Electrical Engineering - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 6)
6.
Three 10 k
resistors are connected in series. A 20 k resistor is connected in parallel across one of the 10 k resistors. The voltage source is 24 V. The total current in the circuit is
resistors are connected in series. A 20 k resistor is connected in parallel across one of the 10 k resistors. The voltage source is 24 V. The total current in the circuit is900 
A
A9 mA
90 mA
800 A
ADiscussion:
23 comments Page 3 of 3.
Coolman said:
6 years ago
Ohms Law I = V/R.
So, As question said 3 resistors in series with each value 10K and second question said 20K resistor parallel connected with one of 10K resistor.
Therefore, we can say total resistance is = R1+R2+(R3//20K), hence R3 resistor parallel connected with 20K,
= 10K + 10K+((10K*20K) / (10K+20K).
= 26.67K ohm.
So now the total current is,
I = V/R.
= 24V / 26.67K.
= 900 micro ampere.
So, As question said 3 resistors in series with each value 10K and second question said 20K resistor parallel connected with one of 10K resistor.
Therefore, we can say total resistance is = R1+R2+(R3//20K), hence R3 resistor parallel connected with 20K,
= 10K + 10K+((10K*20K) / (10K+20K).
= 26.67K ohm.
So now the total current is,
I = V/R.
= 24V / 26.67K.
= 900 micro ampere.
(3)
Narasimha raju s said:
4 years ago
Thank you @Arun.
(2)
Dusmanta said:
4 years ago
across one of the resistors 20 ohms placed in parallel .so solving parallel formula
R1*R2/R1+R2=>20*10/20+10=6.66 .
Other two resisters with parallel solved value =6.66+10+10=26.66.
I=V/R => 24/26.66=900*10^-6.
R1*R2/R1+R2=>20*10/20+10=6.66 .
Other two resisters with parallel solved value =6.66+10+10=26.66.
I=V/R => 24/26.66=900*10^-6.
(3)
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