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.
John said:
7 years ago
What is the use of key=a 50% ?
If this means 50% max of the potentiometer, then the max resistance is 50 ohms. Thus using a voltage divider,
Vout max = 24* 5k/5k + 5k = 12 volts [which is wrong].
Is the key=a 50% a typo? Using the whole value of the potentiometer,
Vout max = 24 *10k/ 5k + 10k = 16 volts.
If this means 50% max of the potentiometer, then the max resistance is 50 ohms. Thus using a voltage divider,
Vout max = 24* 5k/5k + 5k = 12 volts [which is wrong].
Is the key=a 50% a typo? Using the whole value of the potentiometer,
Vout max = 24 *10k/ 5k + 10k = 16 volts.
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.
D.saraswathi said:
9 years ago
Min current = 24/15K = 1.6 mA.
We at point A = 1.6mA * 5K = 8 V.
We at point B = 1.6mA * 15K = 24 V.
So, min voltage (24 - 24) = 0 V.
Max voltage (24 - 8) = 16 V.
We at point A = 1.6mA * 5K = 8 V.
We at point B = 1.6mA * 15K = 24 V.
So, min voltage (24 - 24) = 0 V.
Max voltage (24 - 8) = 16 V.
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)
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
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.
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.
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
Sowjanya said:
1 decade ago
Apply voltage divider rule i.e. vout = v1*(r2/r1+r2).
= 24*(10k/15k) = 16v.
= 24*(10k/15k) = 16v.
Akila said:
7 years ago
vout = v1 * (r2/r1+r2).
= 24 * (10k/15k),
= 16v.
= 24 * (10k/15k),
= 16v.
(2)
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