Three resistors connected in parallel have individual values of 4.0 ohm, 6.0 ohm, and 10.0 ohm. If this combination is connected in series with a 12.0 V battery and a 2.0 ohm resistor, what is the current in the 10.0 ohm resistor?
2007-07-11 00:02:24 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
effective resistance of parallel combination
= (4*6*10)/(4*6+6*10+10*4)
=240/124
=60/31
= 1.935 ohms
Total resistance R in the circuit
=2+1.935=3.935 ohms
Total current in the circuit
=12/3.935
=3.05 amps.
Same current will flow through the parallel combination also. This will be distributed in each parallel branch in the inverse ratio of their resistance. Therefore, the current in 10 ohm resistor
=(total res. of parallel comb.)*total current/resistance
=1.935*3.05/10
=0.59 Amps.
2007-07-11 00:14:41 · answer #1 · answered by Jain 4 · 1⤊ 0⤋
Effective resistance of the 4, 6, and 10 ohm resistors:
(4x6x10)/(4x6 + 4x10 + 6x10) = 1.935
Overall resistance = 2 + 1.935 = 3.935 ohm
Overall current = 12/3.935 = 3.049 amp
Voltage across the 2 ohm = 2x3.049 = 6 099 v
Voltage across the 10 ohm = 12 - 6.099 = 5.901v
Current through the 10 ohm = 5.901/10 = .5901 amp
2007-07-11 00:38:12 · answer #2 · answered by jsardi56 7 · 1⤊ 0⤋
global resistor of 4/6/10 ohms in parallel = 240/124 (aprox. 1.93 ohms)
Current outside battery: I=U/R I=12/(2+1.93)=3.05A
Current in the 10ohm resistor : 3.05* 1/10 * 240/124 = 0.5901A
2007-07-11 00:41:20 · answer #3 · answered by Anonymous · 0⤊ 0⤋
i = v/r = a hundred and twenty / 150 = 0.8 amps There could be some extra contemporary flowing with the aid of the one hundred ohm resistor yet you probably did no longer ask approximately that. in case you had to, you're able to desire to come to a decision that contemporary the comparable way.
2016-11-09 00:07:52 · answer #4 · answered by ? 4 · 0⤊ 0⤋
a. 0.59 A
b. 1.0 A
c. 11.2 A
d. 16.0 A
2007-07-11 00:20:20 · answer #5 · answered by Prabhakar G 6 · 0⤊ 1⤋