Electrical Engineering - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 16)
16.
A 12 k
resistor, a 15 k resistor, and a 22 k resistor are in series with two 10 k resistors that are in parallel. The source voltage is 75 V. Current through the 15 k resistor is approximately
resistor, a 15 k resistor, and a 22 k resistor are in series with two 10 k resistors that are in parallel. The source voltage is 75 V. Current through the 15 k resistor is approximatelyDiscussion:
16 comments Page 2 of 2.
Fatima said:
7 years ago
5 mamp.
Priya said:
7 years ago
1.39 mA.
Aniket kumbhar said:
7 years ago
Please explain the answer clearly.
HAris ShAikh said:
6 years ago
R = 12+15+22 = 49kohm.
R = 10/2 = 5kohm.
TR = 49+5 = 54kohm = 54000 ohm.
V = 75V.
I = V/R.
I = 75/54000 = 1.38mA.
R = 10/2 = 5kohm.
TR = 49+5 = 54kohm = 54000 ohm.
V = 75V.
I = V/R.
I = 75/54000 = 1.38mA.
Jeju said:
4 years ago
RT = 12+15+22 = 49.
RT of parallel = (10x10) /10+10) = 5
So, 49+5 =5 4
I = V/R. = 75/54 = 1.388888889 round off so the total current is 1.4 mA.
RT of parallel = (10x10) /10+10) = 5
So, 49+5 =5 4
I = V/R. = 75/54 = 1.388888889 round off so the total current is 1.4 mA.
(1)
Asamare said:
8 months ago
The source voltage 75v in the circuit.
VT = 75v
15krI = ?
IT = ?
The circuit is both series and parallel but total voltage is given.
R15kI = VT/R15k.
R15kI = 75/15000,
= 5mA.
VT = 75v
15krI = ?
IT = ?
The circuit is both series and parallel but total voltage is given.
R15kI = VT/R15k.
R15kI = 75/15000,
= 5mA.
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