Electrical Engineering - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 8)
8.
The parallel combination of a 470 
resistor and a 1.5 k resistor is in series with the parallel combination of five 1 k resistors. The source voltage is 50 V. The percentage of the load current through any single 1 k resistor is
resistor and a 1.5 k resistor is in series with the parallel combination of five 1 k resistors. The source voltage is 50 V. The percentage of the load current through any single 1 k resistor isDiscussion:
22 comments Page 3 of 3.
Priya said:
1 decade ago
470ohm and 1500ohm are in parallel hence effective resitance is=470*1500/1970
=357.87ohm
Parallel combination of 5 1kohm resistors=1000/5=200ohm
Now toatal effective resistance=357.87+200=557.87ohm
Hence current in the circuit=50/557.87=.089amp
Now current throuh 5 1kohm resistors can be divided into two parallelcombination one with resistance 1kohm and other with 1/4kohm...........
Hence by current divider rule current through 1kohm resistor=(.089*1/4)/(1+1/4)=.018amp
Hence % = (.018/.089)*100 = 20%.
=357.87ohm
Parallel combination of 5 1kohm resistors=1000/5=200ohm
Now toatal effective resistance=357.87+200=557.87ohm
Hence current in the circuit=50/557.87=.089amp
Now current throuh 5 1kohm resistors can be divided into two parallelcombination one with resistance 1kohm and other with 1/4kohm...........
Hence by current divider rule current through 1kohm resistor=(.089*1/4)/(1+1/4)=.018amp
Hence % = (.018/.089)*100 = 20%.
(1)
Anand said:
1 decade ago
Can anybody explain me about the answer?
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