1 decade ago
Please explain clearly.
Sandi said:
1 decade ago
Total current I= 24/15k = 1.6m amp
voltage drop across R1= 5*1.6m= 8v
So voltage drop across R2=10*1.6=16v ans
voltage drop across R1= 5*1.6m= 8v
So voltage drop across R2=10*1.6=16v ans
Arun Jain said:
1 decade ago
Some time in multiple type Q. we use our common sense,
In the give Q. the min. voltage is 0V, so we cant calculate the min. voltage. and for calculation of max. voltage we can use voltage divider rule, So
Vmax = Vout = (10k * V1)/(10k + R1)
Vmax = (10k * 24)/(15k)
Vmax = 16V.
In the give Q. the min. voltage is 0V, so we cant calculate the min. voltage. and for calculation of max. voltage we can use voltage divider rule, So
Vmax = Vout = (10k * V1)/(10k + R1)
Vmax = (10k * 24)/(15k)
Vmax = 16V.
Rajesh,bdcet,jharsuguda said:
1 decade ago
min current=24/15=1.6ma , than 1.6*5=8v,
So that max volatage drops=24-8=16
So that max volatage drops=24-8=16
SAGAR SALWE said:
1 decade ago
Apply the kcl law over here that will be (Vout-24)/5K+ (Vout)/10K = 0.
By this we can calculate the Vout.
By this we can calculate the Vout.
Sowjanya said:
1 decade ago
Apply voltage divider rule i.e. vout = v1*(r2/r1+r2).
= 24*(10k/15k) = 16v.
= 24*(10k/15k) = 16v.
Parthipan said:
9 years ago
The given key value is 50%. So apply the voltage divider rule by kept the pot value as 5 Kohm.
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